Crackiitjee.in.Phy.ch18

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Resistance Resistance in Series:

“The Connected resistances are said to be in series when a potential difference that is applied across the combination is the sum of the resulting potential differences across the individual resistances".

The current through each resistor in series is necessarily the same but this is

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not the sufficient condition for resistors in series. The condition becomes sufficient when the current through each resistor is the same as the current through the terminals of the network.

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A combination of three resistances R1, R2 and R3 in series. The current through each resistor is the same as the current i through the terminals A or B of the

ac

networks. Suppose that V1, V2 and V3 are the potential differences across R1' R2

cr

and R3 respectively. If V = VA - VB be the potential difference across the network, then according to the definition of series combination. V = V1 + V2 + V3

== iR1 + iR2 + iR3 == i (R1 + R2 + R3)

Important Points: Total potential (V) is divided in such a way that :

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V1 : V2 : V3 == R1 : R2 : R3

Or,

If R1= R2 =R3 then V1= V2 = V3= V/3 and Req =3R

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Similarly, if there are n resistors in series,

cr

ac

ki

Req=R1 +R2+ .. + Rn

If R1= R2 =R3' then V1= V2 = V3= V/3 and Req =3R , Similarly, if there are n resistors in series, Req=R1 +R2+ .. + Rn

RESISTANCES IN PARALLEL

The connected resistances are said to be in parallel when a potential difference that is applied across the combination is the same as' the resulting potential difference across the individual resistances. It shows

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three resistances in parallel V = potential difference across each resistor = potential difference across the combination between A and B The currents in three resistances are: i1 =

V V V ,i = and i3 = R1 2 R2 R3

.in

The total current is

Or, V = V + V + V Req

R1

R2

R3

R2

R3

cr ac

For n resistors in parallel,

itj

R1

ki

Req

ee

Or, 1 = 1 + 1 + 1

1 =1 +1 + +1 ... Req R1 R2 Rn

For two resistors in parallel,

1 = R1 R2 Req R1 + R2

(i) here equivalent resistance is less than each of the individual resistance.

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(ii) i1 : i2 : i3 = 1 : 1 : 1 R1 R2 R3

Example. What is the equivalent resistance between two points A and B?

Solution: Taking R2 and R3 in parallel, the resulting

will be as following.

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Now, equivalent resistance will be in series combination.

. . Req = R1 + R + R4

ki

= (2 + 2+ 2) Ω = =6Ω

ac

ELECTRIC CURRENT AND KIRCHHOFF'S LAW

Internal Resistance: Every source of emf has a resistance of its own, known

cr

as the internal resistance (r). For example, in case of a cell, the components inside it offer a resistance to the flow of current. The value of the internal resistance of a cell depends on: (a) the surface area of its electrodes, (b) the separation between them and (c) the nature, concentration and temperature of its electrolyte.

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A simple electric circuit When a current i flows through C, the source C does work to transfer positive charge from one end A to the other end B. lf E is the emf of the cell and r its internal resistance, then a part v

of

E is used up in transferring

charge from A to B. From Ohm's law, this part is given by v = ir

... (i)

The remaining part V = (E - v) drives the charge through the external resistor R; this part is given by ... (ii)

V = iR

Now, E = V + v

... (iii)

Equations (i) to (Hi) give

cr

ac ki it

ER V R " = E R+r (R + r) E-V = E-V Also, i = ..... (iv) R r E - V = (E - V) or r = R i V E - V = ir ... (v)

V=

je e. in

I

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Notice from Equation (iv) that if i = 0 E = V, i.e., if no current is drawn from the ell (i.e., in an open circuit), the potential difference V across its terminals equals its emf E. If ≠ 0, V is less than E.

From (v), E=V, when r = O, V < E. lf r ≠O

Kirchhoff's Rules: Kirchhoff found two rules for determining the current and

ee

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resistance in a complicated circuit. These laws are as follows:

1. In an -electric circuit-the algebraic sum of the currents meeting at any

ki

itj

junction in the circuit is zero, that is ∑I = 0

ac

When applying this law, the current going towards the junction is taken as

cr

positive while that going away is negative. In the following

11 + 13 - 12 - 14 -Is = 0

(a) and (b)

or 11 + 13 = 12 + 14 + Is

Thus the sum of the currents flowing toward the junction is equal to the sum of the currents flowing away, from the junction. Therefore, when a steady current flows in a circuit then there is neither any accumulation of charge at any point in the circuit nor any depletion of charge from there. Thus, Kirchhoff's first rule expresses the conservation of charge.

