# Important Questions for Class 12 Chapter 3 - Current Electricity

**Important questions based on NCERT syllabus for Chapter 3 - Current Electricity:**

*Question-1*: A battery has an emf of 10 V and internal resistance is observed to be 3 Ω and is connected to a resistor. If the current flowing in the circuit is 0.5 A, calculate the resistance of the resistor? Also, calculate the terminal voltage of the battery when the circuit is closed.

*Solution*: Given Data :

Given that Emf of the battery, E = 10 V

Battery has an Internal resistance of R = 3 Ω

Current flowing in the circuit, I = 0.5 A

Let the Resistance of the resistor be = R

The relation for current using Ohm’s law is,

I=ER+r

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

V = 0.5 × 17

V = 8.5 V

Therefore, the resistance of the resistor calculated is 17 Ω and the terminal voltage is found to be

8.5 V.

*Question-2*: The car has a storage battery with an emf of 12 V. 0.4Ω is the internal resistance of the battery, what is the maximum current that can be drawn from the battery?

*Solution*:

Given that Emf of the battery, E = 12 V

Battery has an Internal resistance of R = 0.4 Ω

The Maximum current drawn from the battery is given by = I

According to Ohm’s law,

E = IR

I=E/R

I=12/0.4 = 30 A

The maximum current drawn from the given battery is 30 A.

*Question-3*: The length of a wire is 15 m and uniform cross – section is 6.0 x 10 – 7 m 2. Negligibly small current is passed through it with a resistance of 5.0 Ω. Calculate the resistivity of the material at the temperature at which the experiment is conducted.

*Solution*:

Given that the length of the wire , L = 15 m

Area of cross – section is given as , a = 6.0 x 10 – 7 m^2

Let the resistance of the material of the wire be , R , ie. , R = 5.0 Ω

Resistivity of the material is given as ρ

R=ρLA

ρ=RxAL=5×6×10–715=2×10^–7

Therefore, the resistivity of the material is calculated to be 2×10^–7