INTRODUCTION - Chapter 3 Current Electricity Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 3.1:

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

Answer:

Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law,

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

Question 3.2:

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:

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.

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:

(a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

(b) Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, R = 6 Ω

The relation for current using Ohm’s law is,

Potential drop across 1 Ω resistor = V₁

From Ohm’s law, the value of V₁ can be obtained as

V₁ = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = V₂

Again, from Ohm’s law, the value of V₂ can be obtained as

V₂ = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = V₃

Again, from Ohm’s law, the value of V₃ can be obtained as

V₃ = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

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:

(a) There are three resistors of resistances,

R₁ = 2 Ω, R₂ = 4 Ω, and R₃ = 5 Ω

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

Therefore, total resistance of the combination is.

(b) Emf of the battery, V = 20 V

Current (I₁) flowing through resistor R₁ is given by,

Current (I₂) flowing through resistor R₂ is given by,

Current (I₃) flowing through resistor R₃ is given by,

Total current, I = I₁ + I₂ + I₃ = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

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 

Answer:

Room temperature, T = 27°C

Resistance of the heating element at T, R = 100 Ω

Let T₁ is the increased temperature of the filament.

Resistance of the heating element at T₁, R₁ = 117 Ω

Temperature co-efficient of the material of the filament,

Therefore, at 1027°C, the resistance of the element is 117Ω.

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:

Length of the wire, l =15 m

Area of cross-section of the wire, a = 6.0 × 10⁻⁷ m²

Resistance of the material of the wire, R = 5.0 Ω

Resistivity of the material of the wire = ρ

Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10⁻⁷ Ω m.

Question 3.7:

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Answer:

Temperature, T₁ = 27.5°C

Resistance of the silver wire at T₁, R₁ = 2.1 Ω

Temperature, T₂ = 100°C

Resistance of the silver wire at T₂, R₂ = 2.7 Ω

Temperature coefficient of silver = α

It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Question 3.8:

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

Answer:

Supply voltage, V = 230 V

Initial current drawn, I₁ = 3.2 A

Initial resistance = R₁, which is given by the relation,

Steady state value of the current, I₂ = 2.8 A

Resistance at the steady state = R₂, which is given as

Temperature co-efficient of nichrome, α = 1.70 × 10⁻⁴ °C ⁻¹

Initial temperature of nichrome, T₁= 27.0°C

Study state temperature reached by nichrome = T₂

T₂ can be obtained by the relation for α,

Therefore, the steady temperature of the heating element is 867.5°C