Class Assignment Current Electricity

Important questions of Current Electricity 1. In a metre bridge, the null point point is found at a distance of 40 cm from A. If a resistance of 12 Ω is connected in parallel with S, the null point occurs at 50.0 cm from A. Determine the values of R and S. 12 Ω

Ans: Initially the null point is found at a distance of 40 cm from end A, hence

ʼ = Fˋ =  = , ʽ {Fˋ  

When a resistance of 12

⇒ Ω

 R= S 

S

R

…(i)

is connected in parallel with S, the effective

W ) Ω. As now null point occurs at 50.0 cm from A, resistance becomes ( W ʼ Fˋ   ʽ hence = = , R= …(ii) ⇒ ʽ ʽ {$ʽ {Fˋ    S = W , Comparing (i) and (ii), we get, S = 6 Ω,or ⇒  W

A

B

=

R=

 S= X6Ω=4Ω  

2. In a metre bridge, the null point is found at a distance of 60.0 cm from A. If now a resistance of 5 connected in series with S, the null point occurs at 50 cm. Determine the values of R and S.

Ω

is

S

R

Ans: Initially the null point is formed at a distance distance of 60 cm from the end A, hence hence

ʼ = Fˋ =  = , ʽ {Fˋ  

⇒

R=

S 

…(i)

A

B

When a resistance of 5 Ω is connected in series with S, the effective eff ective resistance becomes (S + 5) Ω. As now null point occurs at 50 cm, hence

ʼ = Fˋ = '" =1, {ʽ {Fˋ '"

Comparing (i) and (ii), we get,

⇒

R=S+5

 S =S + 5, 

⇒

…(ii) S = 10 Ω,

or

R=

 S =  X 10 = 15 Ω  

3. In metre bridge, the null point is found at a distance distance of l 2 cm from A. If now a resistance of X is connected in parallel with S the null point po int occurs at l 2cm. Obtain a formula for X in terms of l 1, l2 and S. X

Ans.: Initially the null point is found at a distance distance of l 2 cm from end A, hence

ʼ = ˊFˋ = ˊ ʽ {ˊFˋ {ˊ

…(i)

R

S

When a resistance of X is connected in parallel with S, the effective

˂ʽ ). As now null point occurs at l cm, hence 2 ˂ʽ ʼ = ˊFˋ = ˊ …(ii) ˂ʽ {˂$ʽ {ˊFˋ {ˊ ˂ʽ = ˊ X {ˊ Comparing (i) and (ii), we get, ˂ {ˊ ˊ ʽ ˊ {ˊ  9 ʽ{ˊ  - 1, ˊ =  or X = ˊ %ˊ  = ⇒   ˂ {ˊˊ {ˊˊ  %ˊ ˊ # resistance becomes (

A

B

4. Write any two factors on which internal resistance of a cell depends. The reading on a high resistance V voltmeter, when a cell is connected across it, is 2.2 V. When Whe n the terminals of the cell are also connected to a resistance of 5 Ω as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell. R=5Ω

K

Ans: The internal resistance of a cell depends on nature and electrolyte as well as the concentration of the electrolyte. As per data given, emf of the cell ε = 2.2 V External resistance R = 5 Ω, and terminal potential difference of the cell V = 5 Ω or

Internal resistance of cell r =

ˀ . R =  X 5 = ˂ = 1.1 Ω ˀ  

5. Use these rules to write the expressions for the currents I1, I2 and I3 in the circuit diagram shown. Ans: In the network given here as per Kirchhoff’s first law, we have, In loop ECDEF, we have - 1 – 3I2 – 2I3 + 4 = 0, ⇒ Again in loop EABFE, we have, - 2 – 4I 1 – 2I3 + 4 = 0, ⇒ Multiplying (ii) by 4 and (iii) by 3 and then adding them, we get 14I3 + 12I2 + 12I1 = 18, 14I3 + 12 (I1 + I2) = 18 ⇒ But I1 + I2 = I3, hence we have, 14I3 + 12I3 = 26I3 = 18

