Current Electricity Notes JEE Main and Advanced

Current electricity

CHAPTER

18

CURRENT ELECTRICITY

LEARNING OBJECTIVES (i) Current through a given area of a conductor is the net charge passing per unit time through the area. Current is a scalar although we represent current with an arrow. Currents do not obey the law of vector addition. That current is a scalar also follows from it’s definition. The current I through an area of cross-section is given by the scalar product of two vectors:     I  J.A , where J and A are vectors.  Current density j gives the amount of charge flowing per second per unit area normal to the flow, J  nqvd where n is the number density (number per unit volume) of charge carriers each of charge q, and vd is the drift velocity of the  charge carriers. For electrons q = – e. If J is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nevd A. (ii) Ohm’s law: The electric current I flowing through a substance is proportional to the voltage V across its ends, i.e., V  I or V = RI, where R is called the resistance of the substance. The unit of resistance is ohm: 1 = 1 VA–1. Homogeneous conductors like silver or semiconductors like pure germanium or germanium containing impurities obey Ohm’s law within some range of electric field values. If the field becomes too strong, there are departures from Ohm’s law in all cases.   Equation E  J another statement of Ohm’s law, i.e., a conducting material obeys Ohm’s law when the resistivity of the material does not depend on the magnitude and direction of applied electric field. (iii) The resistance R of a conductor depends on its length  and constant cross-sectional area A through the relation,  , where , called resistivity is a property of the material and depends on temperature and pressure. Electrical resistivity of A substances varies over a very wide range. Metals have low resistivity, in the range of 10–8  m to 10–6  m. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale. (iv) In the temperature range in which resistivity increases linearly with temperature, the temperature coefficient of resistivity á is defined as the fractional increase in resistivity per unit increase in temperature. (v) (a) Total resistance R of n resistors connected in series is given by R = R1 + R2 +..... + Rn R

1 1 1 1 (b) Total resistance R of n resistors connected in parallel is given by R  R  R  ......  R . 1 2 n

(vi) Kirchhoff’s Rules – (a) Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. (b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero. Kirchhoff’s junction rule is based on conservation of charge and the outgoing currents add up and are equal to incoming current at a junction. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule. (vii) The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4. The null-point condition is given by R1 R 3  R2 R4

using which the value of one resistance can be determined, knowing the other three resistances. (viii) The potentiometer is a device to compare potential differences. Since the method involves a condition of no current flow, the device can be used to measure potential difference; internal resistance of a cell and compare emf’s of two sources. INTRODUCTION If there is a life’s blood in our technology, it’s most assurely electricity–electricity coursing along wire veins, delivering power and information, trickling through a metal nervous system just as it tricklets through our own. The ordered flow of charge is called electric current, whether we’re talking about electrons propelled down a wire by a battery or protons hurled through space by an exploding star. And currents carry energy. Much of the energy we “consume” is delivered by electricity conveniently on tap at wall outlets everywhere, refrigerators, Computers, heart-lung machines, all plug into the electric stream and draw energy from it. GyaanSankalp

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Current electricity Electric circuits, while very useful, can also be hazardous. To reduce the danger inherent in using circuits, proper electrical grounding is necessary. Appliances are connected to a wall socket via a three-prong plug that provides safe electrical grounding. The third prong connects the metal casing directly to a copper rod driven into the ground or to a copper water pipe that is in the ground. This arrangement protects against electrical shock in the event that a broken wire touches the metal casing, for charge would flow through the casing, through the third prong, and into the ground, returning eventually to the generator. No charge would flow through the person’s body, since the copper rod provides much less electrical resistance than does the body. Serious and sometimes fatal injuries can result from electric shock. The severity of the injury depends on the magnitude of the current and the parts of the body through which the moving charges pass. The amount of current that causes a mild tingling sensation is about 0.001A. Currents on the order of 0.01-0.02A can lead to muscle spasms, in which a person “can’t let go” of the object causing the shock. Currents of approximately 0.2A are potentially fatal because they can make the heart fibrillate, or beat in an uncontrolled manner. Substantially larger currents stop the heart completely. However, since the heart often begins beating normally again after the current ceases, the larger currents can be less dangerous than the smaller currents that cause fibrillation. ELECTRICCURRENT The electric current through a section is defined as the rate of charge flow past the section. Thus, if q charge passes across a certain section in time t, then the current i through that q t For nonuniform flow to charge, the average current < i > over a certain time interval t, will be i

section (fig.) is given by

q

q t

i

where q is the total charge flown in time t, and instantaneous current, will be

 q  dq i  lim    t 0  t  dt The SI unit of current is ampere, which consists of a charge flow of 1C per second. Thus, 1 A 1C/1s Conventional direction of flow of current is opposite to the flow of electrons or in a conductor the direction of flow of positive charge gives the direction of current. Electric current is a scalar quantity. If in a conductor n electrons pass in t second from any point them total charge passing through that point in t second is q = ne Current through the conductorI = Thus for 1 Ampere current

n=

q ne = t t 1 1.6 x 10 19

= 6.25 × 1018 electronsecond

Ixt e For a given conductor current does not change with change in cross-sectional area. In the following figure i1 = i2

Number of electron flowing in conductor in time t

Current due to translatory motion of charge : If n particle each having a charge q, pass through a given area in time t then i 

nq t

n=

+

+ +

i2

i1 = i 2

+

+

i1

+

If n particle each having a charge q pass per second per unit area, the current associated with cross-sectional area A is, i = nqA If there are n particle per unit volume each having a charge q and moving with velocity v, the current thorough, cross section A is, i = nqvA. Current due to rotatory motion of charge : If a point charge q is moving in a circle of radius r with speed v (frequency , angular speed  and time period T) then corresponding current i  qv 

2

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q qv q   T 2 r 2 

Current electricity Current density (J) Current density at any point inside a conductor defined as a vector having magnitude equal to current per unit area surrounding that point. Remember area is normal to the direction of charge flow (or current passes) through that point.  di J nˆ dA

1.

Current density is given by

2

If the cross-sectional area is not normal to the current, but makes an angle  with the direction of current then     di J  di  JdA cos   JdA  i   JdA dA cos 

4.

i   If current density J is uniform for a normal cross-section A then J  A   Current density J is a vector quantity. It’s direction is same as that of E . It’s S.I. unit is amp/m² and dimension [L–2A]

5.

In case of uniform flow of charge through a cross-section normal to it as i = nqvA  J 

3.

6.

i = nqv A

   E Current density relates with electric field as J  E   , where  = conductivity and

 = resistivity or specific resistance of substance. ELECTRIC CURRENT IN CONDUCTOR : (1) Conductors are substances through which electric charges can flow easily. (2) They are characterised by presence of a large number of free electrons (1029 electrons per m³). (3) The number density of free electrons in a conductor is same throughout the conductor. This is because free electrons experience repulsive force between them and conductor allows movement of free electrons. Thus, free electrons are evenly scattered throughout the volume of conductor. (4) These free electrons transport electric charge so are called as conduction electrons. (a) Behaviour of conductor in absence of applied potential difference : (i) The free electrons present in a conductor gain energy from temperature of surrounding and move randomly in a conductor. (ii) The speed gained by virtue of temperature is called as thermal speed of an electron. 1 3 3kT mv2rms  kT so thermal speed v rms  where m is mass of electron 2 2 m At room temperature T = 300 K, vrms = 105 m/sec. (iii) The average distance travelled by a free electron between two consecutive collisions is called as mean free path . ( ~ 10 A°) total distance travelled mean free path   number of collisions

(iv) The time taken by an electron between two successive collisions is called as relaxation time . ( ~ 10–14 s)

–

e

total time taken relaxation time   number of collisions Random motion of electrons   (v) In absence of applied potential difference electrons have random motion. The average displacement and average velocity is zero. There is no flow of current due to thermal motion of free electrons in a conductor. L (b) Behaviour of conductor in presence of applied potential difference : - - - LP HP (i) When two ends of a conductors are joined to a battery then one end is at - - higher potential and another at lower potential. This produces an electric field E inside the conductor from point of higher to lower potential i.e. E = V/L.

(v) The thermal speed can be written as vT 

V GyaanSankalp

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Current electricity (ii) The field exerts an electric force on free electrons causing acceleration of each electron. 



e E so acceleration a  F  m a  e E m (iii) Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through it. Drift velocity is very small it is of the order of 10–4 m/s as compared to thermal speed (105m/s) of electrons at room temperature. 















e E Using we have v d  (vd ~ 10–4 m/sec)  vu  a t m (iv) The direction of drift velocity for electrons in a metal is opposite to that of applied field E. (5) Relation between current and drift velocity Let n be number density of free electrons and A be area of cross-section of conductor. Number of free electrons in conductor of length L = nAL Total charge on these free electrons q = neAL Time taken by drifting electrons to cross conductor t = L/vd

v q  neAL d  neAv d or I = neAvd. t L The current flowing through a conductor is directly proportional to the drift velocity (I vd)

 Current

I

  eE   ne²    nevd  ne      E  m  m  A or J= E

(6) The current density J  so J E where   

ne 2  is specific conductivity of conductor which depends on temperature and nature of material. m



J   E is microscopic form of ohm’s law.. (7) The drift velocity depends on nature of metal through , applied potential difference, length of conductor. eE eV   m mL (8) When a steady current flows through a conductor of non-uniform cross-section drift velocity varies inversely with area of crossvd 

1  section  vd   A (9) If diameter (d) of a conductor is doubled, then drift velocity of electrons inside it will not change. (10) The rise of temperature causes increase in vrms and hence a decrease in  and relaxation time  causing a decrease in drift velocity. (11) Mobility of a charge carrier is defined as drift velocity acquired per unit electric field. vd e e m     E m m ne 2 ne The unit is m2 V–1 s–1 and dimensions are M–1 T2 A1 The mobility depends on applied potential difference, length of conductor, number density of charge carriers, current in conductor, area of cross-section of conductor. Example 1 : A typical wire for laboratory experiments is made of copper and has a radius 0.815 mm. Calculate the drift velocity of electrons in such a wire carrying a current of 1A, assuming one free electron per atom. Mobility

µ

Q  qnAvd relates the drift velocity to the number density of charge carriers, which equals the number density t of copper atoms na. We can find na from the mass density of copper, its molecular mass, and Avogadro’s number.

Sol. Equation I 

The drift velocity is related to the current and number density of charge carries : vd 

I nqA

If there is one free electron per atom, the number density of free electrons equals the number density of atoms na : n = na. 4

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Current electricity The number density of atoms na is related to the mass density m. Avogadro’s number NA, and the molar mass M. For copper r =  = 8.93 g/cm3 and M = 63.5 g/mol

na 

m N A (8.93 g / cm3 ) (6.02  1023 atoms / mol)  M 63.5 g / mol

n a  8.47  1022 atoms / cm3  8.47  1028 atoms / m 3 The magnitude of the charge is e, and the area is q = e related to the radius r of the wire A = r2

Substituting numerical values yields vd : v d  vd 

1 1  nqA n a er 2

1C /s 28

3

(8.47  10 m ) (1.6  10 19 C)  (0.000815m) 2

 3.54  10 5 m / s

Example 2 : In a certain particle accelerator, a current of 0.5mA is carried by a 5 MeV proton beam that has a radius of 1.5 mm. (a) Find the number density of protons in the heat. (b) In the beam hits a target, how many protons hit the target in 1s ? Sol. (a) The number density is related to the current, charge, cross-sectional area, and speed : n 

I qAv

We find the speed of the protons from their kinetic energy :

1 2 1.6  1019 J mv  5 MeV  5  106 eV   8  1013 J 2 1eV Use m = 1.67 × 10–27 kg for the mass of a proton, and solve for the speed : K

v

2K  m

(2) (8  10 13 J) 1.67  10 27 kg

Substitute to calculate n :

 3.10  107 m / s

n

I 0.5  103 A  qAv (1.6  10 19 C / proton)  (1.5  10 3 m) 2 (3.10  107 m / s)

n = 1.43 × 1013 proton/m³ (b) The number of protons that hit the target in 1s is related to the total charge Q that hits in 1s and the proton charge q :

Q q The charge Q that strikes the target in some time t is the current times the time : N

Q  t  (0.5mA) (1s)  0.5 mC The number of protons is then : N

Q 0.5  103 C   3.13  1015 protons q 1.6  1019 C / proton

Check the result : The number of protons hitting the target in time t is also the number in the volume Avt . Then N  nAvt . Substituting n = (I/qAv) then gives N  nAvt  (1/ qAv) (Av) t  Q / q , which is what we used in part (b) Example 3 :    4 A  j . Find the rate of charge flow through a cross-sectional area S such that The current density at a point is j   2  10 2 m       (i) S  (2cm 2 ) j , (ii) S  (4cm 2 ) i and S  (2iˆ  3jˆ cm 2 )

Sol. The rate of charge flow = current   i  j.dS GyaanSankalp

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Current electricity (i) Current i  (2  104 A / m 2 ) ˆj.(2  104 m 2 ) ˆj  2A (ii) Current i  (2  104 A / m 2 ) ˆj.(4 cm 2 ) ˆi  0

[Using ˆj.jˆ  1] [Using ˆj.iˆ  0]

(iii) Current i  (2  10 4 A / m 2 ) ˆj.(2iˆ  3ˆj)  104 m 2  6A Example 4 : Two boys A and B are sitting at two points in a field. Both boys are sitting near assemblence of charged balls each carrying charge +3e. A throws 100 balls per second towards B while B throws 50 balls per second towards A. Find the current at the mid point of A and B. Sol. Let mid point be C as shown 100e Charge moving to the right per unit time = 100 × 3e = 300e A C B Charge moving to the left per unit time = 50 × 3e = 150e Movement of charge per unit time is 300e –150e = 150e towards right 50e I = 150e = 150  1.6  10–19 A = 2.4 × 10–17 A.