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2. In any closed mesh of a circuit, the sum of the products of the current and the resistance in each part of the mesh is equal to the algebraic sum of the e.m.f.'s in that mesh, that is, ∑(IR) = ∑E

In applying this law, when we traverse in the direction of current then the product of the current and the corresponding resistance is taken as positive, and the e.m.f. is taken as positive when we traverse from the negative to the positive electrode of the cell through the electrolyte.

.in

Applying Kirchhoff's" rule to mesh 1, (or closed mesh ABFGA)

ee

l1 R1 - l2 R2 = E1 - E2

ki

l2 R2 + (l1 + l2)R3 = E2

itj

Similarly, for the mesh 2, (or closed mesh BCDFB) we have

cr ac

From these equations, we can determine the values of currents in different parts of the circuit.

Grouping of Cells: A cell is a source of electric current. A single cell cannot give a strong current.

Therefore, to get strong current two or more cells are to be combined. The combination of cells is called a 'battery'. Cells can be combined in three ways: (a) In series, (b) In parallel and (c) In mixed grouping.

1. In series: In this combination, the negative pole of the first cell is

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connected to the positive pole of the second cell, the negative pole of the second. to the positive pole of the third, the negative pole of the third to the positive pole of the fourth, and so on

Suppose, n cells each of e.m.f E and internal resistance r are connected in series. These cells are sending current in an external resistance R. Then, Total e.m.f of the cells = n E

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Total internal resistance = n r

ee

Therefore, Total resistance of the circuit = (nr + R)

nE (nr + R)

cr

ac

Important Points:

ki

itj

If the current in the circuit be l, then = I =

1. If r« R, then from eq. (i) I = nE R (approx,) that is, if the internal resistance of the connected cells is much smaller than the external resistance, then the current given by these cells will be nearly n times the current given by one cell. Hence, when the internal resistance often connected cells is much smaller than the external resistance, then the 'cells should be connected in series to obtain it strong current.

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2. If nr » R, then, I = nE = E (approx), that is, if internal resistance R r of the connected cells is much greater than the external resistance, then nearly the same current is obtained by single cell. Hence, there is no advantage of connecting cells in series. 3. If in series grouping of n cells, n1 cells are reversed then Eeq = (n- n1)E - n1E

= (n - 2n1)E and req = nr

So I = (n - 2n1) E

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R + nr

2. In Parallel: In this combination, the positive poles of all the cells are connected to one point, and the negative poles to another point . Suppose m cells, each of e.m.f. E and internal resistance r, are connected in parallel

ki

and this battery of m cells is connected to an external resistance R. Since

ac

the

cell are connected in parallel, the e.m.f. of the battery will also be E. If the

cr

equivalent internal resistance of the cell be R1,

then

1 = 1 + 1 + ..... upto m terms = m or R = r 1 R1 r r r m

Therefore, Total resistance of the circuit = (r/m + R). If the current in the external circuit be l,

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Then, l =

E = mE (r/m) + R r + mR

Important Points: i.

If rR, that is, if the internal resistance of the cells in

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ii.

larger than the external resistance, then the current will

ac

be I = nE (approx.). This current is nearly m times the R

cr

current given by r

a single cell. Hence, when the internal resistance of the cell is much larger than the external resistance, then the cells should be connected in parallel.

Mixed Grouping: - In this combination, a certain number of cells are connected in series, and all such series. Combinations are then connected mutually in parallel.

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Suppose n cells are connected in each series, and such m rows are connected in parallel. Let the e.m.f. of each cell be E and the internal resistance be r. This battery of cells is sending current in an external resistance R.

The total e.m.f. of n cells connected in one series is nE. Since all the series are connected in parallel the e.m.f. of the battery as a whole will also be nE.