  I3 = A = A  

⇒

2X

 + 4I = 2,  1

⇒

I1 =

 A, 

or

r2 = 3 Ω

I3 E3 = 4 V

r3 = 2 Ω

…(i) ….(ii) ….(iii)

E1 = 2 V

I1

r1 = 4 Ω

I2 E2 = 1 V

I 3 = I1 + I2 2I3 + 3I2 = 3 2I3 + 4I1 = 2 r1 = 4 Ω

B

A C E

Substituting the value of I 3 in equation (iii), we get

E1 = 2 V

I1

E2 = 1 V

I2

r2 = 3 Ω D

I3 E3 = 4 V

F

-=A   

I2 = I3 – I1 =

6. State Kirchhoff’s rules. Apply Kirchhoff’s rules to the loops ACBPA and ACBQA to write the expressions for the currents I 1, I2 and I3 in the network.

r3 = 2 Ω

E1 = 6 V P I1 0.5 Ω

A

B

1Ω Ans: In the network given, as per Kirchhoff’s first law, we have, I2 Q I1 + I2 = I3 …(i) In loop ARBE1A, we have, I3 E2 = 10 V C - 12I3 – 0.5I1 + 6 = 0, 12I3 + 0.5I1 = 6 ….(ii) ⇒ R = 12 Ω Again in loop ARBE 2, we have - 12I3 – 1.I2 + 10 = 0 12I3 + I2 = 10 ….(iii) Multiplying (ii) by 2 and then adding to (iii), we get ⇒

36I3 + (I1 + I2) = 22

⇒

36I3 + I3 = 37I3 = 22,

I2 =

 A, 

or

 A 

12 X

7. State Kirchhoff’s rules. Apply these rules to the loops PRSP and PRQP to write the expressions for the currents I 1, I2 and I3 in the given circuit. Ans: In the network given as per Kirchhoff’s first law, we have, I 1 + I2 = I3 In loop PRQP, we have, - 20I3 – 60I1 + 4 = 0 or 20I3 + 60I1 = 4, or 20 (I1 + I2) + 60I1 = 4 80I1 + 20I2 = 4 …(ii) ⇒ Again in loop PRSP, we have, - 20I3 – 200I2 + 5 = 0 20I3 + 200I2 = 5 or 20 (I1 + I2) + 200I2 = 5 20I1 + 220I2 = 5 ….(iii) ⇒ On solving equations (ii) and (iii), we get,

I3 =

 + I2 = 10     A I1 = I3 – I2 = =  

Substituting the value of I 3 in equation (iii), we get, ⇒

or

I1 =

200 Ω

S 5V

60 Ω

20 Ω

I2

I3 P

…(i)

R

I1

200 Ω

S

Q

4V I2

I1

5V 20 Ω I2

I3 P

I1

X

60 Ω Q

4V

 A, I =  A and I =  A 3  2  

8. The figure shows experimental set up of a metre bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from

R

Y B G

A

D

C

the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y. Ans: When two unknown resistances X and Y are inserted as shown in the figure, then null point D is obtained at 40 cm from end A i.e.. l = 40 cm. or

˂ =  = , ˃ { 



Y=X

⇒

….(i)

When a resistance of 10 Ω is connected in series with X, the null point should shift away from point A. As null point shifts by 10 cm, hence l 1 = 40 + 10 = 50 cm. or

˂ =  = , ˃  

X + 10 = Y

⇒

…(ii)

Solving (i) and (ii), we get X = 20 Ω and Y = 30 Ω. If a resistance of 10 Ω is connected in series with resistance Y then the null point with be obtained at a distance l2 from point A, such that

˂ = ˊ ˃ ˊ

or

 =  = ˊ ,   ˊ

⇒

l2 =

 cm = 33.3 cm 

R

E J

A

B

9. Two students X and Y perform an experiment on potentiometer separately using the circuit diagram shown here. Keeping other things unchanged ε (i) X increases the value of resistance R S (ii) Y decreases the value of resistance S in the set up. How would these changes affect the position of the null point in each case and why?