TRY IT YOURSELF Q.1 Current of 4.8 amperes is flowing through a conductor. The number of electrons per second will be – (A) 3 × 1019 (b) 7.68 × 1021 (c) 7.68 × 1020 (d) 3 × 1020 Q.2 When the current i is flowing through a conductor, the drift velocity is v. If 2i current is flowed through the same metal but having double the area of cross-section, then the drift velocity will be – (A) v/4 (b) v/2 (c) v (d) 4v Q.3 Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1mm diameter, then the drift velocity (approx.) will be (Density of copper = 9 × 103 kg m–3 and atomic weight = 63) (A) 0.3 mm/sec. (b) 0.1 mm/sec. (c) 0.2 mm/sec. (d) 0.2 cm/sec. Q.4 In hydrogen atom, the electron makes 6.6 × 1015 revolutions per second around the nucleus in an orbit of radius 0.5 × 10–10 m. It is equivalent to a current nearly – (A) 1A (b) 1 mA (c) 1 µA (d) 1.6 × 10–19 A 18 + 18 Q.5 In a neon discharge tube 2.9 × 10 Ne ions move to the right each second while 1.2 × 10 electrons move to the left per second. Electron charge is 1.6 × 10–19 C. The current in the discharge tube – (A) 1A towards right (b) 0.66 A towards right (C) 0.66 A towards left (d) zero Q.6 A non-conducting ring of radius R has two positive point charges lying diametrically opposite to each other, each of magnitude Q. The ring rotates with an angular velocity  . If I is the equivalent current then, (A) I = Q (B) I = 2Q/ (C) I = Q/ (D) i = zero Q.7 A conductor of area of cross-section A having charge carriers, each having a charge q is subjected to a potential V. The number density of charge carriers in the conductor is n and the charge carriers along with their random motion are moving with average  velocity v. A current I flows in the conductor. If j is the current density, then:   (A) | j |  nqV , in the direction of charge flow.. (B) | j |  nqV , in the direction opposite to charge flow..   (C) | j |  nqV , in the direction perpendicular to charge flow.. (D) | j |  nqV , in the direction of charge flow.. Q.8 The electron drift speed is small and the charge of the electron is also small but still, we obtain a large current in a conductor. This is due to: (A) the conducting property of the conductor (B) the small resistance of the conductor (C) the small electron number density of the conductor (D) the enormous electron number density of the conductor

ANSWERS (1) (A) (5) (B)

(2) (C) (6) (C)

(3) (B) (7) (D)

(4) (B) (8) (D)

RESISTANCE : 1. When current flows in a conductor then conductor opposes the flow. Due to the applied electric field the flow of free electrons is obstructed due to temperature effect resulting in irregular velocity and because of collision with ions. This property of conductor is called its Resistance. 2. The value of resistance of a conductor is equal to the ratio of potential difference to the current. 3. If potential difference between the ends of the conductor is V and the current flowing through it is I then the value of resistance

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Current electricity  V of the conductor is : R    I 4. 5. 6.

7.

8.

Unit of resistance is ohm and Ohm = Volt@Ampere The symbol of ohm is . If potential difference between the ends of the conductor is 1 volt and 1 ampere current flows through it then resistance of conductor is given by 1 ohm. International ohm is defined has resistance of mercury column of length 106.3 cm, 1 mm square area of cross-section and mass 14.4521 gram at 0°C. 1 International ohm = 1.0048 ohm 1 ohm= 1 volt / 1 ampere = 108 e.m.u. potential difference / 10–1 e.m.u. current or 1 ohm = 109 e.m.u. resistance units and dimension of resistance are following & ohm = volt / ampere= joule / coulomb/ ampere = (newton × metre) / (ampere × second) / ampere = kilogram × metre2 × second–3 ampere2 Dimension of resistance = ML2 T–3 A–2 Resistance of a conductor depends on (a) Directly proportional to length (L) of the conductor, that is R  L (b) Inversely proportional to the area of cross-section of the conductor, that is R 

1 1 or R  2 , where r is the radius of the A r

conductor. (c) Nature of material of the conductor. (d) Temperature of the conductor. Resistance of metal conductor increases with increase in temperature. 9.

Thus R 

L or R =  A

 L   where  is a constant called as 'specific resistance' or 'resistivity'. A

 RA    =   L   r 2  If conductor is in the form of a wire of radius r, then  = R  L    10. Specific resistance of a material is equal to the resistance of unit length of that material with unit area of cross-section and current perpendicular to the cross-section or specific resistance of a material is equal to the resistance of a cube of that material with a side of unit length and current perpendicular to the cross-section. 11. If R is the resistance of the conducting wire,

R=

m V  m    = 2 =  2    A I ne A ne 

 If mean free path of electron is  and its root mean square velocity is vrms , then  = v rms

 R=

mvrms ne 2 A

m 1 ne 2  RA    Resistivity of the conductor  = = and conductivity of the conductor  m  ne 2  12. For different substances their resistivity is also different e.g. silver  min imum  1.6  108 -m and fused quartz  maximum  1016   m insulator (Maximum for fused quartz)

 alloy  semi  conductor  conductor (Minimum for silver)

13. On bending a straight wire from one or more points does not change its resistance, because the value of resistance of a wire depends on number of free electrons (n) and relaxation time () which does not change on bending the wire. GyaanSankalp

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Current electricity 1.

Effect of temperature on Resistance and Resistivity % Resistance of a conductor depends on temperature. On increasing temperature random velocity of free electron increases. If the number of charge carrying electrons remain constant as in conductors then by increase in random velocity, resistivity increases. In metals changein resistance with temperatureis given by following relation : Rt = R0 (1 + t + t2) Where Rt and R0 are resistances respectively at t°C and 0°C and &  are constants. Value of  is very small and usually it is neglected. Thus

2.

3.

FG H

6.

Rt  R0 = R x t 0

or

Constant  is called the temperature coefficient of resistance of the material. If R0 = 1 ohm] t = 1°C then  = (Rt – R0) Thus on increasing the temperature of a conductor of 1ohm resistance by 1°C increase in its resistance is equal to the temperature coefficient of resistance of the material of conductor. The value of temperature coefficient of resistance may be positive or negative. On the base of calculations it is found that for most metals value of  is approximately 1/273 per °C. Rt Thus putting the value of in above equation

IJ K

FG H

IJ K

FG IJ H K

t t T = R0 273  = R0 R0 273 273 273 Where T is the absolute temperature of the conductor. t°C Thus Rt  T Means resistance of a pure metal conductor is directly proportional to its absolute temperature. Straight line graph is obtained between Rt and temperature t. The value of resistivity or specific resistance changes with temperature. This change is due to change in resistance of the conductor with temperature. Dependence of resistivity on temperature is given by following equation : t = 0 (1 + t) Resistivity or specific resistance of pure metal increases with increase in temperature and of insulator and semiconductor decreases with increase in temperature. For alloys it increases with increase in temperature but less than metals. Variation of resistivity with temperature for metal, semiconductor and super conductor is shown in figure : Rt = R0 1 

4. 5.

Rt = R0 (1 + t)

superconductor Metal 

Temperature

7.

semiconductor





Temperature

Tc Temperature

For applying pressure on pure metals, their resistivity decreases but on applying tension force resistivity increases. Variation of resistance of some electrical material with temperature Material Temp. coefficient resistance () Variation of resistance with temp. rise Metals Positive Increases Solid non-metal Zero Independent Semi-conductor Negative Decreases Electrolyte Negative Decreases Ionised gases Negative Decreases Alloys Small positive value Almost constant Stretching of wire : If a conducting wire stretches, it’s length increases, area of cross-section decreases so resistance increases but volume remain constant. Suppose for a conducting wire before stretching it’s length =  1 , area of cross-section = A1, 1 radius = r1, diameter = d1, and resistance R1   A

1

After stretching length =  2 , area of cross-section = A2, radius = r2, diameter = d2, 8

GyaanSankalp

Current electricity and resistance R 2  

2 A2

Ratio of resistances before and after stretching 2

2

4

4

A  r  d  R1 1 A 2  1         2    2    2  [Volume remain same V = A  = A  ] 1 1 2 2 R 2  2 A1   2   A1   r1   d1 

1.

R1  1   If length is given then R    R 2   2 

2.

If radius is given then R 

2

2

1 r4



R1  r2   R 2  r1 

4

Colour coding of resistance To know the value of resistance colour code is used. These code are printed in form of set of rings of strips. By reading the values of colour bands, we can estimate the value of resistance. The carbon resistance has normally four coloured rings or bands say A, B, C and D as shown in following figure.

A B C D Colour band A and B : Indicate the first two significant figures of resistance in ohm. Band C : Indicates the decimal multiplier i.e. the number of zeros that follows the two significant figures A and B. D : Indicates the tolerance in percent about the indicated value or in other words it represents the percentage accuracy of the indicated value. The tolerance in the case of gold is ±5% and in silver is ±10%. If only three bands are marked on carbon resistance, then it indicate a tolerance of 20%. Colour code for carbon resistance : Letters as an Colour figure Multiplier aid to memory (A,B) C B Black 0 100 B Brown 1 101 R Red 2 102 O Orange 3 103 Y Yellow 4 104 G Green 5 105 B Blue 6 106 V Violet 7 107 G Grey 8 108 W White 9 109 To remember the sequence of colour code following sentence should kept in memory. B B R O Y Great Britain Very Good Wife OHM’S LAW It the physical condition of the conductor (length, temperature, mechanical strain etc.) remains some, then the current flowing

1. 2.

through the conductor is directly proportional to the potential difference across it’s two ends i.e. i  V  V  iR , where R is a proportionality constant, known as electric resistance. Ohm’s law is not a universal law, the substances, which V V T1 obey ohm’s law are known as ohmic substance. 1 T2 Graph between V and i for a metallic conductor is a 2 straight line as shown. At different temperature V-i curves are different. i

2 1 i Here tan 1> tan 2 So R1 > R2 i.e., T 1 > T2

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9

Current electricity 3.

The device or substances which don’t obey ohm’s law e.g. gases, crystal rectifiers, thermionic valve, transistors etc. are known as non-ohmic or non-linear conductors. For these V-i curve is not linear.

Static resistance, R st 

V 1  , i tan 

Dynamic resistance R dyn 

i

V 1  i tan 

V

Example 5 : A conductor has its resistance of 100 ohm at 0°C, and its temperature coefficient of resistance is 0.008 per °C at 20°C. Then its resistance at 60°C is nearly – (A) 132 (B) 148 (C) 157 (D) none of the above Sol. (C). From

R T  T  R 0 0

we get,  20 

0 . 1   0  20

This gives  0  0 . 0095

Thus R 60  R 0 (1   0  60)  100 (0.0095  60)  157 ohm Derivation of relation  T  1 We know

0 1   0 T1

R T1  R 0 (1   0 T1 )

............ (1)

R T2  R 0 (1   0T2 )

............ (2)

Also

R T2  R T1 (1   T1 (T2  T1 ))

or

R T2  R T1  R T1  T1 (T2  T1 )

............ (3)

From (1) and (2),

R T2  R T1  R 0 0 (T2  T1 ) Using (3) and (4) we get,

............ (4)

R T1 T1  R 0 0 (1   0T1 )  T1   0

Now using (1) we get Example 6 :

The current i through a non-ohmic conductor varies with voltage V as i  (V  V 2 ) A , where  and  are positive constants and V is in volt. Find the dimensions of  and the resistance (dynamic) of the conductor for i = i0. Sol. From, the principle of homogeneity of dimensions, we have Here,

[i]  [V 2 ]

Work [MLT 2 ] [L]   [ML2 T 3 A 1 ] V = potential difference = Ch arg e [AT]

 [A]   [ML2T 3A 1 ]2 or   [M 2 L4 T 6 A3 ]

Given, that current

i  V  V 2

At i = i0 let V = V0 



di    2V dV

i0  V0  V02

or

1 2  The dynamic resistance of the non-ohmic conductor is Since, negative V0 is meaningless,

Rd 

10



V0 

V0 

   2  4i 0

   4i   2

1 1 1   slope of i versus V graph (di / dV) V  V0  2  4i0

GyaanSankalp

2 0

Current electricity Example 7 : A wire has a resistance of 16.0 ohm. It is melted and drawn to a wire half its initial length. What will be the new resistance of the wire ? Sol. Here, the factor by which the length is changed is n 

 1   2 2

The new resistance R’ is given by

 1 R   R (n 2 )  16     4  2

Example 8 : The resistivity of silver at 0°C is 1.6 × 10–8 -m. If its temperature coefficient of resistance is 4.1 × 10–3 °C–1, find the resistivity of silver at 80°C. Sol. Here,  0  1.6 × 10–8 -m,   4.1 × 10–3 °C–1 and t = 80°C Using

  0 [1   (t)] , the resistivity  at a temperature 80°C will be   1.6  10 8   m [1  (4.1  10 3 C 1 ) (80C)]  2.1  10 8   m

TRY IT YOURSELF Q.1 If a 0.1% increase in length due to stretching, the percentage increase in its resistance will be – (A) 0.2% (B) 2% (C) 1% (D) 0.1% Q.2 The resistivity of iron is 1 × 10–7 ohm-m. The resistance of a iron wire of particular length and thickness is 1 ohm. If the length and the diameter of wire both are doubled, then the resistivity in ohm-m will be – (A) 1 × 10–7 (B) 2 × 10–7 (C) 4 × 10–7 (D) 8 × 10–7 Q.3 The temperature coefficient of resistance for a wire is 0.00125/°C. At 300 K its resistance is 1ohm. The temperature at which the resistance becomes 2 ohm is – (A) 1154 K (B) 1100 K (C) 1400 K (D) 1127 K Q.4 The resistance of a wire is 20 ohms. It is so stretched that the length becomes three times, then the new resistance of the wire will be – (A) 6.67 ohms (B) 60.0 ohms (C) 120 ohms (D) 180.0 ohms Q.5 On increasing the temperature of a conductor, its resistance increases because – (A) Relaxation time decreases (B) Mass of the electrons increases (C) Electron density decreases (D) None of the above Q.6 The specific resistance of a wire is , its volume is 3m³ and its resistance is 3 ohms, then its length will be– 1 (C)  3

1 3 Q.7 A piece of wire of resistance 4 ohms is bent through 180° at its mid point and the two halves are twisted together, then the resistance is – (A) 8 ohms (B) 1 ohm (C) 2 ohms (D) 5 ohms Q.8 Dimensions of a block are 1cm × 1cm × 100cm. If specific resistance of its material is 3 × 10–7 ohm-m, then the resistance between the opposite rectangular faces is – (A) 3 × 10–9 ohm (B) 3 × 10–7 ohm (C) 3 × 10–5 ohm (D) 3 × 10–3 ohm Q.9 In the above question, the resistance between the square faces is – (A) 3 × 10–9 ohm (B) 3 × 10–7 ohm (C) 3 × 10–5 ohm (D) 3 × 10–3 ohm Q.10 Resistance of tungsten wire at 150°C is 133 . Its resistance temperature coefficient is 0.0045/°C. The resistance of the wire at 500°C will be – (A) 180  (B) 225  (C) 258  (D) 317  Q.11At what temperature will the resistance of a copper wire become three times its value at 0°C (Temperature coefficient of resistance for copper = 4 × 10–3 per °C) (A) 400°C (B) 450°C (C) 500°C (D) 550°C