Similarly, the total internal resistance of cells in a series is nr. Such m rows are

.in

connected in parallel. Hence,if the internal resistance of the whole battery be R1

ee

then

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itj

1 = 1 + 1 + m nr ... upto m terms = or R1 = R1 nr nr nr m

cr ac

Therefore, total resistance of the circuit = $` nr j + R . . If the current in the m external circuit be l, then I = mnE …… (i) nr + mR

In eq. (i) for the value of

l to be maximum, the value of (nr + mR)

should be minimum.

Now

Nr +mR = { √nr - √mR }2 + 2√(mnRr)}

Therefore, for (nr +mR) to be minimum, {√nr - √mR} = 0

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Or √nr = √mR

or

nr = mR

or

R=

nr m

But, nr/m is the internal resistance of the whole battery. Thus, in mixed grouping the current in the external Circuit will be maximum when the internal resistance of the battery is equal to external resistance. By substituting nr/m = R in eq. (i), we can see that the maximum current in the external circuit will be nE/2R or mE/2r.

.in

Moving Coil Galvanometer Moving coil galvanometer is a device used to detect small current flowing in an electric

ee

circuit. With suitable modifications, it can be used to measure current and potential

itj

difference.

cr ac

ki

Conversion of galvanometer into an Ammeter

An ammeter is an instrument which is used to measure current in a circuit in ampere (or miIli-ampere or micro- ampere). Hence, it is always connected in series in the circuit. Since, the galvanometer coil has some resistance of its own, therefore, to convert a galvanometer into an ammeter, its resistance is to be decreased so, to convert a galvanometer into ammeter a low resistance, called shunt (S) is connected in parallel to the galvanometer as shown in .

Here, ig = And RA =

S i S+G

^ Sh^Gh

(G + S)

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Where RA = Resistance of ammeter S = Resistance of Stunt G = Resistance of Galvanometer

Conversion of Galvanometer into Voltameter

A voltameter is an instrument which is used to measure the potential difference between two points of an electric circuit directly in volt (or milli-volt or micro-volt). Hence, it is connected in parallel across is to be measured. When it is

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connected. Since, the resistance of coil of galvanometer of its own is low, hence, to convert a galvanometer into a voltmeter, high resistance R in series is connected with the galvanometer.

V R+G

ki

Here, ig =

ac

Where R + G = Rv = resistance of voltameter

cr

Example. A millimeter of range 10 m A and resistance 9 X is joined in a circuit as shown. The metre gives full – scale deflection for current I when A and B are used as its terminals i.e., current enters at A and leaves at B (C is left isolated). The value of I is (A)100mA

(B) 900mA

(C) 1A

(D) 1.1A

Solution. According to loop rule, -9X10-0.9X10+0.1X(I-10) = 0 Or, I – 10 = 90 + 9 = 990 0.1

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Therefore, I = 1000mA = 1A

Example. In the circuit shown in

reading of votameter is V1 when only S1 is

closed, reading of voltmeter is V 2 when only S2 is closed. The reading of voltmeter is V3 when both Sl and S2 are closed then (A)V2>V1>V3

(B)V3>V2>V1

(C)V3>V1>V2

(D)V1>V2>V3

Solution. When S1 is closed, I=

f 4R

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v1 = I X 3R = f X 3R 4R 3f v1 = 4

When S2 is closed, f 7R

ki

I=

cr

ac

v2 = I X 6R = f X 6R 7R 6f v2 = 7

When S1 and S2 both are closed, I=

f 3R

v3 = I X 2R = v3 =

f X 2R 3R

2f 3

Example: In the circuit shown in

the reading of ammeter is the same with both

switches open as with both closed. Then find the resistance R. (ammeter is ideal)

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Solution.

Step-I: Discuss the circuit when both switches open: According to loop rule: 1.5 - 300 I-100 I - 50 I = 0

Therefore, I = 1.5 = 15 = 1 A 450 4500 300

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Step – II: Discuss the circuit after closing the switch.

