Current Electricity Class 12 Physics Chapter 3:  NCERT Solutions

SimplyAcad is providing you the best NCERT Solutions Class 12 Current Electricity Chapter 3 for getting a better command on the concepts of the chapter and scoring maximum marks in your physics paper.

As you all know, Physics is one of the toughest subjects which demands proper time and dedication towards it. Chapter 3 of your NCERT textbook Current Electricity helps you to learn about different principles of electricity through theoretical knowledge as well as practical applications.

Current Electricity is an extremely important chapter from exam point of view as well, so start your preparations with the given solutions without any delay, these are beneficial for students who are stuck between the questions of the exercise.

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

Question 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?

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.

Question2. A battery of emf $10 \, \text{V}$ and internal resistance $3 \, \Omega$ is connected to a resistor. If the current in the circuit is $0.5 \, \text{A}$, what is the resistance of the resistor and the terminal voltage of the battery when the circuit is closed?

To solve this problem, we need to determine two things: the resistance of the external resistor and the terminal voltage of the battery when the circuit is closed. **Given:** – Electromotive force (EMF) of the battery, $E = 10 \, \text{V}$ – Internal resistance of the battery, $r = 3 \, \Omega$ – Current in the circuit, $I = 0.5 \, \text{A}$ **1. Finding the Resistance of the External Resistor (R):** The total resistance in the circuit includes both the internal resistance of the battery and the external resistor. According to Ohm’s Law and the definition of EMF: \[ I = \frac{E}{R + r} \] Rearranging to solve for the external resistance $R$: \[ R = \frac{E}{I} – r \] Substituting the given values: \[ R = \frac{10 \, \text{V}}{0.5 \, \text{A}} – 3 \, \Omega \] \[ R = 20 \, \Omega – 3 \, \Omega \] \[ R = 17 \, \Omega \] **2. Finding the Terminal Voltage of the Battery:** The terminal voltage $V$ is the voltage across the external resistor, which can be found using Ohm’s Law: \[ V = I \cdot R \] Substituting the known values: \[ V = 0.5 \, \text{A} \times 17 \, \Omega \] \[ V = 8.5 \, \text{V} \] Thus, the resistance of the external resistor is $17 \, \Omega$, and the terminal voltage of the battery is $8.5 \, \text{V}$.

Question3. (a) Three resistors of 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.

(a) Total Resistance in Series:** When resistors are combined in series, the total resistance is the sum of the individual resistances. Given resistances: – \( R_1 = 1 \, \Omega \) – \( R_2 = 2 \, \Omega \) – \( R_3 = 3 \, \Omega \) The total resistance \( R_{\text{total}} \) is: \[ R_{\text{total}} = R_1 + R_2 + R_3 \] \[ R_{\text{total}} = 1 + 2 + 3 = 6 \, \Omega \] **(b) Potential Drop Across Each Resistor:** The combination of resistors is connected to a battery with an EMF \( E = 12 \, \text{V} \) and negligible internal resistance. To find the potential drop across each resistor, we first calculate the total current \( I \) using Ohm’s law. Total resistance of the circuit, \( R_{\text{total}} = 6 \, \Omega \). The current \( I \) through the circuit is: \[ I = \frac{E}{R_{\text{total}}} \] \[ I = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A} \] Now, we find the potential drop across each resistor: 1. **Potential Drop Across the 1 Ω Resistor (\(V_1\)):** \[ V_1 = I \times R_1 \] \[ V_1 = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \] 2. **Potential Drop Across the 2 Ω Resistor (\(V_2\)):** \[ V_2 = I \times R_2 \] \[ V_2 = 2 \, \text{A} \times 2 \, \Omega = 4 \, \text{V} \] 3. **Potential Drop Across the 3 Ω Resistor (\(V_3\)):** \[ V_3 = I \times R_3 \] \[ V_3 = 2 \, \text{A} \times 3 \, \Omega = 6 \, \text{V} \] Therefore, the potential drops across the 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V, respectively.

Question4. (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.

