## Current Electricity class 12 ncert solutions: Ncert Class 12 Physics Chapter 3 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 3 |

Chapter Name | Current Electricity class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 3 Current Electricity? Now you can download Ncert Class 12 Physics Chapter 3 Exercise Solutions pdf from here.

### 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?

**Solution 3.1:** To determine the maximum current that can be drawn from a car battery, we use Ohm’s Law:

\(I = \frac{E}{R}\)

where:

- ( I ) is the current,
- ( E ) is the voltage (emf of the battery),
- ( R ) is the total resistance.

In this case, the total resistance is just the internal resistance of the battery since we’re looking for the maximum current (which occurs when the external load resistance is zero, or when the battery is short-circuited).

Given:

- Emf of the battery, E = 12 V
- Internal resistance, R = 0.4 Ω

Now, substitute the values into Ohm’s Law:

\(I = \frac{12 \, \text{V}}{0.4 \, \Omega} = 30 \, \text{A}\)

So, the maximum current that can be drawn from the 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?

**Solution 3.2:** To solve this problem, we’ll break it down into two parts:

**Part 1: Finding the Resistance of the Resistor**

Given:

- EMF of the battery, E = 10 V
- Internal resistance of the battery, r = 3 Ω
- Current in the circuit, I = 0.5 A

According to Ohm’s Law, the total voltage drop in the circuit (which equals the EMF of the battery) is the sum of the voltage drops across the internal resistance r and the external resistor R.

The total voltage drop in the circuit is given by: E = I(R + r)

Rearranging to find R:

\(R = \frac{E}{I} – r\)

Substitute the given values:

\(R = \frac{10 \, \text{V}}{0.5 \, \text{A}} – 3 \, \Omega = 20 \, \Omega – 3 \, \Omega = 17 \, \Omega\)

So, the resistance of the resistor ( R ) is **17 Ω**.

**Part 2: Finding the Terminal Voltage of the Battery**

The terminal voltage \(V_{\text{terminal}}\) of the battery is given by:

\(V_{\text{terminal}} = E – I \times r\)

Substituting the known values:

\(V_{\text{terminal}} = 10 \, \text{V} – (0.5 \, \text{A} \times 3 \, \Omega) = 10 \, \text{V} – 1.5 \, \text{V} = 8.5 \, \text{V}\)

So, the terminal voltage of the battery when the circuit is closed is **8.5 V**.

Therefore, the resistance of the resistor is 17 Ω, and the terminal voltage is 8.5 V.

### Question 3.3:** **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}.

**Solution 3.3:** To solve this problem, we’ll use the formula that relates the resistance of a conductor at different temperatures:

\(R_T = R_0 \left[ 1 + \alpha (T – T_0) \right]\)

where:

- \( R_T \) is the resistance at the temperature T,
- \( R_0 \) is the resistance at the reference temperature \( T_0 \),
- \( \alpha \) is the temperature coefficient of resistance,
- \( T \) is the unknown temperature we need to find,
- \( T_0 \) is the reference temperature.

Given values:

- \( R_0 = 100 \, \Omega \) resistance at \(( T_0 = 27.0 \, °C )\),
- \( R_T = 117 \, \Omega \) resistance at temperature ( T ),
- \( \alpha = 1.70 \times 10^{-4} \, °C^{-1} \).

We need to find ( T ). Plugging the values into the equation:

\(117 = 100 \left[ 1 + 1.70 \times 10^{-4} \times (T – 27.0) \right]\)

Now, let’s solve for T.

- Divide both sides by 100:

\(1.17 = 1 + 1.70 \times 10^{-4} \times (T – 27.0)\)

- Subtract 1 from both sides:

\(0.17 = 1.70 \times 10^{-4} \times (T – 27.0)\)

- Divide both sides by \( 1.70 \times 10^{-4} \):

\(\frac{0.17}{1.70 \times 10^{-4}} = T – 27.0\)

1000 = T – 27.0

- Finally, add 27.0 to both sides:

T = 1000 + 27.0

T = 1027 °C

So, the temperature of the element is ( 1027 °C ).

### Question 3.4:** **A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^{–7} m^{2}, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

**Solution 3.4:** To find the resistivity of the material, we can use the formula for the resistance of a uniform wire:

\(R = \rho \frac{L}{A}\)

Where:

- R is the resistance of the wire (5.0 Ω),
- \( \rho\) is the resistivity of the material,
- L is the length of the wire (15 m),
- A is the cross-sectional area of the wire 6.0 × 10
^{–7}m^{2}.