G

Ans: (i) When student X increases the value of resistance R, the current flowing in the potentiometer wire AB and consequently the potential gradient decreases. Consequently, the used length of potentiometer increases i.e.., the null point shifts towards right (towards point B). (ii) When student Y decreases the value of resistance S in the circuit of auxiliary cell ε, then current I flowing through S increases and as a result, the potential difference across cell V = ε- SI decreases. As a result, position of null point shifts towards left (towards point A). R = 1.5 Ω 10. 12 cells, each of emf 1.5 V and internal resistance 0.5 Ω, are arranged in m rows each contacting n cells connected in series, as shown. Calcul ate the values of n and m for which this combination would send a maximum current through an external resistance of 1.5 Ω

m rows

(n cells in each row) Ans: The equivalent internal resistance of each cell row having n cells in series, each on internal resistance r = nr and the net equivalent resistance of whole combination of cells, having m rows in parallel req = nr/m. For maximum current through an external resistance R, the net equivalent internal resistance should be exactly equal to the external resistance. Thus, we have

n r/m = R i.e.,

 X 0.5 Ω = 1.5 Ω, 

⇒

n=3m

Again total number of cells in the combination = n m = 12,

Solving (i) and (ii), we get n = 6 and m = 2

11. For the circuit shown here, calculate the potential difference between the points B and D. Ans: The given circuit and the current distribution in it can be represented as shown in figure given below:

2 V, 2 Ω B

A 1 V, 1Ω

1 V, 1Ω

2Ω D

C 3V 3Ω

Applying Kirchhoff’s second law for the mesh ADBA, we have - I1. (1) – I2. (2) – I1 (2) + 2 – 1 = 0, 2I2 + 3I1 = 1 …..(i) ⇒ Again applying Kirchhoff’s second law for mesh DCBD, we have - (I1 – I2) (3) + 3 – 1 – (I1 – I2). (1) + I2. (2) = 0, - 6I2 + 4I1 = 2 ….(ii) ⇒ On solving equations (i) and (ii), we get, or

' # I 1 = # A and I2 = - # A

Potential difference between the points B and D, i.e..,

#  V B – V D = - I2 X (2 Ω) + #X 2 = + # V = + 0.154 V.

A

2V

l1

2Ω

l1

1V

B (l1 – l2)

l2

1Ω l2

1Ω D

C (l1 – l2)

3Ω

3V

12. The I – V characteristics of a resistor, are observed to deviate from a straight line for higher values of current as shown below. Why? Ans: For higher values of current the I-V curve is being deviated from a straight line and I current is falling. It shows that for higher current resistance of the given r esistor has increased. The rise in resistance R for higher currents is due to the fact that the temperature of resistor increases on passing stronger current and consequently its resistance increases.

V

13. Carbon and silicon are known to have similar lattice structures. However, the four bonding electrons of carbon are present in the second orbit while those of silicon are present in its third orbit. How does this difference result in a difference in their electrical conductivities? Ans: Due to difference in orbit of bonding electrons, the ionization energy of silicon is considerably reduced as compared to carbon. As a result of it, the electrical conductivity of silicon is considerably enhanced. In fact, silicon is a semiconductor but carbon is an insulator. 14. A cell of emf (ε) and internal resistance (r) is connected across a variable external resistance (R) Plot graphs to show variation of (i) ε with R, and (ii) terminal potential difference of the cell (V) with R. Ans: (i) We know that the emf ε of a cell is equal to terminal potential difference of the cell ε when the cells is an open circuit and no current is being drawn from it. So, emf of a cell 0 does not depend on the external resistance R and hence ε – R graph is as shown here: (ii) When the cell circuit is closed through an external resistance R, current flowing I=

, ʼG

and terminal potential difference V = RI =

ʼ = _8D ʼG {#

R

V

ε

Thus, as R increases V increases and for large values of R the value of V approaches the limiting value ε. Hence, V – R graph is as shown here:

R Y

X

15. The given figure shows the experimental set up of a metre bridge. The null point is found to be 60 cm away from the end A with X and Y in position as shown. When a resistance of 15 Ω is connected in series with Y, the null point is A found to shift by 10 cm towards the end A of the wire. Find the position of null point if a resistance of 30 Ω were connected in parallel with Y. Ans: Initially as per null point condition of metre bridge, we have, On joining 15

Ω

B

C

D

˂ = Fˋ =  =  ⇒ ˃ {Fˋ  

X=

 Y. 

resistance in series with Y the null point shifts to 10 cm towards A i.e.., l = 60 – 10 = 50 cm.

