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In chapter 3 of Physics Class 12th **“Current Electricity”**, students will
learn cell EMF, resistivity, internal resistance, conductance, drift velocity, graphing
of internal resistance and resistivity, conversion of complex circuits into simple
series, potentiometer, and various problems based on the concepts related to the
chapter.

##### Question 1:

A storage battery of a car has an e.m.f. of 12V. If the internal resistance of the battery is 0.4 $\mathrm{\Omega}$, what is the maximum current that can be drawn from the battery ?

##### Answer:

Given

E = 12V, r = 0.4 $\mathrm{\Omega}$, I = ?

##### Question 2:

A battery of e.m.f. 10V and internal resistance 3$\mathrm{\Omega}$ is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?

##### Answer:

Given

E = 10V, r = 3$\mathrm{\Omega}$, I = 0.5, R = ?

Since V = E – Ir

or V = 10 – 0.5 × 3

or V = 8·5 volt

##### Question 3:

Three resistors 1 $\mathrm{\Omega}$, 2 $\mathrm{\Omega}$ and 3 $\mathrm{\Omega}$ are combined in series. What is the total resistance of the combination ?

If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.

##### Answer:

Given

R_{1} = 1 $\mathrm{\Omega}$, R_{2} = _{2} $\mathrm{\Omega}$, R_{3.} = 3 $\mathrm{\Omega}$

(a) Total resistance of series combination

R_{s} = R _{1} + R_{2} + R_{3.}

or R_{s} = 1 + 2 + 3 = 6 $\mathrm{\Omega}$

(b) Since E = I(R + r)

∴ V_{1.} = IR_{1} = 2 × 1 = 2 V

V_{2.} = IR_{2} = 2 × 2 = 4 V

V_{3.} = IR_{3} = 2 × 3 = 6 V

##### Question 4:

Three resistors 2 $\mathrm{\Omega}$, 4 $\mathrm{\Omega}$ and 5 $\mathrm{\Omega}$ are combined in parallel. What is the total resistance of the combination ?

If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

##### Answer:

Given R _{1} = 2 $\mathrm{\Omega}$ , R _{2} = 4 $\mathrm{\Omega}$ , R _{3} = 5 $\mathrm{\Omega}$

(a) Total resistance in parallel combination

##### Question 5:

At room temperature (27.0$\xb0$C) the resistance of a heating element is 100 $\mathrm{\Omega}$. What is the temperature of the element if the resistance is found to be 117 $\mathrm{\Omega}$, given that the temperature coefficient of the material of the resistor is 1.70 × 10
^{–4} $\xb0$C ^{–1} ?

##### Answer:

Given

R _{27} =100 $\mathrm{\Omega}$, R_{t} = 117 $\mathrm{\Omega}$, α = 1.70 × 10 ^{–4} $\xb0$C ^{–1}, t = ?

##### Question 6:

A negligibly small current is passed
through a wire of length 15 m and uniform crosssection
6.0 × 10 ^{–7} m^{2} and its resistance is measured
to be 5.0 $\mathrm{\Omega}$. What is the resistivity of the material
at the temperature of the experiment ?

##### Answer:

l =15 m, A = 6.0 × 10 ^{–7} m^{2.}, R = 5.0 $\mathrm{\Omega}$

Applying the relation,

##### Question 7:

A silver wire has a resistance of 2.1 $\mathrm{\Omega}$ at 27.5$\xb0$ C, and a resistance of 2.7 $\mathrm{\Omega}$ at 100$\xb0$ C. Determine the temperature co-efficient of resistivity of silver.

##### Answer:

R_{t1} =R_{27.5} = 2.1 $\mathrm{\Omega}$ and R _{
t2} = R_{100} = 2.7 $\mathrm{\Omega}$

##### Question 8:

A heating element using nichrome
connected to a 230 V supply draws an initial
current of 3.2 A which settles after a few seconds
to a steady value of 2.8 A. What is the steady
temperature of the heating element if the room
temperature is 27.0 $\xb0$C ? Temperature coefficient
of resistance of nichrome averaged over the
temperature range involved is 1.70 × 10 ^{–4} $\xb0$C ^{–1}.