We can rearrange this formula to solve for the resistivity \( \rho \):

\(\rho = \frac{R \times A}{L}\)

Now, substitute the given values into the equation:

\(\rho = \frac{5.0 \, \Omega \times 6.0 \times 10^{-7} \, \text{m}^2}{15 \, \text{m}}\)

Now, calculate the resistivity:

\(\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15}\)

\(\rho = \frac{30.0 \times 10^{-7}}{15}\)

\(\rho = 2.0 \times 10^{-7} \, \Omega \text{m}\)

Thus, the resistivity of the material is \( 2.0 \times 10^{-7} \, \Omega \text{m} \).

### Question 3.5:** **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.

**Solution 3.5:** To determine the temperature coefficient of resistivity ((\alpha)) of silver, you can use the formula for the temperature dependence of resistance:

\( R(T) = R_0 [1 + \alpha (T – T_0)]\)

Where:

- R(T) ) is the resistance at temperature T
- \( R_0\) is the resistance at reference temperature \( T_0\)
- \(\alpha\) is the temperature coefficient of resistivity
- T and \(T_0\) are the temperatures in °C

Given:

- \( R_0 = 2.1 \, \Omega at T_0 = 27.5 \, ^\circ\text{C} \)
- \(R(T) = 2.7 \, \Omega at T = 100 \, ^\circ\text{C}\)

First, rearrange the formula to solve for (\alpha\):

\( \alpha = \frac{R(T) – R_0}{R_0 (T – T_0)} \)

Substitute the given values:

\( \alpha = \frac{2.7 \, \Omega – 2.1 \, \Omega}{2.1 \, \Omega \cdot (100 \, ^\circ\text{C} – 27.5 \, ^\circ\text{C})}\)

Calculate the difference in resistance:

\(2.7 \, \Omega – 2.1 \, \Omega = 0.6 \, \Omega\)

Calculate the difference in temperature:

\(100 \, ^\circ\text{C} – 27.5 \, ^\circ\text{C} = 72.5 \, ^\circ\text{C} \)

Now, substitute these into the formula:

\(\alpha = \frac{0.6 \, \Omega}{2.1 \, \Omega \cdot 72.5 \, ^\circ\text{C}}\)

Calculate the value:

\(\alpha = \frac{0.6}{2.1 \cdot 72.5} \approx \frac{0.6}{152.25} \approx 0.00393 \, ^\circ\text{C}^{-1} \)

Rounding to the significant figures in the given answer:

\( \alpha \approx 0.0039 \, ^\circ\text{C}^{-1} \)

So, the temperature coefficient of resistivity of silver is indeed approximately \(0.0039 \, ^\circ\text{C}^{-1}\).

### Question 3.6:** **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 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^{–4} °C^{–1}.

**Solution 3.6:** In the given problem,

The supply voltage, V = 230 V

The initial current drawn, I_{1} = 3.2 A

Consider the initial resistance to be R_{1}, which can be found by the following relation:

\(R _{ 1 } = \frac{ V }{ I }\) = 71.87 Ω

Value of current at steady state, I_{2} = 2.8 A

Value of resistance at steady state = R_{2}

R_{2} can be calculated by the following equation:

\(R _{ 2 } = \frac{ 230 }{ 2.8 } = 82.14 Ω\)

- The temperature coefficient of nichrome averaged over the temperature range involved is 1.70 x 10
^{–4}° C^{–1} - Value of initial temperature of nichrome , T
_{1}= 27.0 ° C - Value of steady-state temperature reached by nichrome = T
_{2}

This temperature T_{2} can be obtained by the following formula:

\(\alpha = \frac{R_2 – R_1}{R_1 \left(T_2 – T_1\right)}\)

\(T _{ 2 } – 27 = \frac{ 82.14 – 71.87 }{ 71.87 \times ( 1.7 \times 10^{-4}) }\)

\(T_{ 2 } – 27 = 840.5\)

T_{2} = 840.5 + 27 = 867.5 ° C

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

### Question 3.7:** **Determine the current in each branch of the network shown in the figure:

**Solution 3.7:** The current flowing through various branches of the network is shown in the figure given below:

For the closed circuit ABDA, potential is zero i.e.,

10I_{2} + 5I_{4} − 5I_{3} = 0

2I_{2} + I_{4} −I_{3} = 0

I_{3} = 2I_{2} + I_{4} … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I_{2} − I_{4}) − 10(I_{3} + I_{4}) − 5I4 = 0