˂ = Fˋ =  = 1, ˃ {Fˋ 

Hence,

⇒

X = Y + 15 or

When a 30 Ω resistance is joined in parallel to Y = 30 Y’ =

 Y = Y + 15 

⇒

Y = 30

,

Ω

and

X = 45 Ω

, the combined resistance

Ω

˂ = 15 Ω. Now if null point be obtained at l’ cm from point A, then  ˂ =  = ˊ ⇒ l’ = 75 cm. ˃  {ˊ

6V

16. The plot of the variation of potential difference across a combination of three identical cells in series versus current is as shown below. What is the emf of each cell? Ans: In open circuit, when no current is being drawn from the cells, the terminal potential 1A difference is equal to emf of cells. Or emf of the series combination of three cells = 6 V or emf of each cell ε = (6V/3) = 2 V. 17. You are given n resistors each of resistance r. These are first connected to get minimum possible resistance. In the second case, these are again connected differently to get maximum possible resistance. Compute the ratio between the minimum and maximum value of resistances so obtained. Ans: For getting minimum possible resistance, we join n resistors in parallel and then R min = n/r. For getting maximum possible resistance, we join n resistors in series and then R max = nr or

Rmin/Rmax =

G

ˌ ˌG

= 1/n2.

18. A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V. Calculate the energy liberated per second in the new combination. Ans: Resistance of heater coil R = V 2/P = ((200)2/100) = 400 Ω On cutting it into two identical parts, resistance of each part = R/2 = (440/2) = 200 Ω On joining these two parts in parallel, equivalent resistance, R 0 = 200 X 200/200 + 200 = 100 As again voltage = V = 200 V, or Energy liberated per second in new combination = V2/R0 = 200 X 200/100 = 400 W. 19. The following graph shows the variation of terminal potential difference V, across 6.0 a combination of three cells in series to a resistor, versus the current, i: 3.0 (i) Calculate the emf of each cell. V (ii) For what current i will the power dissipation of the circuit be max imum? O

1.0

2.0

Ω

i (ampere)

Ans: (i) From the graph it is clear that when no current is being drawn from the cells (i.e.., i = 0), voltage is 6.0 volt. As the battery is a combination of 3 cells and in an open circuit terminal potential difference is equal to emf, hence emf of each cell = ε = (6.0/3) V = 2.0 V (ii) From graph, terminal potential difference V is zero when current drawn is i s = 2.0 A. It represents the short circuit condition. Thus, ε = r i, where r = internal resistance of a cell r = ε/is = 2.0 V/2.0 A = 1 Ω ⇒ Power dissipation of the circuit will be maximum when external resistance is equal to internal resistance (R = r) i.e.., current drawn is i = ε/r + R = ε/2r, = 2V/2 X 1Ω = 1 A. 20. In the circuit of fig. a metre bridge is shown in its balanced state. The metre bridge wire has a resistance of 1 ohm/cm. calculate the value of the unknown resistance X and the current drawn from the battery of negligible internal resistance

l

X

Ans:

˂  As per balance condition = ,  

⇒

6Ω

G

60 cm

40 cm

˂ = 4 Ω X= 

A

B

J

Resistance of AJ part of wire = 40 X 1 = 40 Ω and that of JB part = 60 x 1 = 60 Ω 6V As galvanometer shows no deflection, bridge is balanced. Combined effect of series combination of X + 6 = 4 + 6 = 10 Ω. Combination effect of series combination of AJ = 40 + 60 = 100 Ω or

Resultant resistance between the points A and B,

or

Current drawn from the battery,

˂ =  =  Ω      = ˂ = 0.66 A. I= = ʼ È 

R=

21. For the potentiometer circuit shown in the given figure, points X and Y represent the two terminals of an unknown emf ε. A student observed that when the jockey is moved from the end A to the end B of the potentiometer wire, the direction of the deflection in the galvanometer r emains in the same direction. What may be the two possible faults in the circuit that could result in this observations? If the galvanometer deflection at the end B is (i) more, (ii) less, than that at the E K end A which of the two faults, listed above, would be there in the circuit? A Give reasons in support of your answer in each case.