(A) 1 / 

(B) 3 / 

(1) (A) (5) (A) (9) (D)

(2) (A) (6) (B) (10) (C)

(D) 

ANSWERS (3) (D) (7) (B) (11) (C)

(4) (B) (8) (B)

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11

Current electricity POTENTIAL DIFFERENCE EMF and Batteries To maintain a steady current in a conductor, we need a constant supply of electric energy. A device that supplies electrical energy is called a source of emf (The letters emf stand for electromotive force, Examples of emf sources are a battery, which converts chemical energy into electrical energy, and a generator, which converts mechanical energy into electrical energy. A source of emf does work on the charge passing through it, raising the potential energy of the charge. The work per unit charge is called the emf, E, of the source. The unit of emf is the volt, the same as the unit of potential difference. An ideal battery is a source of emf that maintains a constant potential difference between its two terminals, independent of the rate of flow of charge between them. The potential difference between the terminals of an ideal battery is equal in magnitude to the emf of the battery. c

a

Figure shows a simple circuit consisting of a resistance R connected to an I ideal battery. The straight lines indicate connecting wires of negligible resis+ tance. The source of emf maintains a constant potential difference equal to E R E between points a and b, with point a being at the higher potential. There is – negligible potential difference between points a and c or between points d and b because the connecting wire is assumed to have negligible resistance. d b The potential difference from c and d is therefore equal in magnitude to the emf E, and the current in the resistor is given by I = E/R . The direction of the current in this circuit s clockwise, as shown in the figure. Note that inside the source of emf, the charge flows from a region of low potential to a region of high potential, so it gains potential energy. When charge Q flows through the source of emf E, its potential energy is increased by amount QE . The charge then flows through the resistor, where this potential energy is converted into thermal energy. The rate at which energy is supplied by QE  EI t In the simple circuit of figure, the power put out by the source of emf equals that dissipated in the resistor. A source of emf can be thought of as a charge pump that pumps the charge from a region of low electrical potential energy to a region of high electrical potential energy. In a real battery, the potential difference across the battery terminals, called the terminal voltage, is not simply equal to the emf of the battery. V Consider the circuit consisting of a real battery and a resistor in fig. If the current is varied by varying the resistance R and the terminal voltage is meaE sured, the terminal voltage is found to decrease slightly as the current increases, just as if there were a small resistance within the battery. Thus, we can consider a real battery of consist of an ideal battery of emf E plus a small resistance r, called the internal resistance of the battery. I The internal resistance of cell depends on. (1) Distance between electrodes (r  d) larger is the separation between electrodes more is the length of electrolyte through which ions have to move so more is internal resistance. (2) Conductivity or nature of electrolyte (r  1/) (3) Concentration of electrolyte (r  c) (4) Temperature of electrolyte (r  1/T) (5) Nature and area of electrodes dipped in electrolyte (r  1/A) Terminal Potential Difference : The potential difference between the two electrodes of a cell in a closed circuit i.e. when current is being drawn from the cell is called terminal potential difference. (a) When cell is discharging : r E When cell is discharging current inside the cell is from cathode to anode.

the source of emf is the power output :

Current



E or rR

P

E = R + r = V + r

I

I

or V = E – r

R

When current is drawn from the cell potential difference is less than emf of cell. Greater is the current drawn from the cell smaller is the terminal voltage. When a large current is drawn from a cell its terminal voltage is reduced. (b) When cell is charging : r E When cells is charging current inside the cell is from anode to cathode. VE or V = E + r r During charging terminal potential difference is greater than emf of cell.

Current 12



GyaanSankalp

I

I –

+ V

Current electricity (c) When cell is in open circuit : E  0 So V= E Rr In open circuit terminal potential difference is equal to emf and is the maximum potential difference which a cell can provide. (d) When cell is short circuited :

In open circuit R =   I 

E E and V = IR = 0  Rr r In short circuit current from cell is maximum and terminal potential difference is zero. (e) Power transferred to load by cell :

In short circuit R = 0 so

P  2R 

I

Pmax

2

E R

E2 4r

(r  R)2 P

s o P = Pmax if P = Pmax if

dP 0 dR r=R

Power transferred by cell to load is maximum when r = R and Pmax =

E 2 E2  4r 4R

r=R R

KIRCHHOFF’S RULE 1. When any closed-circuit loop is traversed, the algebraic sum of the changes in potential must be equal to zero. 2. At any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents out of the junction. Kirchhoff’s first rule, called the loop rule, follows directly from the conservation of energy. If we have a charge q at some point where the potential is V, the potential energy of the charge is qV. As the charge traverses a loop in a circuit, it loses or gains energy as it passes through resistors, batteries, or other devices, but when it arrives back at its starting point, its energy must again be qV. That is, the net change in the potential must be zero. I2 Kirchhoff’s second rule, called the junction rule, follows from the conservation of charge. Figure shows the junction of three wires carrying currents I1, I2 I1 and I3. Since charge does not originate or accumulate at this point, the conservation of charge implies the junction rule, which for this cases gives a I1 = I2 + I3

I3

SINGLE-LOOP CIRCUITS As an example of using Kirchhoff’s loop rule, consider the circuit shown in fig. containing two batteries with internal resistances r1 and r2 and three external resistors. We wish to find the current in terms of the emfs.

a

+

–

b + R2 – c

I – Battery 1

r1

+ E – d 2 Battery + r2 2 –

+ g + – E1

f

–

+

Changes in potential a b Drop IR1 b c Drop IR2 c d Drop E2 d e Drop Ir2 e f Drop IR3 f g Increase E1

e

Assume that I is clockwise, as indicated in figure, and apply Kirchhoff’s loop rule as we traverse the circuit in the assumed direction of the current, beginning at point a. The potential decreases and increases are given in the figure. Note that we encounter a potential difference as we traverse the source of emf between points c and d and a potential increase as we traverse the source of emf between f and g. Beginning at point a, we obtain from Kirchhoff’s loop rule. GyaanSankalp

13

Current electricity IR1  IR12  E 2  Ir2  IR 3  E1  Ir1  0 Note : You can use opposite sign convention also. Solving for the current I, we obtain I

E1  E 2 R1  R 2  R 3  r1  r2

If E2 is greater than E1, we get a negative value for the current I, indicating that we have assumed the wrong direction for I. MULTI-LOOPCIRCUIT To analyze circuits containing more than one loop, we need to use both of Kirchhoff’s rules, with Kirchhoff’s junction rule applied to points where the current splits into two or more parts. The general methods for the analysis of multiloop circuits : 1. Draw a sketch of the circuit. 2. Choose a direction for the current in each branch of the circuit, and label the currents in the circuit diagram. Add plus and minus signs to indicate the high and low potential sides of each resistors, capacitor, or source of emf. 3. Replace any combination of resistors in series or parallel with its equivalent resistance. 4. Apply the junction rule to each junction where the current divides. 5. Apply the loop rule to each loop until you obtain as many equations as unknowns. 6. Solve the equations to obtain the values of the unknowns 7. Check your results by assigning a potential of zero to one point in the circuit and use the values of the currents found to determine the potentials at other points in the circuit. Example 9 : a – + b Suppose the elements in the circuit in figure have the values E1 = 12V, + I E2 = 4V, r1 = r2 = 1, R1 = R2 = 5, and R3 = 4, as shown in figure. (a) 5 Find the potentials at points a through g in the figure, assuming that – – the potential at point f is zero. (b) Find the power input and output in c 1 the circuit. + 4V + – d g Sol. Picture the problem : To find the potential difference, we first need to + + 12V – 1 find the current I in the circuit. The voltage drop across each resistor is – then IR. To discuss the energy balance, we calculate the power into or out of each element using Equations, P = VI and P = V²/R f (a) 1. The current I in the circuit is found using equation – 0V + I

E1  E 2 R1  R 2  R 3  r1  r2

12V  4V 8V   0.5 A 5  5  4  1  1 16 2. We now find the potential at each labeled point in the circuit : I

Vg  Vf  E1  0  12V  12V Va  Vg  Ir1  12V  (0.5A) (1)  11.5V Vb  Va  IR1  11.5V  (0.5A) (5)  9V Vc  Vb  IR 2  9V  (0.5A) (5)  6.5V Vd  Vc  E 2  6.5V  4V  2.5V Ve  Vd  Ir2  2.5V  (0.5A) (1)  2.0V Vf  Ve  IR 3  2.0V  (0.5A) (4)  0 (b) 1. First, calculate the power delivered by the emf source E1 :

PE1  E1I  (12V) (0.5A)  6W 2. Part of this power is dissipated in the resistors, both internal and external : PR  I 2 R1  I2 R 2  I2 R 3  I 2 r1  I 2 r2

14

GyaanSankalp

Current electricity PR  (0.5A) 2 (5  5  4  1  1)  4.0 W 3. The remaining 2W of power goes into charging battery 2 :

PE 2  E2 I  (4V) (0.5A)  2W Example 10 : Calculate the currents I1, I2 and I3 in the circuit shown in figure. Sol. Junction rule at C yields I1 + I2 – I3 = 0 i.e., I1 + I2 = I3 ....(1) while loop for meshes a and b yields respectively : –14 – 4I2 + 6I1 – 10 = 0 i.e., 3I1 – 2I2 = 12 ....(2) and, 10 – 6I1 – 2I3 = 0 i.e., 3I1 + I3 = 5 ....(3) Substituting I3 from Equation (1) in (3) 4I1 + I2 = 5 Solving equations (2) and (4) for I1 and I2, we find I1 = 2A and I2 = –3A And hence equation (1) yields, I3 = –1A The fact that I2 and I3 are negative implies that actual direction of I2 and I3 are opposite to that shown in the circuit. You can solve above problems using nodal analysis. (Set zero potential reference of any junction, mark potential of other junctions and use Kirchhoff’s first rule.] COMBINATION OF CELL In series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive. If dissimilar plates of cells are connected together their emf’s are added to each other while if their similar plates are connected together their emf’s are subtractive. (A) Series grouping In series grouping anode of one cell is connected to cathode of r E r E E r other cell and so on. In identical cells are connected in series I

1. 2.

Equivalent emf of the combination Eeq = nE Equivalent internal resistance req = nr

3.

Main current = Current from each cell = i 

4.

R

nE R  nr If out of n cells in series m cells be connected in an opposite manner, then the current will given by m  (n  m)    Ei   E j   i 1  j1 i n

 ri  R i 1

5. 6.

Potential difference across external resistance V = iR Potential difference across each cell V' = V/n

7.

 nE  .R Power dissipated in the external circuit =   R  nr 

8.

 E2  P  n max  4r  Condition for maximum power R = nr and  

2

 dP   0, Maximum power transfer theorem   dR

Total power consumed in the circuit E²/(R + r) [and not E²R/(R + r)²] and will be maximum (= E²/r) when R = min = 0 with I = (E/r) = max. If R = r, P = (E²/2r) with I = (E/2r) 10. It is a common misconception that “current in the circuit will be maximum when power consumed by the load is maximum.” Actually current I = E/(R + r) is maximum (= E/r) when R = min = 0 with PL = (E/r)2× 0 = 0 = min. while power consumed by the load E²R/(R + r)2 is maximum (E2/4r) when R = r and I = E/2r  max (= E/r) 15 GyaanSankalp 9.

Current electricity (B) Parallel grouping In parallel grouping all anodes are connected at one point and all cathode are connected together at other point. If n identical cells are connected in parallel. 1. Equivalent emf Eeq = E 2. Equivalent internal resistance Req = r/n

4.

E Rr/n Potential difference across external resistance = p.d. across each cell = V = iR

5.

Current from each cell

6.

 E  .R Power dissipated in the circuit P    R  r / n 

7.

 E2  P  n  4r  Condition for max. power is R = r/n and max  

3.

Main current i 

i 

E

r

E

r

E

r

I R

i n 2

8.

This type of combination is used when nr >> R. Dissimilar cells in parallel Let two cells of emf E1 and E2 and internal resistances r1 and r2 be connected in parallel to an external resistance R. Applying loop rule to loop d m n c d – IR –1 r1 + E1 = 0 ......(1) Applying loop rule to loop a m n b a – IR –2 r2 + E2 = 0 ......(2) Applying first rule at junction  I = I1 + I2. Multiply equation 1 by r2 and eqn. 2 by r1 and put I2 = I – I1 we get –IR r2 – I1 r1r2 + E1 r2 = 0 and –IR r1 – (I–I1) r2r1 + E2 r1 = 0 On adding these we get, E1 r2 + E2 r1 = IR (r1 + r2) + Ir1 r2 = I (r1 + r2)  R  r1r2  r1  r2   or I 

E eq (E1r2  E 2 r1 ) / (r1  r2 )  rr R  req R 12 r1  r2

1

E1 r1

c

d

a

2

E 2 r2

b

R m

n

E eq

req

equivalent cell R

r1r2 Two dissimilar cells in parallel are equivalent to a single cell of internal resistance req  r  r and emf 1 2

E1r2  E 2 r1 r r E E   12  1  2 r1  r2 r1  r2  r1 r2  (C) Mixed Grouping If n identical cell’s are connected in a row and such m row’s are connected in parallel as shown. 1. Equivalent emf of the combination Eeq = nE 2. Equivalent internal resistance of the combination req = nr/m E E r r nE mnE  3. Main current flowing through the load i  R  (nr / m ) mR  nr 4. Potential difference across load V = iR 5. Potential difference across each cell V' = V/n 6. Current from each cell i' = i/n I E eq 

7. 8. 16

E r

R nr E2 and Pmax  (mn) m 4r In mixed grouping' as both current in the circuit and power transferred to the load are maximum under same condition it is preferred over series or parallel grouping of cells.