(a) Total Resistance in Parallel:** For resistors connected in parallel, the reciprocal of the total resistance \( R_{\text{total}} \) is the sum of the reciprocals of the individual resistances: Given resistances – \( R_1 = 2 \, \Omega \) – \( R_2 = 4 \, \Omega \) – \( R_3 = 5 \, \Omega \) The total resistance \( R_{\text{total}} \) is given by: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] Substituting the given values: \[ \frac{1}{R_{\text{total}}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} \] Calculating each term: \[ \frac{1}{R_{\text{total}}} = 0.5 + 0.25 + 0.2 \] \[ \frac{1}{R_{\text{total}}} = 0.95 \] Thus, the total resistance \( R_{\text{total}} \) is: \[ R_{\text{total}} = \frac{1}{0.95} \approx 1.05 \, \Omega \] **(b) Current Through Each Resistor and Total Current:** The EMF of the battery \( V = 20 \, \text{V} \). The current through each resistor can be calculated using Ohm’s law: – **Current through \( R_1 \) (\( I_1 \)):** \[ I_1 = \frac{V}{R_1} \] \[ I_1 = \frac{20 \, \text{V}}{2 \, \Omega} = 10 \, \text{A} \] – **Current through \( R_2 \) (\( I_2 \)):** \[ I_2 = \frac{V}{R_2} \] \[ I_2 = \frac{20 \, \text{V}}{4 \, \Omega} = 5 \, \text{A} \] – **Current through \( R_3 \) (\( I_3 \)):** \[ I_3 = \frac{V}{R_3} \] \[ I_3 = \frac{20 \, \text{V}}{5 \, \Omega} = 4 \, \text{A} \] The total current \( I \) drawn from the battery is the sum of the currents through each resistor: \[ I = I_1 + I_2 + I_3 \] \[ I = 10 \, \text{A} + 5 \, \text{A} + 4 \, \text{A} = 19 \, \text{A} \] Therefore, the current through each resistor is 10 A, 5 A, and 4 A, respectively, and the total current drawn from the battery is 19 A.

Question5.At room temperature (27.0 °C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1.

Solution :Given: Temperature coefficient of filament,

α = 1.70 × 10-4 per °C

Let T1 be the temperature of element, R1 = 100Ω (Given: T1 = 27°C)

Let T2 be the temperature of element, R2 = 117Ω

To find T2 = ?

The formula is: R2 = R1[1 + α(T2-T1)]

⇒ R2 – R1 = R1α(T2 – T1)

⇒ R2 – R1 = R1α (T2 – T1)

Question6. 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?

Solution :Given, the length of wire l = 15 metre

Area of cross section of wire a = 6.0 × 10-7 metre square

Resistance of material of wire,R = 5 Ω

Resistivity of material of wire = ρ

Question7. 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 :Resistance of silver wire at 27.5 °C R1 = 2.1 Ω

Resistance of silver wire at 100 ° C R2 = 2.7 Ω

Let the temperature coefficient of silver wire is given by α.

The formula for α is: R2 = R1[1 + α(T2-T1)]

⇒ R2 – R1 = R1α(T2 – T1)

⇒ R2 – R1 = R1α (T2 – T1)

Question8. 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 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.

Solution :Given Supply Voltage = 230 Volt

And it draws current I1 = 3.2 Ampere

Resistance is given by R1 = V/I

R1 = 230V/3.2A

R1 = 71.87 Ω

Steady value of current I2 = 2.8Ampere

Resistance R α is given by R2 = 230V/2.8A

R2 = 82.14 Ω

Temperature coefficient of nichrome wire α = 1.70 × 10-4 per °C

Initial temperature of nichrome T1 = 27 ° C

Steady Temperature attained by nichrome = T2

The formula is: R2 = R1[1 + α(T2-T1)]

⇒ R2 – R1 = R1α(T2 – T1)

⇒ R2 – R1 = R1α (T2 – T1)

Question9. Determine the current in each branch of the network shown in Fig 3.30: – \( I_1 \):

