Class 12 Physics NCERT Solutions Current Electricity Important Questions

Question 3.1:

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

Answer 3.1:

In the given question,

The EMF of the battery is given as E = 12 V

The internal resistance of the battery is given as R = 0.4 Ω

The amount of maximum current drawn from the battery is I

According to Ohm’s law,

E = IR

A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer 3.2:

Given:

The EMF of the battery (E = 10 V)

The internal resistance of the battery (R = 3 Ω)

The current in the circuit (I = 0.5 A)

Consider the resistance of the resistor to be R.

The current in the circuit can be found out using Ohm’s Law as,

Question 3.3:

a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 3.3:

a) We know that resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are combined in series.

The total resistance of the above series combination can be calculated by the algebraic sum of individual resistances as follows:

Total resistance = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus calculated Total Resistance = 6 Ω

b) Let us consider I to be the current flowing the given circuit

Also,

The emf of the battery is E = 12 V

The total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law, the relation for current can be obtained as

Question 3.4:

a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

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

Answer 3.4:

A ) Resistors r 1 = 2 Ω , r2 = 4 Ω and r 3 = 5 Ω are combined in parallel

Hence the total resistance of the above circuit can be calculated  by the following formula:

Question 3.5:

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1.

Answer 3.5:

Given that the room temperature, T = 27 ° C

The heating element has a resistance of R = 100 Ω

Let the increased temperature of the filament be T 1

At T 1, the resistance of the heating element is  R 1 = 117 Ω

Temperature coefficient of the material used for the element is 1.70 x 10– 4 C – 

Question 3.6:

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer 3.6:

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

The 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, i.e., R = 5.0 Ω

The resistivity of the material is given as ρ