Condition for maximum power R 

GyaanSankalp

Current electricity 9. Total number of cell = mn 10. If N (= mn) and r/R = (m/n) are given then m = no. of rows and n = no. of cells contained in each row can be calculated for maximum current through R. It may be mentioned here, that the values of m and n should necessarily be positive integers. If, upon solving for m and n, the values come out to be fractions, then get the two set of integral values of m and n, one immediately lesser and the other greater than the obtained fractions. Next, check the value of i for these two set of m and n values, and then decide the values of m and n for i to be maximum. i2=4.0A a b Example 11 : R' Figure shows a part of an electrical network. The current flows along the directions as shown in the diagram. Find the value of resistance R’. Sol. Let us consider the loop along the sense abcda. R=5 i1=2.0A Applying Kirchhoff’s second law to the loop abcda, yields R  (4.0A)  (2.0A)(0)  12V  5 (3A)  0

i3=3A d

c 12  (15V)  0.75  E=12V or R   4.0A The value of the resistance is 0.75  and the direction of current flow is from b to a with a magnitude 4.0A. Example 12 : Two cells P and Q connected in series have each an emf of 1.5 V and internal resistances 1.0 and 0.5 respectively. Find the current through them and the voltages across their terminals. Sol. For a single closed loop, consisting of cells and resistors the current i flowing through it is given by

i

E i r  ri

E 1=1.5V

r1=1.0

P

1.5  1.5V  i = 1.5A 1.0  0.5 The voltage across the cell P is

i

Vp  E1  ir1  1.5V  1.5 (1.0)V  zero

Q

r2=0.5 E2=1.5V VQ  E 2  ir2  1.5V  1.5 (0.5)V  0.75V andQ is Example 13 : An accumulator is connected first to an external resistance R1 and then to another external resistance R2 for the same time. At what value of the internal resistance of the accumulator will the amount of heat dissipated in the external resistances be the same in the two cases ? Sol. When an accumulator of emf E and internal resistance r is connected across a load resistance R, the heat dissipated in the external

circuit

H  I2 Rt 

E 2 Rt (R  r)

 E   as I  (R  r)   

2

According to given problem : E 2 R1 t (R1  r)2



And as R 2  R1 (given)

E2R 2 t (R 2  r)2

i.e.,

(R 2  R1 ) (r 2  R1R 2 )  0

r 2  R1R 2  0

Example 14 : Twelve cells each having the same emf and negligible internal resistance are kept in a closed box. Some of the cells are connected in the reverse order. This battery is connected in series with an ammeter, an external resistance R and two cells of the same type as in the box. The current when they aid the battery is 3 ampere and when they oppose, it is 2 ampere. How many cells in the battery are connected in reverse order ? Sol. Let n cells are connected in reverse order. Then emf of the battery is E’ = (12n – n)E – nE = (12 – 2n) E In case (i)

I

E   2E  3 or R

E' + 2E = 3R

or

(14 – 2n) E = 3R

........ (1)

In case (ii)

I

E   2E  2 or R

E' – 2E = 3R

or

(10 – 2n) E = 2R

........ (2) GyaanSankalp

17

Current electricity 14  2n 3  or n=1 10  2n 2 One cell is connected in reverse order. Example 15 : How will you connect 24 cells each of internal resistance 1 so as to get maximum power output across a load of 10?

Dividing (1) and (2)

Sol. For maximum power output So R 

n pr r 2 m m

(a) If m = 1 then n 

(b) If m = 2 then n 

or

R r  and n × m = p = 24 n m m2 

p  24 m

p 24   12 m 2

pr 24  1  R 10

so

so

or

P1 

P2 

m  2.4  1.56

RE 2 R r     n m

2

10E 2  10 1     12 2

2



10E 2  10    1 24

2

 5E 2

 5.6 E 2

So to get maximum output power cells must be arranged in two rows having 12 cells in each row.

TRY IT YOURSELF Q.1 By a cell a current of 0.9 A flows through 2 ohms resistor and 0.3A through 7 ohm resistor. The internal resistance of the cell is – (A) 0.5 (B) 1.0 (C) 1.2 (D) 2.0 Q.2 A cell of emf 1.5V having a finite internal resistance is connected to a load resistance of 2 ohm. For maximum power transfer the internal resistance of the cell should be – (A) 4 ohm (B) 0.5 ohm (C) 2 ohm (D) None of these Q.3 The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in external circuit is also r ohm. The p.d. across the cell will be – (A) E/2 (B) 2E (C) 4E (D) E/4 Q.4 A cell of e.m.f. E is connected with an external resistance R, then p.d. across cell is V. The internal resistance of cell will be – (E  V) R (E  V) R (V  E) R (V  E) R (B) (C) (D) E V V E Q.5 When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is – (A) 0.5 ohm (B) 1.0 ohm (C) 1.6 ohm (D) 2.0 ohm Q.6 The terminal potential difference of a cell when short-circuited is (E = E.M.F. of the cell) (A) E (B) E/2 (C) zero (D) E/3 Q.7 n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is –

(A)

nE nE E nE (B) (C) (D) R  nr nR  r R  nr Rr Q.8 A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if – (A) R = r (B) R < r (C) R > r (D) R = r/2

(A)

Q.9 The current flowing through the segment AB of the circuit shown in figure is (A) 1 amp from A to B (B) 1 amp from B to A (C) 2 amp from A to B (D) 2 amp from B to A

18

GyaanSankalp

Current electricity Q.10 Two cells of the same emf  and different internal resistances r 1 and r2 are connected in series to an external resistances R . The value of R for which the potential difference across the first cell is zero is given by (see figure) : r1 (A) R = r 2

(B) R = r1 + r2

(1) (A) (5) (B) (9) (A)

(2) (C) (6) (C) (10) (C)

(C) R = r1 – r2

(D) R = r1 = r2

ANSWERS (3) (A) (7) (A)

(4) (B) (8) (A)

WHEATSTONE BRIDGE Figure shows the fundamental diagram of wheatstone bridge. The bridge has four resistive arms, together with a source of emf (a battery) and a galvanometer. The current through the galvanometer depends on the potential difference between the point c and d. The bridge is said to be balanced when the potential difference across the galvanometer is 0 V so that there is no current through the galvanometer. This condition occurs when the potential difference from point c to point a, equals the potential difference from point d to point a; or by referring to the other battery terminal, when the voltage from other point c to point b equals the voltage from point d to point b. Hence, the bridge is balanced when I1R1 = I1R2 …(i) if the galvanometer current is zero, the following conditions also exist: ε I1  I 3  …(ii) R1  R 3

a I2

I1 R1 c

R2 G



d

R3

R4

I3 Standard arm

Unknown I4

b

ε …(iii) R2  R4 Combining Eqs. (i), (ii) and (iii) and simplifying, we obtain and

I2  I4 

R1 R2  R1  R 3 R 2  R 4 from which we get

R1R4 = R2R3

or

…(iv)

R1 R 3  R2 R4

…(v)

Equation (v) is the well known expression for balance of the wheatstone bridge. If three of the resistances have known values, the fourth may be determined from Equation (v). Hence, if R4 is the unknown resistor, its resistance can be expressed in terms of remaining resistors

R4  R3

R2 R1

…(vi)

COMBINATION OF RESISTANCE R1 R2 I I I R3 (a) Series combination 1. Resistances are connected in series as follows& V1 V2 V3 2. Same current pass through each resistance. V 3. Potential difference between the ends of various – + resistances depend on the value of resistance. 4. The sum of potential differences developed between the ends of resistances is equal to the total voltage applied in the circuit. Means, V = V1 + V2 + V3 5. Equivalent resistance of the circuit is : R = R1 + R2 + R3 Means equivalent resistance of resistances connected in series is equal to the sum of different resistances. 6. If n equal resistances of R' are connected in series then total resistance will be, R = nR' 7. On connecting resistances in series, total resistance in the circuit increases and current in the circuit decreases. GyaanSankalp

19

Current electricity (b) Parallel combination& 1. Resistances are connected in parallel as follows&

R1 I1

2. 3. 4.

I2

Potential difference between the ends of each resistance is same. Current through different resistances is inversely proportional to the resistance. The sum of currents through each resistance is equal to the total current flowing through the circuit. Means I = I 1 + I2 + I3

I3

R2 R3

I +

V

–

1 1 1 1    R R1 R 2 R 3

5.

If equivalent resistance of circuit is R then

6. 7.

That is for parallel combination of resistances reciprocal of equivalent resistances is equal to the sum of reciprocal of different resistances. Total resistance of the circuit is less than the value of smallest resistance in the circuit. If n resistances each of value R' are connected in parallel then equivalent resistances of the circuit will be : R = R'/n On connecting resistances in parallel, the total resistances of the circuit decreases. Current divide inversely proportional to the

8.

resistance. I1 =

V V & I2 = R1 R2

and

1 1 1   R R1 R 2

I1 R 2 R 1R 2  I  R , equivalent resistance R = R1  R 2 2 1 Thus V = IR from which

F GH

I JK

I R 1R 2 R2 R1 IR = R = I and I = I 2 R1  R 2 R1  R 2 R1  R 2 1 R1 Resistance R3 is called the standard arm of the bridge and resistors R2 and R1 are called the ratio arms. Resistances in Mixed Combination If resistance are arranged in series-parallel mixed grouping, we apply method of successive reduction to find equivalent resistance. Example 16 : Find equivalent resistance between a and b. I1 =

If not able to reduce the given network in series and parallel through simple successive reduction follow the steps : (1) Find same potential points in network and considering same potential point as a single point, redraw the circuit you will find resistance in series and parallel combination Example 17 : Find equivalent resistance between the points A and B? 2R 2R R 3 B A 1 4 2 Sol. Point 1 and 3 are at same potential. Similarly point 2 and 4 are at same potential. Joining resistance between 1 and 2, 2 and 3, 3 and 4 we find that they all are in parallel. 2R

1,3

2R R

20

GyaanSankalp

2,4

Current electricity 1 1 1 1 2 So equivalent resistance R  2R  2R  R  R p

or Rp = R/2 5

Example : 18 Twelve equal resistances R are used to generate shape of a cube. Calculate equivalent resistance across the side of cube?

1

a

4

Sol. By symmetry potential at point 4 and 5 is same. Similarly potential at point 3 and 6 is same. The equivalent circuits can be drawn as :

6 7 b

1

a

4,5

R R

R

R

R R

2

R/2

1

a

R

R

3,6

3

2

8

R R

4,5

a

R 2

R

 b

8

b

7

The equivalent resistance between a and b is R eq

2

R

2R



R/2

R R

1

b

3,6

7R 5

2

7 R R 5  7 R  7 12 R R 5

(2) If not able to see same potential point try to observe if network contains balanced wheatstone bridge. Example 19 : 3 Calculate the effective resistance between A and B in the following network. 2 A

Sol. The circuit can be redrawn as Here

P R 2   Q S 3

(P  Q)(R  S) (2  3)(4  6)  PQR S 23 46

C

6

B

4 P=2

R eq 

7

C

so bridge is balanced

So the resistance between C and D is not useful. Equivalent resistance = (P + Q) | | (R + S)

D C

Q=3 7

A

B

R=4

S=6

5  10 10 D   15 3 (3) Even if not able to observe balanced wheatstone bridge try to observe symmetry in network and use plane cutting method. Example 20 : Seven resistors each of resistance R, are connected as shown in figure. The equivalent resistance between A and B is – 

A

B

GyaanSankalp

21

Current electricity (A)

4 R 3

(C) 7R

(B)

3 R 2

(D)

8 R 7 A

Sol. (D). RAB = RAC + RCB Line of symmetry

B

R/2

R/2 R

A

(4) If not able to see symmetry try to observe star-delta in the network.

R

R

R

C

R

C

B

R

1

Star-Delta Conversions : The combination of resistances shown in fig. 1 (a) is called a star connection and that shown in fig. 1 (b) is called a delta connection.

1

These two arrangements are electrically equivalent for the resistance measured between any pair of terminals. A 3 2 3 2 star connection can be replaced by a delta, and a delta (a) (b) can be replaced by a star. Fig. 1 (a) Star, (b) Delta connections Star to Delta Conversion If resistances R1, R2 and R3 are known and connected in Star configuration (as in fig. 1 (a))then it can be replaced by a delta configuration with following resistances : R12  R1  R 2 

R R RR R1 R 2 R 23  R 2  R 3  2 3 ; R13  R1  R 3  1 3 ; R3 R1 R2

Example 21 : Consider the unbalanced wheatstone bridge shown in fig. 2. Find its equivalent resistance between A and B (all values are in ohm) –

1

X

3

Y

1

B

A

Sol. Consider the star combination between XYZ. It can be reduced into a delta combination. Then the fig. 2 like fig.3, with R xy  1  3 

1 3 7 1

R xz  1  1 

11 7  3 3

Rxy

X R xz

1 3 3 7 1 Thus the equivalent diagram now looks like fig. 4. The resistance between ends A and B is then easily determined. Its value is –

Rzy

5 3 Thus the answer is RAB = 5/3 ohm. Note : For the circuit of fig. 5, memorize the following formula for equivalent resistance. (Try to prove it by using star-delta conversion).

Z Fig. 3

B

1

7 X

Y 7/3

A

R AB 

22

GyaanSankalp

1

Y

A

R zy  1  3 

 3n  1 R equivalent   r  n  3 

Z Fig. 2

3

7

3

Fig. 4

r

nr

r

nr Fig. 5

r

1

B

Current electricity Delta to Star Conversion Consider fig. 1 (a) and (b) again. If resistances R12, R23 and R13 are known and connected in a delta configuration as in fig. 1 (b), then it can be replaced by a star connection with the following resistances – R1 

R12 R13 R 23 R 21 R 31R 32 R2  R3  R12  R13  R 23 ; R12  R13  R 23 ; R12  R13  R 23 (note R13  R 31 , R 23  R 32 , R12  R 21 )

Example 22 : Each resistance in the network is of r ohm. Then calculate the equivalent resistance between the terminals A and B.

r

1

r

r

Sol. Consider the delta connection between points 123. Converting it into a star connection, the fig. 6 now looks like fig. 7, with resistances –

2

r

r

A rr r r r 3 r  , R 2  , R3  rrr 3 3 3 Fig. 6 Thus fig. 7, reduces to a balanced wheatstone bridge (fig. 8), whose equivalent resistance is

R eq 

8 r 7

1

4r/3

2

B

r

R1 

4r/3

r

r

r/3

3 r B A nr r Fig. 7 Fig. 8 Note : The Star-delta conversion is useful if you do not wish to use Kirchhoff’s laws for solving equivalent resistance determination problems. However, if numerical values are not given, at first sight, it leads to tedious algebra, when then leads to simpler r expressions. If you have patience verify the following two results. R r

(i)

The equivalent resistance for the circuit of fig. 9 is R eq

r (3R  r)  3r  R

2

r 6

4

R

r

Fig. 9 A

2

4

B

2

12

8 Fig. 10

(ii) The equivalent resistance for the circuit of fig. 10 is (all resistances are in ohm), RAB = 8 ohm Example 23 : Consider the unbalanced wheatstone bridge shown in fig. 11. Find the equivalent resistance between the points A and B. (all resistances are in ohm)

b 1 a

b

Sol. Consider one of the delta combination, say abc. Then converting it into equivalent star combination, we find fig. 12. A direct use of the conversion formula give –

R1 

1 2 1  1 2  5 4

R1=1/4

B 5

d B

R2=5/8

d

2

A

3

a A

3

7 c Fig. 11

R3=5/4 c Fig. 12 GyaanSankalp

23

Current electricity 5 1 5 25 5  , R3   8 8 8 4 Thus RAB, now can be calculated from the simplified figure 12. It gives R2 

R AB 

5 13 13 5 429 544  ||    8 4 4 8 184 184

or

R AB  2.96 

(5) If network contain infinite resistance then use simple concept – 1 = . Example 24 : The circuit diagram shown consists of a large number of elements (each element has two resistors R1 and R2). The resistances of the resistors in each subsequent element differs by a factor of k = 1/2 from the resistances of the resistors in the previous elements. The equivalent resistance between A and B shown in figure is –

A

R1

k ²R 1

kR1

R2

k4R1

k ²R 2

kR2

k4R2

B

(A)

(C)

R1  R 2 2

(B)

(R1  R 2 )  6R1R 2 2

(R1  R 2 )  R12  R 22  6R1R 2

(D) None of these 2 Sol. (C). When each element of circuit is multiplied by a factor k then equivalent resistance also becomes k times. Let the equivalent resistance between A and B be x.