##### Answer:

Given

V = 230 V, I = 2.8 A

##### Question 9:

Determine the current in each branch of the following network :

##### Answer:

Applying Kirchhoff’s 2nd law of closed circuits ABD and BCD,

10I_{1} + 5I_{3} – 5I_{2} = 0

or 2I_{1} + I_{3} – I_{2} = 0 ....(1)

5 (I _{1} – I _{3}) – 10 (I _{2} + I _{3}) – 5I _{3}
= 0

5I_{1} – 10I_{2} – 20I_{3} = 0

or I_{1} – 2I_{2} – 4I_{3} = 0 ...(2)

Multiply (1) by 4 and add (2),

I_{1} – 2I_{2} – 4I_{3} = 0

8I_{1} – 4I_{2} + 4I_{3} = 0

9I_{1} – 6I_{2} = 0

3I_{1} – 2I_{2} = 0 ....(3)

Again applying Kirchhoffs law to closed ckt ADCBA

10I + 5I_{2 + 10} (I_{2 + I}3) = 10

10 (I_{1 + I}2) + 5I_{2 + 10} (I_{2 + I}3)= 10

10I_{1} + 25I_{2} + 10I_{3} = 10

2I_{1} + 5I_{2} + 2I_{3} = 2 ...(4)

Multiplying (1) by 2 and subtracting from (4),

4I_{1} – 2I_{2} + 2I_{3} = 0

2I_{1} + 5I_{2} + 2I_{3} = 2

(4)

2I_{1} – 7I_{2} + 0 = – 2

or 6I_{1 – 21I}2 = – 6 ...(5)

Multiplying (3) by 2 and subtracting from (5),

6I_{1} – 4I_{2} = 0

6I_{1} – 21_{I2} = – 6

– + –

##### Question 10:

In a meter bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 $\mathrm{\Omega}$. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips ?

Determine the balance point of the meter bridge if X and Y are interchanged. What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ?

##### Answer:

Given l_{1} = 39.5, l_{2} = 100 – 39.5 = 60.5 cm For a metre bridge, we have

To minimise resistance of the connections which are not accounted for in the bridge formula, we use thick copper strips.

On interchanging X and Y,

P : Q : : 1.53 : 1 or l_{1} : l_{2} : : 1.53 : 1

Thus the new balance point will be 60.5 cm from A.

(c) The galvanometer will show no current.

##### Question 11:

A storage battery of e.m.f. 8.0 V and internal resistance 0.5 $\mathrm{\Omega}$ is being charged by a 120 V d.c. supply using a series resistor of 15.5 $\mathrm{\Omega}$. What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit ?

##### Answer:

Total resistance of the circuit. = 15.5 + 0.5 = 16 $\mathrm{\Omega}$

Current through the cell drawn from the battery

E = V + Ir = 8 + 7.5 × 0.5 = 8 + 3.75 = 11.75 volt. The series resistor limits the current drawn from the external source. In its absence, the current will be dangerously high.

##### Question 12:

In a potentiometer arrangement, a cell of e.m.f. 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm. What is the e.m.f. of the second cell ?

##### Answer:

Given

##### Question 13:

The number density of free electrons
in a copper conductor estimated is 8.5 × 10^{28} m^{–3}.
How long does an electron take in drifting from
one end of a wire 3.0 m long to its other end ?
The area of cross-section of the wire is 2.0 ×
10^{ –6} m^{2} and it is carrying a current of 3.0 A.

##### Answer:

n = 8.5 × 10^{28} m^{3}; A = 2 × 10^{ –6} m^{2}, I =
3.0A;

l = 3 m

I = $\mathrm{\upsilon}$_{d}neA