In loop ABCDEA, -IR + 1.5 – 300I1=0 Or 300I1 + IR = 1.5

ki

In loop BCGFB,

ac

-100I + (I1 – I)R =0

I1R = (100+R)I

Therefore, I1 =

cr

Or (I1 – I)R = 100I

^100 + R h I

R

From eqn (1) & (2) 300

(100 + R) I + IR = 1.5 R

300

(100 + R) 1 + R = 1.5 R 300 300

R = 600X

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Example. The battery in the diagram is to be charged by the generator G The generator has a terminal voltage of 120 volts when the charging current is10 amperes. The battery has an emf of 100 volts and an internal resistance of 1 ohm. In order to charge the battery at 10 amperes charging current, the resistance R should be set at: (A) 0.1X (B) 0.5 X (C) 1.0 X

.in

Solution. VA - VB = - { algebraic sum of rise up and drop up of voltage } 120 = -{-IR-1 X I -100}

cr ac

Therefore, R = 1 X

itj

20 = 10R + 10

ki

Or

ee

120 = IR + I +100

Example. The resistance of thin silver wire is 1.0 X at 20°C. The wire is placed in a liquid bath and its resistance rises to 1.2 X . What is the temperature of the bath? α for silver is 3.8X10-3 per °C.

Solution.

R(T) = R0 *1+α(T-T0)] Hence, R(T) = 1.2 X , R0 = 1.0 X Α = 3.8 X 10-3 per°C and T0 = 20°C

Substituting the values, we have 1.2 = 1.0[1+3.8 X 10-3(T-20)]

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Or

3.8 X 10-3 (T-20) = 0.2

Solving this, we get T = 72.6°C

Example. A resistance R of thermal coefficient of resistivity = α is connected in parallel with a resistance = 3R, having thermal coefficient of resitivity = 2α. Find the value of αeff.

Solution. The equivalent resistance at 0°C is

R10 R20 R10 + R20

….(1)

The equivalent resistance at t°C is

…(2)

R1 = R10(1+αT) R2 = R20(1+2αT) R = R0(1 + αefft)

…(4)

…(5)

cr

And

…(3)

ki

But,

R1 R2 R1 + R2

ac

R=

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R0 =

Putting the value of (1), (3), (4) and (5) in eqn (2) αeff = 5 a 4

Example. (a) The current density across a cylindrical conductor of radius R varies according to the equation J = J0 (1 - r ) , where r is the distance from the axis. Thus the current density is a R maximum Jo at the axis r = 0 and decreases linearly to zero at the surface r = R

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Calculate the current in terms of Jo and the conductor's cross sectional area is A = nR 2 (b) Suppose that instead the current density is a maximum J 0 at the surface and decreases linearly to zero at the axis so that J = J = J0 r . Calculate the current. R

Solution. (a) We consider a hollow cylinder of radius r and thickness dr. The cross-sectional area of considered element is dA = 2πrdr The current in considered element is

dI = 2rJ0 (1 - r ) rdr R

or

R

Therefore,

I = 2rJ0 0

# (1 - Rr ) rdr

I = I = J0 rR = J0 A 3 3

ac

ki

2

itj ee .in

dI = JdA = J0 (1 - r ) 2rrdr R

R

cr

(b)

2rJ0 # r2 I= dr R 0 R

I=

2rJ0 ; r3 ER R R 0

I=

2rJ0 r3 = 2A J R R 3 0

Example. A network of resistance is constructed with Rl & R2 as shown in the . The potential at the points 1, 2,3,……….., N are V1, V2, V3,…….., VN respectively each having a potential k time smaller than previous one. Find: 1) R1 and R2 in terms of K R2 R3

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2) Current that passes through the resistance R2 nearest to the V0 in terms V0, k and R3.

Solution.

1) According to kcL,

I = I1 + I2

Or

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Or

V0 - V0 V V0 V0 - 0 0 k2 k = k + k R1 R2 R1

(k - 1)V0 (k - 1)V0 = V0 + kR1 kR2 k2 R1

Therefore,

R1 = (k - 1) R2 k

Also,

I’ = I’1 + I’2

2

Or

ac

ki

V0 - V0 V0 V0 0 0 N-1 k N-2 k N-1 = k N-1 k + R1 R2 R1 + R3

cr

After Solving,

R2 = k R3 k - 1

2)

V0 0 V 1 -0 Here, I1 = = k R2 R2

I1 =

V0 = kR2

(k - 1)V0 V0 = k k2 R3 k( )R3 k-1

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