A

R1 R2

kR1 kR2

k²R1

A

R1

k²R2

A

R2

B

R1

k ² R2

kR2

+

kX

So equivalent circuit becomes k

1 (R  R 2 )  X= 1 2

R12

 R 22

R1

A

 6R1R 2 x

2

(6) Even if not able to see star-delta use Kirchhoff’s rule. Example 25 : What is the equivalent resistance between points A and C in the circuit shown in figure. Sol. Distribution of current in various branches of the circuit in accordance with Kirchhoff’s I law is shown in figure. Applying Kirchhoff’s II law to meshes a and b we have respectively : (I  I1 ) R  I 2 R  I1R  I1R  0 i.e., 3I1 – I2 = I

B

GyaanSankalp

(I–I1)

I A

B

I1 R

(I–I 1–I2)

N

I2 R

a

b

R (I–I1–I2)

R M I1

2 1 I I2  I ......... (4) .......... (3) and 5 5 Now if (Req)AC is the equivalent resistance between points A and C V = (Req)AC = I1R + I1R + (I1 + I2) R = 3I1R + I2R ......... (5) Substituting the values of I1 and I2 from eqs (3) and (4) in eq. (5)

24

kX

R2

......... (1)

And (I  I1  I2 ) R  (I  I1  I2 ) R  (I1  I 2 ) R  I 2 R  0 3I1 + 4I2 = 2I ......... (2) Solving eqs. (1) and (2) for I1 and I2 I1 

k3R2

B

B x

For

k ² R1

kR1

D

(I1+I 2)

C I

Current electricity 1 7 2  I (R eq )AC  3   I R  IR  IR 5  5 5 (R eq )AC 

7 R 5

TRY IT YOURSELF Q.1 The equivalent resistance between points A and B is : (A) 2R

3 (B) R 4

A

4 (C) R 3

R

3 (D) R 5

(A) R

R (C) 4

2R

A

R (D) 8

B

R

R

Q.2 The equivalent resistance between points A and B in the circuit shown is: R (B) 2

R

B 2R

R

Q.3 Sixteen resistors each of resistance 16 are connected in the circuit as shown. The net resistance between AB is: (A) 1

(B) 2

(C) 3

(D) 4 B

A

Q.4 The effective resistance between the points A and B in the figure is:

(A) 2 

(B) 3 

(C) 6 

(D) 9 

Q.5 The equivalent resistance of a group of resistances is R. If another resistance is connected in parallel to the group, its new equivalent becomes R1 and if it is connected in series to the group, its new equivalent becomes R2. .Hence we deduce : (A) R1 > R (B) R1 < R (C) R2 > R (D) R2 < R A C D Q.6 Each resistor in the network of figure has a resistance of 10 . The resistance between points A and B is : (A) 10  (B) 20  (C) 30  (D) 40  E

Q.7 The equivalent resistance between the points a and b is: (A) r (B) 4r/3 (C) 3r (D) 3r/4

F

B

b

r r

r

a

r Q.8 What is the equivalent resistance of the network shown in figure ? The numbers indicate resistances in ohms . (A) 8  1 (B) (16/3)  (C) 4 A (D) 16/5  2

1

1

1

1

1

1

1

2

2

2

2

2

2

2

B

GyaanSankalp

25

Current electricity

ANSWERS (1) (D) (5) (BC)

(2) (B) (6) (C)

(3) (C) (7) (B)

(4) (A) (8) (B)

MEASURING DEVICES (A) Galvanometer The instrument used to measure strength of current, by measuring the deflection of the coil due to torque produced by a magnetic field, is known as galvanometer. Types of galvanometer

(I) Moving magnet type In this type of galvanometer, magnet is movable and coil is stationary.

(a) (b)

(c) (d)

(II) Moving coil type In this type of galvanometer, the coil and is movable magnet is stationary.

(a) Suspended coil type galvanometer (b) Pivoted coil type (i) In this the coil is suspended by a find (i) In this coil is pivoted on a pivot about phosphor bronze fibre. it is free to rotates. (ii) In this the deflection of coil is measured (ii) In this the deflection of coil is determined by a lamp and scale arrangement by a pointer rotating on a circular scale. (iii) Its sensitivity is high [current of order (iii) It sensitivity is low [current of order 10–9 amp. can be measured by it] 10–6 amp. can be measured]. Moving coil type galvanometer Principle : When a current carrying coil is placed in magnetic, field, it experiences a couple and the coil rotates. The torque of this couple is given by :  = NiAB sin  Where N = no. of turns in the coil, i = current through coil, B = intensity of magnetic field  = angle between magnetic field and plane of the coil. For radial magnetic field = 90°  = NiAB In the state of equilibrium of coil deflecting torque = restoring torque NiAB = C  Where C is restoring torque per unit deflection, (C depends on elastic properties of spring) or i =

C  NAB

or

i

The current is proportional to deflection. The constant

C is called galvanometer constant. NAB

(e) Sensitivity of galvanometer : (i) The deflection of galvanometer coil per unit current is defined as the sensitivity of galvanometer.  i (iii) The deflection obtained by unit current is Si (f) The galvanometer which gives more deflection for a small current is more sensitive. To increase sensitivity the values of N, A, B must be large and the value of C must be small. (g) Figure of merit (K) : (i) The reciprocal of current sensitivity is defined as the figure of merit of galvanometer. (ii) The current required for unit deflection in galvanometer is known as figure of merit (K).

(ii) Si =

1 1 or K = S K i (h) Voltage sensitivity (Sv) : (i) The deflection of galvanometer coil per unit voltage is defined as voltage sensitivity of galvanometer. 26 GyaanSankalp

(iii) Si =

Current electricity (ii) Sv =

1 1 NAB   , where RG is resistance of galvanometer coil. V iR G CR G

(B) Ammeter It is an instrument used to measure currents. It is put in series with the branch in which current is to be measured. An ideal Ammeter has zero resistance. A galvanometer with resistance G and current rating ig can be converted into an ammeter of rating I by connecting a suitable resistance S in parallel to it. The resistance connected in parallel to the ammeter is called a shunt. An ideal ammeter have zero resistance. S

Thus S(I – ig) = igG 

ig G I  ig

(C) Voltmetre It is an instrument to find the potential difference across two points in a circuit. It is essential that the resistance Rv of a voltmeter be very large compared to the resistance of any circuit element with which the voltmeter is connected. Otherwise, the voltmeter itself becomes an important circuit element and alters the potential difference that is measured. Rv >> R For an ideal voltmeter Rv =  .

i g (G  R v )  V



Rv 

R1

R1

V

Rv

V G ig

(D) Post Office Box It is so named because it has a shape of box and was designed to find resistance of electric cables and telegraph wires. It was used in post offices to determine resistance of transmission lines. It is based on principle of wheatstone bridge. Unknown resistance is S 

G

Q R and P

S A

P

B

1000 100 10

D

Q

C

10 100 1000

I R

r 2S KG KC , where r is radius and L is length of wire. L G B' A' In PO box we first press cell key and then press galvanometer key to Post-office Box eliminate induced effects. It is used to find unknown resistance, specific resistance of a wire, internal resistance of cell, resistance of galvanometer etc. (E) Metre Bridge R S B This is the simplest form of wheatstone bridge and is specially useful for comparing resistances more accurately. The G construction of the metre bridge is shown in the Figure. It A C consists of one metre resistance wire clamped between two l1 100–l1 metallic strips bent at right angles and it has two points for specific resistance is  

connection. There are two gaps; in one of them a known resistance and in second an unknown resistance whose value is to be determined is connected. The galvanometer is connected with the help of jockey across BD

Metre scale 

K1 GyaanSankalp

27

Current electricity and the cell is connected across AC. After making connections, the jockey is moved along the wire and the null point is from two resistances of the wheatstone bridge, wire used is of uniform material and cross-section. The resistance can be found with the help of the following relation :  1 R  S  100  1  or

1 1 R  R S or S 100  1 100  1

where σ is the resistance per unit length of the wire and 1 is the length of the wire from one end where null point is obtained. The bridge is most sensitive when null point is somewhere near the middle point of the wire. This is due to end resistances. (F) Potentiometer It is a device used to measure unknown potential difference accurately. The potential drop across any section of wire of uniform cross-section A Ep and composition is proportional to length of that section if a constant Rh current flows through it. L A

C

If  is the current in potentiometer wire AB of uniform cross-sectional area A, length L and specific resistance  then unknown potential difference across AC is V 

B

V

 and known potential difference across A

L A At balance point unknown potential difference = known potential difference

AB is E p 

 Ep  V Ep V    or or V= x so V  .  L  L where x = Ep/L = potential gradient i.e. fall of potential per unit length of potentiometer. Potential gradient The fall of potential per unit length of potentiometer wire is called potential gradient.

or

Ep

r = internal resistance of driving cell; Rh = resistance of rheostat, Re = external series resistance, R is resistance of potentiometer wire, L is length of potentiometer wire. The current through primary circuit I 

Ep

r Rh Primary circuit

A

B Re

A L

r  R h  Re  R

Ep IR  R    L r  R h  Re  R  L 

The potential gradient

x

(1) If Rh = 0 and Re = 0

x = xmax =

Ep R (r  R) L

(2) x = xmin =

Ep

 R   R  R h  Re  R  L 

V current  resistance of potentiometer wire  R  I   L L length of potentiometer wire where R/L is resistance per unit length of potentiometer wire. (3) x 

L R   I  specific resistance of material  or so x  = A L A A acrea of cross-section (5) The potential gradient depends only on primary circuit and is independent of secondary circuit. (6) Keeping the thickness of potentiometer wire constant if the length is changed from L1 to L2 then ratio of potential gradient will

(4) R 

be

x1 L 2  x 2 L1

(7)If two wires of length L1 and L2, resistances R1 and R2 are joined in series with a battery of emf Ep and a rheostat than the ratio of potential gradients can be calculated as 28

GyaanSankalp

Current electricity  E p  R1 x1    R1  R 2  L1

and

 Ep  R2 x2    R1  R 2  L 2

or

Ep Rh

x1 R1 L 2  . x 2 R 2 L1

A

L1 R1

L2 R2

B Sensitivity of Potentiometer Smaller the potential difference that can be measured with a potentiometer more is the sensitivity of the potentiometer. The sensitivity of potentiometer is inversely proportional to potential gradient (S  1/x). The sensitivity can be increased by (a) Increasing length of potentiometer wire (b) For a potentiometer wire of fixed length potential gradient is decreased by reducing the current in circuit. Uses of potentiometer Ep (a) Determination of unknown emf or potential difference K Rh A

If unknown emf E1 is balanced at length 1 then A

 E0  E1 = x1 =    1   0

B X Y

E0

If unknown potential difference V is balanced for length  than

R V= I R

E  V  x   0    0 

G

Z

 L ' If the length of potentiometer wire is changed from L to L’ then the new balancing length is  '      L If length is increased L’ > L so ’ >  and if length is decreased L’ < L so ’ < . So change in balancing length  = (’ – ) If the current flowing through resistance R is  then E  V  R  x    0    0 

so



x  E 0    R   0  R

For determination of current we use a coil of standard resistance. (b) Comparison of emfs of two cells

Ep

K

Rh

Let E1 emf be balanced at length 1 and E2 emf be balanced at length 2 then E1 = x1 and

E2 = x2 so

C

A

E1 1  E2  2

X

E1

Y

If two cells joined in series support each other

G Z

E2

then the balancing length is 1 so E1 + E2 = x1.

then the balancing length is 2 so E1 – E2 = x2.

If two cells joined in series oppose each other E1  E 2 1  E1  E 2  2

B

E1 1   2 or E     2 1 2

Ep

K Rh

(c) Determination of internal resistance of cell Keeping K1 open the balancing length 1 gives emf of cell so E = x1. Keeping K1 closed the balancing length 2 for some resistance R gives potential difference so V = x2. EV Internal resistance r  V

x  x 2 R 1 x 2

   R  1 2 R    2

C

A

B

E G R.B.

R

K1 GyaanSankalp

29

Current electricity (d) Calibration of voltmeter

K1

Ep

Rh

The voltmeters do not given accurate measurements because they do not have infinite resistance.

C

A

E0

1

+ V–

2

The error in measurement is found by comparison with readings of potentiometer.