B

ε

G

Ans: Two possible faults in the circuit can be: 1. Positive terminal of unknown emf X Y source ε is not connected to point A, where positive terminal of battery E has been connected. 1. Emf of battery E is less than the unknown emf ε. (i) If on sliding the jockey from the end A to end B galvanometer deflection gradually increases then it means that connections of unknown emf ε are wrong because then in accordance with Kirchhoff’s laws potential is gradually increasing From A to B. (ii) If on sliding the jockey from the end A to end B galvanometer deflection decreases but null point is not obtained then it means that the emf of battery E is less than unknown emf ε and hence deflection is in one direction only. 22. Four cells of identical emf E, internal resistance r, are connected in series to a variable resistor. The graph shows the variation of terminal voltage of the combination with the curr ent output. 5.6 (i) What is the emf of each cell used? (ii) Calculate the internal resistance of each 4.2 cell. (iii) For what current from the cells, does maximum power dissipation 2.8 occur in the circuit? 1.4

0 0.5 1.0 1.5 2.0 Ans: (i) From the graph when I = 0, terminal potential difference = 56 V. When no current is being drawn from the cells, terminal potential difference is equal to emf of the combination. Hence 5.6 V = 4 E, E = (5.6/4) V = 1.4 V ⇒ (ii) From figure when current drawn is I = 2.0 A, terminal potential difference V = 0. As V = 4E – I. (4r), hence 0 = 5.6 – 2 X 4r, r = (5.6/8) = 0.7 Ω ⇒ (iii) Maximum power dissipation takes place when external resistance = total internal resistance i.e.. R = 4r = 4 X 0.7 Ω = 2.8 Ω, or Current for maximum power I = (4E/4r + R) = (4 X 1.4/4 X 0.7 + 2.8) = 1A.

23. A 20 V battery of internal resistance 1 Ω is connected to three coils of 12 Ω, l 6 Ω and 4 Ω in parallel, a resistor of 5 Ω and a reversed battery 5Ω (emf 8 V and internal resistance 2 Ω) as shown in fig. Calculate (i) the current in the circuit (ii) the current in resistor of 12 Ω coil, and (iii) potential difference across each battery.

l 20 V, 1 Ω 12 Ω

8 V, 2 Ω

6Ω 4Ω

Ans: Combined resistance R 1 of parallel combination of 12 1/R1 =

 ++=    

, 6 Ω and 4 Ω is given by

Ω

R1 = 2 Ω,

⇒

Further R2 = 5 Ω, r1 = 1 Ω and r2 = 2 Ω

Hence, total resistance in the circuit = R 1 + R2 + r1 + r1 = 2 + 5 + 1 + 2 = 10 Ω Total emf = ε1 – ε2 = 20 – 8 = 12 V,

(i) Current in the circuit I =

ˀ = 1.2 A  Ω

(ii) Combined resistance of 6 Ω and 4 Ω in parallel = 6 X 4/6 + 4 = 2.4 Ω As resistances of 12 Ω and 2.4 Ω are in parallel and main current is 1.2 A, hence current flowing through 12 Ω resistor is

I’ =

 = & X 12 = 0.2 A  #&&

(iii) The potential difference across Ist battery, V 1 = ε1 – Ir1 = 20 – (1.2) X 1, = 20 – 1.2 = 18.8 V and the potential difference across 2 nd battery joined in reverse manner V2 = ε2 + Ir2 = 8 + (1.2) X 2 = 8 + 2.4 = 10.4 V. 24. Find the value of the unknown resistance X in the circuit of fig. if no current flows through the section AO. Also calculate the cur rent drawn by the circuit from the battery of emf 6 V and negligible internal resistance.