5I_{2} + 5I_{4} − 10I_{3} − 10I_{4} − 5I_{4} = 0

5I_{2} − 10I_{3} − 20I_{4} = 0

I_{2} = 2I_{3} + 4I_{4} … (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I_{2} + 10I_{1} − 5I_{4}

3I_{2} + 2I_{1} − I_{4} = 2 … (3)

From equations (1) and (2), we obtain

I_{3} = 2(2I_{3} + 4I_{4}) + I_{4}

I_{3} = 4I_{3} + 8I_{4} + I_{4}

− 3I_{3} = 9I_{4}

− 3I_{4} = + I_{3} … (4)

Putting equation (4) in equation (1), we obtain

I_{3} = 2I_{2} + I_{4}

− 4I_{4} = 2I_{2}

I_{2} = − 2I_{4} … (5)

It is evident from the given figure that,

I_{1} = I_{3} + I_{2} … (6)

Putting equation (6) in equation (1), we obtain

3I_{2} + 2(I_{3} + I_{2}) − I_{4} = 2

5I_{2} + 2I_{3} − I_{4} = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I_{4}) + 2(− 3 I_{4}) − I_{4} = 2

− 10I_{4} − 6I_{4} − I_{4} = 2

17I_{4} = − 2

I_{4} = -2/17Ampere

Equation (4) reduces to

I_{3} = − 3(I_{4})

I_{3} = -3 × -2/17

I_{3} = 6/17Ampere

I_{2} = -2(I4)

I_{2} = -2 × -2/17

I_{2} = 4/17Ampere

I_{2 }– I_{4} = 6/17Ampere

I_{3} + I_{4} = 6/17Ampere

I_{1} = I_{2} + I_{3} = 4/17 + 6/17 = 10/17 Ampere

Therefore, current in branch AB = 4/17Ampere

In branch BC = 6/17Ampere

In branch CD = -4/17Ampere

In branch AD = 6/17Ampere

In branch BD = -2/17Ampere

Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.

### Question 3.8:** **A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

**Solution 3.8:** Given:

- The EMF of the given storage battery is E = 8.0 V
- The Internal resistance of the battery is given by r = 0.5 Ω
- The given DC supply voltage is V = 120 V
- The resistance of the resistor is R = 15.5 Ω
- Effective voltage in the circuit = V
^{1} - R is connected to the storage battery in series.

Hence, it can be written as

- V
^{1}= V – E - V
^{1}= 120 – 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

\(I = \frac{ V ^{ 1 }}{ R + r }\)

\(I = \frac{ 112 }{ 15.5 + 5 }\)

\(I = \frac{ 112 }{ 16 }\)

We know that Voltage across a resistor R given by the product,

I x R = 7 × 15.5 = 108.5 V

We know that,

DC supply voltage = Terminal voltage + voltage drop across R

Terminal voltage of battery = 120 – 108.5 = 11.5 V

A series resistor, when connected to a charging circuit, limits the current drawn from the external source.

The current will become extremely high in its absence. This is extremely dangerous.

### Question 3.9:** **The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10^{28} m^{–3}. How long does an electron take to drift 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.

**Solution 3.9:** Given that Number density of free electrons in a copper conductor, n = 8.5 x 10^{28} m^{–3}

Let the Length of the copper wire be l

Given , l = 3.0 m

Let the area of the cross-section of the wire be A = 2.0 x 10^{–6} m^{2}

The value of the current carried by the wire, I = 3.0 A, which is given by the equation,

I = n A e V_{d}

Where,

e = electric charge = 1.6 x 10^{–19} C

\(V _{ d } = Drift velocity = \frac{ Length of the wire \left ( l \right ) }{ time taken to cover l \left ( t \right )}\)

\(I = n A e \frac{ l }{ t }\)

\(t = \frac{ n \times A \times e \times l }{ I }\)

\(t = \frac{ 3 \times 8.5 \times 10 ^{ 28 } \times 2 \times 10 ^{ – 6 } \times 1.6 \times 10 ^{ – 19 }}{ 3.0 }\)

\(t = 2.7 \times 10 ^{ 4 } sec\)