Solution: Current flowing through the outer circuit – \( I_2 \): Current flowing through branch AB – \( I_3 \): Current flowing through branch AD – \( I_4 \): Current flowing through branch BD For the closed circuits, the potential is zero: 1. **For the closed circuit ABDA:** \[ 10I_2 + 5I_4 – 5I_3 = 0 \] Simplifying, \[ 2I_2 + I_4 – I_3 = 0 \quad \text{(Equation 1)} \] \[ I_3 = 2I_2 + I_4 \] 2. **For the closed circuit BCDB:** \[ 5(I_2 – I_4) – 10(I_3 + I_4) – 5I_4 = 0 \] Simplifying, \[ 5I_2 – 10I_3 – 20I_4 = 0 \quad \text{(Equation 2)} \] \[ I_2 = 2I_3 + 4I_4 \] 3. **For the closed circuit ABCFEA:** \[ -10 + 10I_1 + 10I_2 + 5(I_2 – I_4) = 0 \] Simplifying, \[ 3I_2 + 2I_1 – I_4 = 2 \quad \text{(Equation 3)} \] Using the equations derived from Kirchhoff’s laws: – From Equation 1 and 2: \[ I_3 = 2(2I_3 + 4I_4) + I_4 \] \[ I_3 = 4I_3 + 8I_4 + I_4 \] \[ -3I_3 = 9I_4 \] \[ -3I_4 = I_3 \quad \text{(Equation 4)} \] – Using Equation 1 and 4: \[ I_3 = 2I_2 + I_4 \] \[ -4I_4 = 2I_2 \] \[ I_2 = -2I_4 \quad \text{(Equation 5)} \] Using Equation 6: \[ I_1 = I_3 + I_2 \] Substitute Equations 4 and 5 into Equation 7: \[ 3I_2 + 2(I_3 + I_2) – I_4 = 2 \] \[ 5I_2 + 2I_3 – I_4 = 2 \quad \text{(Equation 7)} \] \[ 5(-2I_4) + 2(-3I_4) – I_4 = 2 \] \[ -10I_4 – 6I_4 – I_4 = 2 \] \[ 17I_4 = -2 \] Solving for \( I_4 \): \[ I_4 = -\frac{2}{17} \] Using \( I_4 \) to find other currents: – \( I_3 = -3I_4 \) – \( I_2 = -2I_4 \) – \( I_1 = I_3 + I_2 \) Total current at C: \[ (I_2 – I_4) + (I_3 + I_4) = \left(\frac{6}{17}\right) + \left(\frac{4}{17}\right) = \frac{10}{17}

Question10. (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor \( Y \) is of \( 12.5 \, \Omega \). Determine the resistance of \( X \). Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if \( X \) and \( Y \) are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

A metre bridge with resistors \( X \) and \( Y \) is represented in the given figure.**(a)** Balance point from end A, \( l_1 = 39.5 \, \text{cm} \) Resistance of the resistor \( Y = 12.5 \, \Omega \) The condition for the balance is given by: \[ \frac{X}{Y} = \frac{l_1}{100 – l_1} \] Substituting the given values: \[ \frac{X}{12.5} = \frac{39.5}{100 – 39.5} \] \[ \frac{X}{12.5} = \frac{39.5}{60.5} \] \[ X = 12.5 \times \frac{39.5}{60.5} \] Calculating the value: \[ X \approx 8.2 \, \Omega \] Therefore, the resistance of resistor \( X \) is \( 8.2 \, \Omega \). The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula. **(b)** If \( X \) and \( Y \) are interchanged, then \( l_1 \) and \( 100 – l_1 \) get interchanged The new balance point of the bridge will be \( 100 – l_1 \) from \ 100 – l_1 = 100 – 39.5 = 60.5 \, \text{cm} \] Therefore, the balance point is \( 60.5 \, \text{cm} \) from A. **(c)** When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