A 2Ω 3Ω

B

4Ω 10 Ω O X C

2.4 Ω

6V

Ans: The circuit can be redrawn as shown in fig, which is a Wheatstone bridge arrangement. As no current flows through arm AO i.e.., 10 Ω resistance, hence the bridge one.. Therefore, applying A balance condition

 = ʼ, we have,  ʽ

 =  ˂

⇒

2Ω

X=6Ω

Now, resistance of 10 Ω is superfluous. The resistance of arm BAC = 2 + 4 = 6 and the resistance of BOC = 3 + 6 = 9 Ω,

Ω

B

4Ω 3Ω

C

10 Ω X

˂ + 2.4 = 3.6 + 2.4 = 6.0 Ω Total resistance of entire network =  ˀ Current drawn by the circuit I =  = 1.0 A.

O 6V

Ω

or or

2.4 Ω

25. When two known resistances R and S are connected in the left and right gaps of a metre bridge, the balance point is found at a distance l 1 from the zero end of the metre bridge wire. An unknown resistance X is now connected in parallel to the resistances S and the balance point is now found at a distance l 2 from the zero end of the metre bridge wire. Obtain a formula for X in terms of l 1, l2 and S. Ans: The arrangement is shown in fig. Applying the formula for balanced metre bridge in first case, we have R/S = (l1/100 – l1), ….(i) In second case in right gap S and X are

ʽ˂ ). Hence, now, we have ʽ˂ ʽ˂ = ˊ. {.ˊ, Dividing (ii) by (i), we get, ˂ ˊ {.ˊ ʽ . X = ˊ %ˊ ˊ%ˊ

arranged in parallel and have anet resistance of (

ʼ

ˊ , ʽ˂ {ʽ$˂ {ˊ ⇒

=

R

S G

26. A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V as shown in the figure. When a cell 3V of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm. A B (i) Calculate unknown emf of the cell. 1.5 V (ii) explain with reason, whether the circuit works, if the driver cell is replaced R with a cell of emf 1 V. (iii) Does the light resistance R, used in the secondary circuit affect the balance point? Justify your answer

Ans: Here emf of first cell ε1 = 1.5 V, l 1 = 60 cm and l2 = 80 cm,



emf of 2nd cell ε2 = ε1  = 1.5 X

or



" = 2.0 V. "

(ii) If the driver cell of 3 V emf is replaced with a cell of emf 1V, the circuit will not work. It is because in order to obtain a null point on the potentiometer wire the fall in potential due to driver cell must be greater than emf of the cell, whose emf is to be determined. (iii) The high resistance R, used in the secondary circuit does not affect the balance point because at the time of null point no current flows in the secondary circuit. 27. A 10 m long wire of uniform cross-section and 20

Ω

resistance is used in a potentiometer. The wire is

connected in series with a battery of 5 V along with an external resistance of 480

Ω

. If an unknown emf E

is balanced at 6.0 m length of the wire calculate. (i) the potential gradient of the potentiometer wire, (ii) the value of unknown emf. Ans: (i) Here resistance of potentiometer wire r = 20 or =

Ω

, ε = 5 V, R = 480 Ω and l = 10 m

G k= {ʼG

Potential gradient of the potentiometer wire,

˂ , {ː

= 2 X 10-2 V/m,

ε=5V

480 Ω

10 m 6m

P

Q

G E

(ii) Value of unknown emf E = kl = 2 X 10 -2 X 6 = 0.12 V.

28. Potentiometer wire PQ of 1 m length is connected to a standard cell E 1. Another cell E 2 of emf 1.02 V is connected as shown in the circuit diagram with a r esistance ‘r’ and switch S. With switch S open, null E1

position is obtained at a distance of 51 cm fr om P. Calculate (i) potential gradient of the potentiometer wire and (ii) emf of the cell E 1. (iii) When switch S is closed,

P

will null point towards P or towards O? Give reason for your answer. Ans: Here ε2 = 1.02 V and with switch open l = 51 cm,

(i) or

(ii) As total length of potentiometer wire L = 1 m = 100 cm,

r E2

Potential gradient k = or

Q

G

S

 =  = 0.02 V cm-1 ˊ ˊ

ε 1 = k L = 0.02 X 100 = 2 V

(iii) When switch S is closed, the null point will remain unaffected because cell E 2 is even now in an open circuit and no current is being drawn from it.