B

3

RB

Rh

K2

error 1 V–V

G

voltmeter reading V

The unknown potential difference V’ is balanced at length 1 then V '  x1 

E0 1 0

If reading of voltmeter is V then error is V – V’ which can be positive, negative or zero. The calibration curve is obtained plotting voltmeter reading V on x axis and error on y axis. K1 Ep (e) Calibration of ammeter V = R and if R = 1 then V =  so potential difference across 1 resistance is equal to current through the resistance. Potential gradient is found by using a standard cell x = E0/0. The current through R or 1 coil is measured by ammeter

A

 E0  (I) and calculated by potentiometer as V’ = ’ × 1 = x1 =    1  

Rh

B

E0

1 2

1 coil

G A

3

0

The error  – ’ can be found and calibration curve is obtained by plotting ammeter reading  on x axis and error on y axis. (f) Measurement of small thermo emf A high resistance box is used in primary circuit to reduce primary current. If galvanometer shows no deflection for some R1 then potential difference across R1 is V = E0 = R1 or I = E0/R1. If balancing length for small thermo emf E is  then E  x 

E0 R .  R1 L

Rh

K2 Ep

K Rh

HRB E0

A

B

G

1 2

Cu

Fe

Cu

3

cold ice

hot sand

Example 26 : A uniform potential gradient is established across a potentiometer wire. Two cells of emf E1 and E2 connected to support and oppose each other are balanced over 1 = 6m and 2 = 2m. Find E1/E2. Sol. E1 + E2 = x1 = 6x and E1 – E2 = 2x E1  E 2 6  E1  E 2 2

30

GyaanSankalp

or

E1 2  E2 1

Current electricity Example 27: Figure shows use of potentiometer for comparison of two resistances. The balance point with standard resistance R = 10 is at 58.3 cm, while that with unknown resistance X is 68.5 cm. Find X. Sol. Let E1 and E2 be potential drops across R and X E 2 IX X so E  IR  R 1

or

58.3cm

A

E X 2R E1

B

G

R=10

G E

E2  2 But E   1 1

68.5cm

X

F

2 68.5 s o X   R  58.3  10  11.75 1

A

B G

8

12

Example 28 : A galvanometer of coil resistance 20 ohm, gives a full scale deflection with a current of 5 mA. What arrangements should be made in order to measure currents upto 1.0 A ? Sol. The upper limiting value of current to be measured is to be increased by a factor.

n

1.0A  200 5A

 The resistance of the shunt required will be

S

G 20   0.1  n  1 200  1

Hence, a shunt of resistance 0.1 should be connected in parallel across the galvanometer coil. Example 29 : A voltmeter having 100 resistance can measure a potential difference of 25V. What resistance R is required to be connected in series, to make it read voltages upto 250 V ? Sol. The upper limiting value of voltage is to be increased by a factor 250V  10 25V R = (n – 1) G = (10 – 1) 100 = 900  Example 30 : The current in a potentiometer wire of 100 cm. length is adjusted to yield a null point at 40 cm. with a cadmium cell of emf 1.018 V. Calculate the potential gradient of the wire. Also, find the balancing length corresponding to an emf of 1.450 V. Sol. From the principle of potentiometer wire, the fall in potential along a 40 cm. length is 1.018 V (the emf of the balancing length).  Potential gradient = fall in potential per unit length of the wire n

1.018V

 24.45 V / m 40  102 m If  be the balancing length corresponding to the cell of emf 1.450V, then =

since,

E 



1.450   1.018 40 cm

or

= 57 cm.

Example 31 : The length of a potentiometer wire is 400cm. Lechlanche cell gets balanced against 150 cm. length of the wire. If the length of the wire is altered to 600 cm, find the balancing length. Sol. Let V0 be the potential drop across the potentiometer wire. If E be the emf of a Lechlanche cell, then from the principle of V0 (150cm) 3  V0 .......... (1) 400cm 8 Next, if be the required balancing length, after the potentiometer wire is increased to 600 cm, then

potentiometer,

E

E

V0 (600cm)

.......... (2) GyaanSankalp

31

Current electricity Dividing eq. (2) by eq. (1), yields  3  600cm 8

or

= 225 cm.

TRY IT YOURSELF Q.1 A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into a voltmeter of range nV volt: (A) nG (B) (n –1) G (C) G/n (D) G/(n–1) Q.2 A voltmeter of variable ranges 3V, 15V, 150 V is to be designed by connecting R1 G resistances R1, R2, R3 in series with a galvanometer of resistance G = 20, as shown in figure. The galvanometer gives full scale deflection when a current of R2 R3 1mA pass through its coil. Then, the resistances R1, R2 and R3 (in kiloohms) 0 should respectively be,: (A) 3 , 12, 135 (B) 2, 98, 12, 135 3V 15V 150V (C) 2.98, 14.98, 149.98 (D) None of these Q.3 An ammeter has a resistance of G ohms and a range of I amp. The value of resistance used in parallel to convert it into an ammeter of range nI amp is: (A) nG (B) (n –1) G (C) G/n (D) G/(n –1) Q.4 A voltmeter can be constructed from a galvanometer by connecting : (A) a low resistance as shunt (B) a high resistance as shunt (C) a low resistance in series (D) a high resistance in series R1 10mV Q.5 A potentiometer wire of length 100 cm has a resistance of 10. It is connected R1 R1 in series with resistance (shown in figure) and a cell of emf 2 V and negligible 40cm resistance. A source of emf 10 mV is balanced against a length of 40cm of 100cm potentiometer wire. What is the value of R1 (A) 526.67  (B) 790  (C) 1580  (D) Zero 2V Q.6 Consider the following circuit where AB is a one metre long potentiometer wire. If both galvanometers G1 and G2 show null deflection, then the E ( ) + – ratio of emf’s E1/E2 is: 60 cm (A) 0.67 (B) 1.5 40cm B A (C) 1.0 (D) cannot be determined E + 1–

Q.7 In the circuit shown in figure, the reading of ammeter A is (A) 0.1 A (B) 0.2 A (C) 0.3 A (D) 0.4 A

G1

Q.8 Two resistances X and Y are to be measured using an ammeter of resistance 0.5 and a voltmeter of resistance 20 K. It is known that X is in the range of a few ohms while Y is in the range of several thousand ohms. V Which of the following two connections would give minimum error in the measurements of resistances X and Y? (A) circuit (a) for both X and Y (B) circuit (b) for both X and Y (C) circuit (a) for X and circuit (b) for Y (D) circuit (b) for X and circuit (a) for Y

ANSWERS (1) (B) (2) (B) (5) (C) (6) (C) HEATING EFFECT OF CURRENT, JOULE'S LAW 1. Work  Heat or Work = a constant × heat or W = JH 32

GyaanSankalp

E + 2–

(3) (D) (7) (D)

(4) (D) (8) (C)

G2

A

A V

Current electricity Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat. 2. Expression for heat produced in a conductor due to current flow through iti B (a) Let the potential difference between the points A and B of a conductor is V, on A account of which a current i flows through it. R (b) As potential difference is the work done per unit test charge.  W = QV But Q = it  W = Vit.

W Vit i 2 Rt   J J J Where R is the resistance of wire. (c) If i is in ampere, V is in volt and R is in ohm, then W = Vit Joule = Vit × 107 ergs. But heat developed H 

Vit Vit calories. = = 0.24 Vit cals = 0.24 i2Rt cals. J 4.2 Joule's laws of heating effects of current (i) The heat developed in a conductor is given by H = i2Rt Joule = 0.24 i2Rt cals. (ii) The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current. i.e. H  i2, when R and t are constants. (iii)The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor. i.e. H  R, when i and t are constants. (iv) The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current. i.e. H  t, when i and R are constants. Electrical energy or work If Q units of charge be carried between two points differing in potential by V, then electrical work done is W = Q × V Joule = Q × V × 107 ergs. Power The rate at which work is done is defined as power  Heat developed H 

QV V2 = Vi = i2 R = t R Rated power (of an electric device) The voltage of operation needed is known as the rated voltage and the corresponding power consumed is called as the rated power. Suppose, R is the electrical resistance of an electric device, with Vr and Pr its is rated voltage and wattage respectively. Clearly,

 Power =

Vr2 ........ (1) R Now, if the device is subjected to a voltage of V, then, the power consumption P will be Pr 

V2 .......... (2) R The resistance R being the property of the electric device, is independent of the applied voltage. P

2

Dividing eq. (2) by eq. (1)

 V P    Pr  Vr 

Thus, if the voltage be  times the rated one, then the actual power consumption will be 2 times the rated one. Thermal Equilibrium of a Heat Emitting Filament Consider a filament of length L and cross-sectional area S (=r2 ) carrying a current i maintaining an equilibrium temperature T (in absolute scale). The thermal power generated in it is P = i²R or V2/R ......... (1) The heat energy dissipated (from Stefan Boltzman’s law) per unit time is H   (2rL) [T 4  T04 ]

.......... (2)

where = Stefen’s constant = 5.67  10 8 W / m 2 -K 4 , T0 = temperature of the surrounding GyaanSankalp

33

Current electricity and = emissivity of the surface of the filament. Form thermal equilibrium, the rate of heat generated due to current = the rate of heat emission due to temperature difference between the filament and the surrounding.

V2   (2rL) (T 4  T04 ) ......... (3) R If 0 and  be the resistivities of the material of the filament at 0°C and t°C respectively then   0 (1  t) where  is the temperature coefficient of the material of the filament.

L 0 (1  t) L  (Ignoring any change in dimensions of the filament due to temperature rise] S r 2

If T = (273 + t), then

R

Eq. (3) changes to

 L2  V2    (2 ) (T 4  T04 ) 0 (1  t)  R 

Electric fuse Commercially it is a device employed to save the different electrical appliances used in a house such as fan, bulb, T.V., tape recorder etc. in circumstances of abrupt increase of currents entering through the main supply. The usual currents supposed to activate household appliances lies within 0-5.0 A. An electric fuse consists of a simple wire made of an alloy of tin and lead, having low melting point and high resistivity. Let a fuse wire be of length  and radius r, made of a substance of resistivity . If i be the current passing through the fuse wire, then the rate of heat generation will be dH 2  i R  i2 dt

    2  r

The heat dissipated to the surrounding varies directly as the surface area of the wire. If P be the thermal power generated per unit area, then the total rate of heat loss is dE  P (2r) dt In the steady state the rate of heat generated due to Joule’s effect becomes equal to the rate of heat loss to the surrounding.    i 2  2   P (2r)  r 

1/ 2

or

 P (2 2 r3 )  i    

Example 32 : Find the heat generated in each of the resistors shown in fig. in a time interval of 1 hour. (Ignore the internal resistance of the battery)

Sol. The resistances 12 and 6are connected in parallel, having an equivalent resistance of

12 2

6

+

–

E = 24V

(12) (6)  4 (12  6) 

This 4 resistance is in series with the 2 resistance yielding an equivalent resistance of 4 + 2 = 6. 24V  4A Current drawn from the battery = 6  Heat produced in the 2 resistance is, H = i²Rt = (4A)² (2) (3600 s) = 115.2 kJ Since 12and 6resistors are in parallel, hence the total current of 4A gets distributed in them in the inverse ratio of their respectively resistances.  Current through the 12 and 6 resistances will be respectively.

 6  4A   1.33 A  12  6 

and

 12  4A   2.67 A  12  6 

The heat generated in the 12 resistor will be and heat generated in the 6 resistor will be 34

GyaanSankalp

H = i²Rt = (1.33 A)² (12) (3600 s) = 76.4 kJ H = (2.67A)² (6) (3600 s) = 154 kJ

Current electricity Example 33 : Two bulbs A and B each with a rated voltage of 220V, have rated powers of 25W and 100W respectively. They are connected in series across a voltage supply of 440V. Find, which of the two bulbs will fuse. Sol. The ratio of resistances of the bulbs A and B is, R A (Pr )B 100 4    R B (Pr )A 25 1

Since, the two bulbs are in series, so the voltage across the two will be in direct proportion to their respective resistances.

 4   1   VA    440V  352 V and VB    440V  88 V 4  1 4  1 The voltage across the bulb A exceeds the rated voltage and hence it will fuse. Example 34 : How much (in present) has a filament diameter decreased due to evaporation if the maintenance of the previous temperature required an increase in voltage by 0.50% ?  L2  V2 4 4  Sol. From eq.  (1  t)  R  (2 ) (T  T0 ) , ignoring any change in dimensions, due to temperature rise,   0 k where k is a positive constant. r Taking log of both sides yields, 2 ln V = ln k – ln r V2 

Differentiating, both sides yields

2

dV dr  V r

 V   r   100      100 or 2    r  V

[ V and r are small]

 r  or    100  2 (0.5%)  1.0% r Thus, the radius (and hence the diameter) has decreased by 1.0% CHEMICAL EFFECT OF CURRENT Faraday's laws of Electrolysis : (i) First law : The total mass of ions liberated at an electrode, during electrolysis, is proportional to the quantity of electricity which passes through the electrolyte. i.e. m  Q But Q = it  m  it Hence the first law may also be stated as follows. The mass of ions liberated at an electrode during electrolysis is proportional to. (a) The current flowing through the electrolyte, and (b) The time for which the current flows. (ii) Second law : If same quantity of electricity is passed through different electrolytes, the masses of the substances (ions) deposited at the respective cathodes are directly proportional to their chemical equivalents (equivalent weights). i.e. m  E (chemical equivalent) (iii) Electro-chemical equivalent (E.C.E.) (a) The electro-chemical equivalent of an element is its mass in grams deposited on the electrode by the passage of 1 coulomb of charge through it i.e. by the passage of 1 ampere current for 1 second. (b) According to Faraday's first law m  i.t or m = Z. i.t. Where Z is the constant of proportionality known as the electro-chemical equivalent of the substance. It is numerically equal to the mass in grams of the element deposited when a unit current flows in unit time. (c) According to Faraday's first law m = ZQ m

z

1 1 If same charge is passed two electrolytes, the m  z 2 2

According to Faraday's second law GyaanSankalp

35

Current electricity m1 Z1  m 2 Z2



E1 Z1  E 2 Z2

or

E2 Z2 = Z1 × E 1

Example 35 : Find the mass of silver liberated in a silver voltameter carrying a current of 1.5A, during 15 minutes. The electro chemical equivalent of silver is 1.12× 10–6 kg/C. Sol. Here, m = ?, i = 1.5 A, Z = 1.12× 10–6 kg/C and t = 15 × 30s Using m = Zit yields mass of silver liberated is

kg   1.5C   m  1.12  106    (450s) = 7.6 × 10–4 kg = 0.76 g  C s 

Example 36 : In a water voltameter, the act of passing a certain amount of current for a certain time produces of 1.2 g of H2 at STP. Find the amount of O2 liberated during that period. Sol. Since the same current flows through both the electrodes of a water voltameter, so the amount of oxygen and hydrogen liberated (for the same charge) will be in direct proportional to their respective equivalent weights by Faradays’ second law of electrolysis, m0 8 i.e., m  1 H

or

 8 m0    m H  8  1.2g  9.6g  1

Thermo e.m.f.