Question11. A storage battery of emf \( 8.0 \, \text{V} \) and internal resistance \( 0.5 \, \Omega \) is being charged by a \( 120 \, \text{V} \) dc supply using a series resistor of \( 15.5 \, \Omega \). What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Sol: \textbf{Step 1: Draw a circuit diagram of the situation} \textbf{Step 2: Find the effective voltage in the circuit.} Given, \begin{align*} \text{Emf of the storage battery, } E & = 8.0 \, \text{V} \\ \text{Internal resistance of the battery, } r & = 0.5 \, \Omega \\ \text{DC supply voltage, } V & = 120 \, \text{V} \\ \text{Resistance in series, } R & = 15.5 \, \Omega \end{align*} The effective voltage in the circuit is given by \[ V_1 = V – E = 120 – 8 = 112 \, \text{V} \] \textbf{Step 3: Find the current in the circuit.} The current flowing in the circuit will be, \[ I = \frac{V_1}{R + r} \] Substituting the values, we get \[ I = \frac{112}{15.5 + 0.5} = 7 \, \text{A} \] \textbf{Step 4: Find the voltage across Resistance.} Voltage across resistor \( R \) will be, \[ IR = 7 \times 15.5 = 108.5 \, \text{V} \] \textbf{Step 5: Find the terminal voltage.} Now, \[ \text{DC supply voltage} = \text{Terminal voltage of battery} + \text{Voltage drop across } R \] \[ \Rightarrow \text{Terminal voltage of battery} = \text{DC supply voltage} – \text{Voltage drop across } R \] \[ \Rightarrow \text{Terminal voltage of battery} = 120 – 108.5 = 11.5 \, \text{V} \] A series resistor in a charging circuit limits the current drawn from the external source. In the absence of a series resistor, the current would be extremely high, which could be dangerous.

Question12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Solution :Given EMF of the cell E1 = 1.25Volt

Balance point of the potentiometer l1 is given 35cm

Since the cell is replaced by another cell of EMF E2

New balance point of potentiometer l2 is 63cm

Question13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 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 m2 and it is carrying a current of 3.0 A.

Solution :Number density of free electrons in a copper conductor, n = 8.5 × 10^28 per metre cube

Length of Copper Wire l = 3 metre

Area of cross section of wire A = 2 × 10-6 metre square

Current drawn by wire is given 3 Ampere

Since I = nAeVd

Here Vd is the drift velocity and “e” is the electron charge 1.6 × x10^-19 Coulomb.

Question14. The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric fields, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)

Solution :Surface Charge density of earth d = 10-9 Coulomb per metre square

Current over the entire globe = 1800 Ampere

Radius of earth r = 6.37 × 106 metre

Surface area of earth A = 4×π×radius×radius

A = 4×π×6.37×106×6.37×106

A = 5.09 × 1014 m2

Charge on the earth surface q = d × A

q = 10-9 × 5.09 ×1014 m2

q = 5.09 × 105 Coulomb

Let the time taken to neutralise earth surface = t

Question 15. (a) Six lead-acid types of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What is the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Solution :A) Number of secondary cells n = 6

Emf of each secondary cell E = 2 Volt

Internal resistance of each cell r = 0.015 Ω

Resistance of resistor R = 8.5 Ω

Let current drawn from supply = I

Hence maximum current drawn from the cell is 0.005Ampere and since a large current is required to start the motor of a car and hence the cell cannot be used to start the motor.

Question 16. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρAl = 2.63 × 10−8 Ω m, ρCu = 1.72 × 10−8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Solution :Resistivity of aluminium, p1 = 2.63 × 10−8 Ω m

Relative density of aluminium, d1 = 2.7

Let l1 be the length of aluminium wire and m1 be its mass.

Resistance of the aluminium wire = R1

Area of cross-section of the aluminium wire = A1

Resistivity of copper, p2 = 1.72 × 10−8 Ω metre

Relative density of copper, d2 = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

We can write,

R1 = (p1 × l1)/A1

R2 = (p2 × l2)/A2

Since R1 = R2

(p1 × l1)/A1 = (p2 × l2)/A2

And l1 = l2

Therefore p1/A1 = p2/A2

A1/A2 = (2.63 × 10-8)/(1.72 × 10-8)

⇒ A1/A2 = 2.63/1.72

Mass of the aluminium wire,

m1 = Volume × Density

= A1 × l1 × d1 = A1 l1 d1 …..1

Mass of the copper wire, m2 = Volume × Density

= A2 × l2 × d2 = A2 l2 d2 …….2

Dividing equation (1) by equation (2), we obtain

m1/m2 = (A1 l1 d1)/(A2 l2 d2)

Since l1 = l2

m1/m2 = A1 d1/A2 d2

Since A1/A2 = 2.63/1.72 calculated above

⇒ m1/m2 = (2.63 × 2.7)/(1.72 × 8.9)

⇒ m1/m2 = 0.46

Since m1 is smaller than m2 Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.

Question 17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?