# 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