THERMOELECTRICITY 1. Thermocouple If two wires of different metals are joined at their ends so as to form two junctions, then the resulting arrangement is called a thermocouple. 2. Seebeck Effect When the two junctions of a thermocouple are maintained at different temperatures, then a current starts flowing through the wires. This is called. Seebeck effect and the e.m.f. developed in the circuit is called thermo emf. If a graph be plotted for the thermoelectric emf e against, the temperature T of the hot junction, then the curve obtained, is in general, something like that shown in fig. Tn Ti TC The emf increases (though not strictly, linearly) with increase in the temperature of the hot junction, reaching a maximum value corresponding to a certain temperature of the hot junction, known as the neutral temperature (270°C). Thereafter, the emf decreases, with an increase in the temperature of the hot junction, eventually reaching to zero, corresponding to a certain temperature, known as the inversion temperature (540°C). Increasing the temperature of the hot junction beyond this value, reverses the direction of emf and hence that of current. Tn  Tc  Ti  Tc

1 2 ; e  aT  bT 2

;

de is called the thermoelectric power or the Seebeck’s coefficient. dT

The quantity

a 2a , Ti  b b Thermoelectric series Seebeck arranged a number of metals in the form of a series called thermoelectric series. The arrangement of some of the metals forming the series are : Sb, Fe, Zn, Ag, Au, Mo, Cr, Sn, Pb, Hg, Mn, Cu, Co, Ni, Bi. In the above series, current flows through the cold junction from the metal which appear earlier to the metal which appears later. Greater the separation of the two metals in the series, greater is the thermo emf generated. Law of Intermediate metals (or successive metals) Suppose that there are A, B, C, ........N metals showing that Seebeck’s effect. Let AB, BC, ...MN and AN be the thermocouples formed from the metals A and B, B and C etc. Further, let the junctions of the thermocouples be the same separately (Say T2 and T1 respectively), then the law of intermediate metals states that, Tn 

1.

2. 3.

e AN  eAB  eBC  eCD  .....e MN 36

de  a  bT dT

GyaanSankalp

Current electricity Law of Intermediate Temperatures : If eT1T2 be the thermo emf produced in a certain thermocouple with its two junctions maintained at temperatures T1 and Tn, then the law of intermediate temperatures states that.

eT1Tn  eT1T2  eT2T3  eT3T4 .....eTn 1Tn where eT1T2 = thermo emf produced in the same thermocouple at junction temperatures T1 and T2 etc. 1. 2. 1. 2.

Peltier effect This effect is the converse of Seebeck effect. If a current is passed through a junction of two dissimilar metals, heat is either absorbed or evolved at the junction. Thomson effect An emf is developed between two parts of a single metal if they are at different temperatures. This is called Thomson effect. Thomson coefficient If de is the potential difference between two points in a metal which have a temperature difference dt, then the ratio is defined as the Thomson coefficient.  =

d 2e de =T dt dT 2

Example 37 : The temperature of inversion of a thermocouple is 620ºC and the neutral temperature is 300ºC. Find the temperature of cold junction ? Sol. Ti – Tn = Tn – TC so Tc = 2Tn – Ti = 600 = – 20ºC Example 38 : Near room temperature the thermoemf of copper constantan thermocouple is 40 µ V/C. A galvanometer of 100 resistance can detect 10–6 A. What is least temperature difference which can be detected ? Sol. Least emf E = R = 10–6 × 100 = 10–4 V emf per unit temperature = 40 µV/ºC = 40 × 10–6 V/ºC smallest temperature difference =

104 4  10 6

= 2.5ºC

Example 39 : 1 T 2 and  = 10µV/ºC2 ,  = µV/ºC2. If temperature of cold junction is zero find neutral 20 2 temperature and temperature of inversion ?

The emf is given by E = T +

Sol. At T = Tn

dE =0 dT

or

Tn =

– –10 = = 200ºC  1 / 20

Ti = 2Tn = 400 ºC

TRY IT YOURSELF Q.1 A current of 30 A is registered when the terminals of a dry cell of emf 1.5 V are connected through an ammeter. Neglecting the meter resistance, find the amount of heat produced in the battery in 10 sec. (A) 250 J (B) 500 J (C) 450 J (D) 900 J Q.2 A house has 15 bulbs, each of resistance 103  connected in parallel to 220V supply. If electricity is changed at Rs. 2/ per BOT unit, the electricity bill in a month of 31 days will be (bulb are operated for 6 hrs. a day) (A) Rs. 270 (B) Rs. 135 (C) Rs. 305 (D) Rs. 415 Q.3 A 500 W heating unit is designed to operate on a 115 V line. If the line voltage drops to 110 V line, the percentage drop in heat output will be (A) 7.6% (B) 8.5% (C) 8.1% (D) 10.2% Q.4 An electric kettle of 1200W has 1 kg water at 20ºC. The water equivalent of the kettle is 500 gm. The time required to heat the water from 20ºC to 100ºC will be (A) 8 minute (B) 7 minute (C) 6 minute (D) 5 minute Q.5 The junctions of a Ni-Cu thermocouple are maintained at 0ºC and 100ºC. The seedbeck emf in the loop is (aNi,cu= 16.3 × 10–8V/ºC and bNi,Cu = – 0.042 × 10–6V/ºC) GyaanSankalp

37

Current electricity (A) 2.82 × 10–3 V (B) 1.42 × 10–3 V (C) 2 × 10–3 V (D) 4 × 102 V Q.6 The temperatures of the junction of a Bi-Ag thermocouple are maintained at 0ºC and 0.001ºC. The thermo emf is [aBi-Ag = –46 × 10–6 V/ºC ; bBi-Ag = – 0.48 × 10–8V/ºC2] (A) 2.2 × 10–3 V (B) 2.6 × 10–8 V (C) 4.6 × 10–8V (D) 2 V Q.7 Consider the following two statements (a) Free-electron density is different in different metals (b) Free-electron density in a metal depends on temperature (A) Both true (B) a true b false (C) b true a false (D) both false Q.8 In producing chlorine through electrolysis 100 KW power at 125 volt is being consumed. If the E.C.E. of chlorineis 0.367 × 10–6kg/ coul, then the mass of chlorine liberated per minute will be (A) 16.3 × 10–4 kg (B) 17.61 × 10–3 kg (C) 18.2 × 10–3 kg (D) 10–4 kg Q.9 Two heater coils A and B are made of the same material. The length and diameter of coil A are double that of coil B. If the two heaters are connected in parallel, then the heat produced in – (A) A is more than that is B (B) B is more than that in A (C) A is the same as that is B (D) the two heaters has no relationship Q.10 In the circuit shown in figure. the heat produced in 5  resistor due to a current flowing in it is 10 cal/s. The heat produced in 4 resistor is 6 4 (A) 4 cal/s (C) 2 cal/s

(B) 1 cal/s (D) 3 cal/s 5

ANSWERS (1) (C) (6) (C)

(2) (A) (7) (A)

(3) (B) (8) (B)

(4) (B) (9) (A)

(5) (B) (10) (C)

USEFUL TIPS The number of electrons crossing when 1A of current flows through a conductor is 6.25 × 1018 per second. One way of writing Ohm’s law is J = E, where j is current density,  conductivity and E, electric field. Ohm’s law cannot be applied to circuits which contain diodes, transistors, capacitors etc. Ohm’s law and Kirchoff’s loop theorem can be applied to AC voltages only for their instantaneous values. They cannot be applied to rms values. 5. A fuse wire should have low melting point and high resistivity. 6. If there is only one battery in the circuit, problems involving solution of current by Kirchoff’s law can be done with one equation. For this the current is judiciously chosen for the convenience of division. 7. If two bulbs of different powers are connected in series to a source, the bulb having less power will glow brighter. 8. If two bulbs of different power are connected to the same source in parallel, the bulb with greater power will be brighter. 9. When a uniform wire conductance ‘c’ is stretched to n times the original length, its conductance becomes c/n2. 10. Two important properties of a metal to make a standard resistor are low temperature coefficient and low linear expansivity. 11. In a potentiometer, if the EMF of the driving cell (cell connected in the main circuit) is increased, the balancing length will decrease. 1. 2. 3. 4.

MISCELLANEOUS SOLVED EXAMPLES Example 1 : The charge q flowing through a wire varies with time t as q = (0.1 + 0.2t + 0.3t²)C where t is in second. Find the current through the wire (i) initially and at (ii) t = 2s. Sol. The instantaneous current is dq d  (0.3t 2  0.2t  0.1) or i = (0.6t + 0.2) C/s dt dt (i) Initially, t = 0,  Initial current is i = 0.6 (0) + 0.2 A = 0.2 A (ii) At t = 2s, the current is i = 0.6 (2) + 0.2 A = 1.4 A Example 2 : i

A potential difference applied to the ends of a wire made up of an alloy drives a current through it such that j    r , where r is the distance of the point from the axis. If R be the radius of the wire, then find the total current through any cross section of the wire. 38

GyaanSankalp

Current electricity Sol. Let c be the centre of the cross-section considered at any section of the wire. If r be the radius of a circular strip concentric with the cross-section considered and, dr be its thickness then (fig.) current through the circular strip of thickness dr will be   di  j.dS  (  r) (2rdr) cos 0 or di  2 (r  r 2 ) dr The total current i can be obtained by summing up the currents flowing through individual circular strips, i.e.,

r C dr

R

R

i   di   2 (r  r 2 ) dr 0

 or i  2   

R

R

 r2   r3        2 0  3 0

2   2 R (3  2R)  6 

Example 3 : Three resistors of resistances 3, 5 and 1 and are joined in series. A potential difference of 24V is applied across the combination. Find the current through the resistors and voltage drops across each of them. Sol. The effective resistance of the three resistances will be R eff  3  5  1  9  The current through the resistors will be i

V 24V   2.67 A R eff 9

The ratio of the voltage drops across different resistors will be as R1 : R2 : R3.  V1 : V2 : V3 : : 3 : 5 : 1 subject to the condition that, V1 + V2 + V3 = 24V

 3  5  1  V1    24V  8V, V2    24V  13.3 V and V3    24V  2.67V 9 9 9 Example 4 : The heat produced in 5resistor due to current flowing through it is 10 calorie/sec. What is heat dissipated in 4 resistance ? (4 ohm and 6 ohm are in series with 5 ohm in parallel) Sol. The resistance 5 and 4 + 6 = 10 are in parallel so P =

V2 cal/sec. RJ

P10 R5 5 2R = or P = × 10 = 5 cal/sec. Now 4 and 6 are in series so P = cal/sec. 10 10 P5 R10 J P4 R4 4 2 3  so P6 = P4 = 1.5 P4 P6 = R 6 = 6 3 2 P4 + P6 = 5 cal/sec or P4 + 1.5 P4= 5 or P4 = 2 cal/sec. Example 5 : A fuse wire of radius 8.0mm is replaced by a wire of radius 2.7 mm made up of the same substance. If the former wire has a maximum current capacity of 5.0A, then find the value for the second wire. 1/ 2

 P (2 2 r3 )  i    Sol. From eq.   



i1  r1    i 2  r 2 

, we know i  r3/2 3/2



 i1   2.7mm   5.0A    8.0mm 

3/2

or

i2 = 0.98 A

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39

Current electricity Example 6 :

R1=3

Three resistors of resistance R1  3, R 2  1 and R 3  8 are connected as shown in fig. across a cell fo 24V and negligible resistance. Find the current drawn from the cell, and the effective resistance across the cell. Sol. The current flowing through the branch containing the resistance R1 and R2 will be

i1 

R3=8

24V  6A (3  1)

and the current flowing through the resistance R3 will be i 2 

R2=1

+ – 24V

24V  3A 8

So, the total current drawn from the cell will be, i  i1  i 2  6A  3A  9A If R be the effective resistance then, by definition 24V = (9A) R R = 2.67  Example 7 : A standard 1 resistance and a low resistance r are connected in series in an electrical circuit. The p.d. across the combination is balanced at 6.30 m length of potentiometer, while the p.d. across 1 resistance is balanced at 6.00 m length. Find the value of the small resistance. Sol. Let I is the current through the resistances, then I (1 + r) = x (6.30) and I (1) = x (6.00) 1  r 6.3  or 1 + r = 1.05 or r = 0.05 ohm 1 6.0 Example 8 : A potentiometer wire of length L and resistance 10 has a battery of 2.5V and a resistance in series in its primary circuit. The null point for a cell of emf 1 volt comes at L/2 distance from one end. If the series resistance in the primary circuit is doubled, then position of the new null point is – (A) 0.5 L (B) 0.6 L (C) 1.0 L (D) None of the above

Sol. (B). For the first case x 

Ep L



RW RW  R

where R is the series resistance. Thus for the cell of 1 volt emf with balancing length L/2, 1 = xL/2

or

1

2.5 10 L   L 10  R 2

or

20 + 2R = 25

or

R = 2.5 

For second case series resistance 2.5 10  L 10  5

R´ = 2R = 5 

 x 

1  x  

where the new balancing length    1 

15L  0.6 L 25

Example 9 : It is desired to measure a small resistance r using the potentiometer. This resistance is connected to a big resistance R and a steady current is passed through it. When p.d. across R is balanced, the balancing length is found to be 320 cm. When p.d. across the series combination of the two is balanced, the balancing length is found to be 360cm. What is the ratio R : r – (A) 8 : 1 (B) 9 : 8 (C) 1 : 8 (D) None of the above Sol. (A). V1  IR  x (320) and V2 = I (R + r) = x (360)

40

Thus

R  r 360  R 320

Thus

R 8  r 1

GyaanSankalp

or

1

r 9  or R 8

r 9 1  1  R 8 8

Current electricity Example 10 : In the network shown, each resistance is equivalent to R. The equivalent resistance between points A and B is – (A) R/3 (B) 2R/3 (C) R (D) 4R/3

C

A

B

Sol. (B). If a potential is applied across A and B, due to symmetry, the points C, O and D will have same potentials. We can therefore join all these points without affecting current in any branch. The network then reduces to

D

O

O

A

A

B

B C,D

C,D

Therefore, the equivalent resistance is R eq 

R R 2R   3 3 3

Example 11 : A fuse with radius 1.0 mm blows at 15A. What is the radius of fuse made of same material which will blow at 30A. Sol. For fuse 2  r3

so

I12 I22



r13 r23

or

r 23 = r 13

I 22 I12

=

13  30  30 15  15

r 23 = 4 s o r = 1.587 mm Example 12 : Two circular rings of identical radii and resistance of 36 each are placed in such a way that they cross each other's centre C1 and C2 as shown in figure. Conducting joints are made at intersection points A and B of the rings. An ideal cell of emf 20 volts is connected across AB. The power delivered by cell is – (A) 80 watt (B) 100 watt (C) 120 watt (D) 200 watt Sol. (B).From the figure, AC1 = AC2 = C1C2 = radius  AC1B = 120° Hence the resistance of four sections are Hence equivalent resistance R across AB is 1 1 1 1 1     R 24 12 12 24 or R = 4.

A C2

C1 B

A 24

12

12

24

B

V 2 (20)2   100 watt  Power = R 4 Example 13 : The thermo emf of a copper constantan thermocouple varies as E = T +

T 2 where  = 40 µV/C and  = 40 × 10–3 µ V/C2. 2

If T = 100ºC and Tc = 0ºC then find thermoemf ? Sol. E = T +

T 2 40  109  (100)2 = 40 × 10–6 × 100 + = 4.2 × 10–3 V 2 2

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41

Current electricity Example 14 : Each of 3 resistors in figure has a resistance of 2 and can dissipate a maximum of 18 W without becoming excessively heated. Find the maximum power the circuit can dissipate. Sol.

Pmax  (i max )2 R 18  i 2max  2 i max  3A Here Req = 3.

So maximum power of circuit = i 2max R eq  32  3  27W Example 15 : E1 The circuit shown in the figure contains three resistors R1 = 100, R2 = 50 & R3 = 20and cells of emfs E1 = 2V & E2. The ammeter indicates a current of 50mA. Determine the currents in the resistors arid the emf of the second cell. The internal resistance of the ammeter and of the cells should be neglected. Sol. Applying KLV in upper loop R2 R1 E1 = (I + 0.05) R1 + IR2 I = – 20 mA mA R3 Current through R1 = 30 mA towards right Current through R2 = 20 mA towards left Applying KLV in lower loop E2 E2 = (I + 0.05) 100 + (0.50) 20 = 4 volts Example 16 : A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. Find the temperature of the wire is raised by the same amount T in the same time t the value of N. 3E , where E is the emf of each cell and R (L) is the resistance of the wire R Also, i² Rt = m.s.T ............. (1) where m is the mass of L length of wire and S is the specific heat of the material of the wire.

Sol. In the first case, i 

In the second case,

i 

NE where R   2L R

andi² R’t = m’s.T Dividing equation (2) by equation (1)

............. (2)

 i   R   m N2 R m N2 1 .  .  2  N=6   .      i R m 9 R m 9 2 Example 17 : copper strips In a practical wheat stone bridge circuit as shown, when one more resistance of 100  is connected is parallel with 100 unknown resistance ‘x’, then ratio 1 /  2 become ‘2’. 1 G is balance length. AB is a uniform wire. Find the value of A x. Sol.  Wheat stone bridge is in balanced condition

x

B

1

2

100

100

x

1

42

GyaanSankalp

2

E

r

copper strips

Current electricity 100x 100 100  x So  1 2



1 2  2

x = 100 

Example 18 : When a galvanometer is shunted with a 4 resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 wire, the further reduction in the deflection will be (the main current remains the same). (A) (8/13) of the deflection when shunted with 4 only (B) (5/14) of the deflection when shunted with 4 only (C) (3/4) of the deflection when shunted with 4 only Rg I/5 I G (D) (3/13) of the deflection when shunted with 4only Sol. (A). When shunted with 4 1 41  Rg   4 [ R g  16 ] 5 5 When further shunted with 2

4I/5

4 1 I   16  (I  I  )   I  3 13

Further reduction in current =

I

1 1 8  1     5 13 13  5 

I'

Rg

G

I–I'

8 13 of the deflection when shunted with 4only.

Hence further reduction in deflection =

Example 19 : 300 nos. of identical galvanic cells, each of internal resistance 9 are arranged a several in-series groups of cells connected in parallel. The arrangement has been laid out so that power output in an externally connected resistance of value 16 is maximum. If n number of cells are connected in every series group that form parallel combination, then find value of n.

E1

E2

En

Sol. Power developed in it is maximum when req = R nr R N/n

(N – total no. of cells) (n – cell in one series group)

RN using values R = 16, N = 300, r = 9, n = 23.1 r As n has to be integer and cofactor of 300, then nearest possible values are 20 and 25. Power is maximum when current through it is maximum n

nE n2r R N It is obvious that when n = 20, I is greater than when n =25. Therefore, n = 20 nos. Example 20 : Relation between current in conductor and time is shown in figure then determine. (a) Total charge flow through the conductor (b) Write expression of current in terms of time (c) If resistance of conductor is R then total heat dissipated across resistance R is

R

I

Sol. (a) q   idt = area of given curve

q

i i0

1 i0 t 0 2 GyaanSankalp

43

Current electricity i t (b) i  t  1  i  i0 0 0

 t 1  t  0 i

3 0  t i R 1   dt  2  t0  i0  t H 2 2  Rt i 2  2 (c) Heat =  i R dt   i R 1   dt ; ; H 00   t   t0  3 3    dt  i  t0  0 2

Example 21 : Consider the potentiometer circuit arranged as in figure. The potentiometer wire is 600 cm long. (a) At what distance from the point. A should the jockey touch the wire to get zero deflection in the galvanometer ? (b) If the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer ? Sol. (a) When jockey is not connected

I

E 16r

r

15r  / cm 600  Let  be the length when we get zero deflection

E E 15r    2 16r 600 (b) Let potential at A is zero

;

Then apply Kirchoff’s law,

x0  14r

x

N

......... (1)

Resistance per unit length  

 E   2   ()

15r

A



E/2

  320 cm.

E  0 (x  E  0) 14E 2  0x  r 2r 22

x I0 

r

E 2 

 14E  E   22  2 3E  r 22r

Example 22 : If two bulbs rated at 600 W, 220 V are put in series in a 110 V line then what is power generated by each bulb ? Sol. In series voltage across each bulb VA = Power consumed by each bulb P' =

R V 110 V= = = 55V RR 2 2

VA2 55  55 VA2 = 2 P= × 600 = 37.5 W VS 220  220 R

Example 23 : The resistivity of a ferric-chromium-aluminium alloy is 51 × 10–8 -m. A sheet of the material is 15 cm long, 6 cm wide and 0.014 cm thick. Determine resistance between (a) opposite ends and (b) opposite faces. Sol. (a) As seen from figure (a) in this case,  = 15 cm = 0.15 m A = 6 × 0.014 = 0.084 cm2 = 0.084 × 10–4 m2  51  108  0.15   9.1  10 3  4 A 0.084  10 (b) As seen from figure (b) here  = 0.014 cm = 14 × 10–5 m A = 15 × 6 = 90 cm2 = 9 × 10–3 m2  R = 51 × 10–8 × 14 × 10–5/9×10–3 = 79.3 × 10–10 

R= 

44

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Current electricity Example 24 : Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connected at point A and zero deflection is observed at length PJ = 10 cm. In second case it is connect at point C and zero deflection is observed at a length 30 cm from P. Then the unknown resistance x is – P (A) 2R (B) R/2 (C) R/3 (D) 3R Sol. (A). In potentiometer wire potential difference is directly proportional to length. Let potential drop unit length a potentiometer wire be K. For zero deflection the current will flow independently in two circles IR = K × 10 ........ (1) IR + Ix = K × 30 ........ (2) Equation (2) – (1) Ix = K × 20 ........ (3) Eq. (1)/(3)

R 1  x 2

G

10 2

10

2

2 10

10 10

2 4

x(assume)

10 2

10

35 volt  6x = 70  x  3

2

4

10V

35 20  3  25 A  I 4 12 Example 26 : In the circuit diagram shown find the current through the 1 resistor.

C

x

r

Sol. The simplified circuit is We have to find I. Let potential of point P be O. Potential at other points P be O. Potential at other points are shown in the figure apply Kirchoff’s current law at x.

Sol.

A

R

2

Example 25 : All batteries are having emf 10 volt and internal resistance negligible. All resistors are in ohms. Calculate the current in the right most 2 resistor.

x  10 x  10 x  20 (x  10)  0    0 4 2 4 2  x – 10 + 2x – 20 + x – 20 + 2x – 20 = 0

Q

J

120V

10

20

P 2

O(assume) 2

2

10V

(v  10)  10 v  0 v  5   0 2 2 1 v  20 v   v 5  0 2 2 v – 20 + v + 2 (v – 5) = 0 = 4v – 20 – 10 = 0 v

30 15  4 2

i

5/ 2 5  amp. 1 2

v 5 

15 15  10 5 5  2 2 2

5V 10V

1 10V

2 2

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45

Current electricity Example 27 : Three identical resistors are connected across a voltage source V so that one of them is in parallel with two others which are connected in series as shown. The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately – (A) the same (B) half as much (C) twice as much (D) four times as much R1

V2 Sol. (D). P1  R

V

R2 2

2

(V / 2) V P3  P2   , R 4R

R3

P1 4 P2

Example 28 : V is 10. To convert it into an ammeter of 1 Amp. a resistor The maximum current in a galvanometer can be 10 mA. It's resistance should be connected in – r E (A) series, 0.1  (B) parallel, 0.1 (C) series, 100 (D) parallel, 100 Sol. (B).IG = 10 mA G = 10  S (I – IG) = IGG ,where S is shunt is parallel A B R=15r S = 0.1  C

G E/2

Example 29 : A hemispherical network of radius a is made by using a conducting wire of resistance per unit length ‘r’. Find the equivalent resistance across OP.

1

P

B Sol. In given circuit A, B, C and D are at same potential by symmetry R eq

R2 

a r 2

O

ar     R eq  4 1  2  R eq 

R2

R1

R2

R1

R2 A,B C,D

R1

Example 30 : In the figure shown the current flowing through 2 R is – (A) from left to right (B) from right to left (C) no current (D) none of these Sol. (B). In figure all resistance are connected in parallel.

46

R1

ar  2   8

R

GyaanSankalp

C

O D

R1  ar

A

A

R  R2  1 4

2R

R

B

A

R

P

R2

2R

R

B

Current electricity So R eq 

2R  (R / 2) and current in all resistance flow from positive terminal of battery (means A end) to negative terminal of 2R  (R / 2)

battery (means B end). Example 31 : The voltmeter shown in figure reads 18V across the 50 resistor. Find the resistance of the voltmeter. Sol.



i

30  18 12 1   A 24 24 2

i

18 (R V  50) 1 A 2 R V  50

RV 

30V

V

50

 25 RV = 18 (RV + 50)

24

900  130 7

R

Example : 32 A circuit consists of a battery, a resistor R and two light bulbs A and B as shown. If the filament in light bulb A burns out, then the following is true for light bulb B : (A) It is turned off (B) Its brightness does not change (C) It gets dimmer (D) It gets brighter Sol. (D).

R  I   VB  Alternative : R

R

A r

i'

i 2

i 2 B

B

A

i' B

r

2

  2 2    r i r P1   i   4 P1    r  ; r  2 4 R   2

P1 

 2r (2R  r)2

2

   P2 = i’²R ; P2    r R  r

P2  2R  r     1 ; P2 > P1 P1  R  r 

So bulb B will become brighter. Example 33 : The given network is a part of a bigger network. Determine the value of current flowing in resistor R6. Sol. The current I in R6 can be calculated by applying I1 =3A Kirchhoff’s current law. However, here it is not necessary to first find the currents through indi10 R1 vidual resistances, R2, R3, R4 and R5. Instead, we R2 R3 can treat the whole network inside the dotted box 5 20 as a single junction. To this junction, currents I1 I2 =5A and I2 are entering the currents I3 and I are leaving. Hence, according to KCL, 7 9 I1 + I2 – I 3 – I = 0 R4 R5 or I = I1 + I3 – I3 = 3 + 5 – 1 = 7A 20

I3 =1A

R6

I

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47

Current electricity Example : 34 In the circuit shown, the charge on the 3 µF capacitor at steady state will be – (A) 6 µC (B) 4 µC (C) 2/3 µC (D) 3 µC

2V 2µF

1 3µF

2

Sol. (B). At steady state, there will be no current in the branches having capacitor only thus equivalent circuit diagram will be as shown in the figure. 1 4 VAB  2 VAB  1   0  VAB  V 3 2 1V thus q = CVAB = 4 µC

1V 2V 2µF 3µF 2

Example 35 : Find current in the branch CD of the circuit (in ampere) 2

2

B

2 A

D 3 C

30V

30  20 amp 3/ 2 From figure current through B D branch = 5 amp.

Sol. Req = 3/2 ;

i

B

D

2

2 2

A

D 20A

A

2

5A 2

D 2

10A 5A

3 D

D

10A 3

30V

Example 36 : The efficiency of a cell when connected to a resistance R is 60%. What will be its efficiency if the external resistance is increased to six times. Sol. Efficiency =  

 

i2R Ei

E

output power input power i 

E ; Rr



R Rr

r

i R

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Current electricity 0.6 

R  3R  3r  5R ; 2R = 3r Rr

 new efficiency,  

6R  0.9  90% 6R  r

A

Example 37 : In the circuit shown all five resistors have the same value 200 and each cell has an emf 3 volts. Find the open circuit voltage and the short circuit current for the terminals A and B. Sol. R = 100V, E = 3V In open circuit

A

B

3 3 3 i . 5  200

So VAB  E  ir  3 

3  400 5  200

VAB = 4.2 V In short circuit

B A zero resistance wire

6V B –6V

600

A 4

B 3V 0  (6) 6 1 0  (3) 3 i1     , i2  600 600 100 400 400 7 A 400 VB – VA = 4.2 V, I = 17.5 mA (B to A)

iAB = i = i1 + i2 =

–3V

2

400

0V

i

Example 38 : Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10-7 m2 carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3, and its atomic mass is 63.5 amu Sol. The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed vd is given by vd = I/neA. Now, e = 1.6 × 10-19 C, A = 1.0 × 10-7 m2, I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of 9.0 × 103 kg. Since 6.0 × 1023 copper atoms have a mass of 63.5 g.

n

6.0  10 23  9.0  10 6 = 8.5 × 1028 m–3 which gives 63.5

vd 

1.5 8.5  10

28

 1.6  1019  1.0  107

= 1.1 × 10-3 m s-1 .

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