PhysicsCBSE Class 12 · Board exam 2024, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2024: question 32(a)

Set in the CBSE Class 12 board exam of 2024, paper 55/1/1, section E, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerAlternating Current

(i) A resistor and a capacitor are connected in series to an ac source $v = v_{m} sin ω t$ . Derive an expression for the impedance of the circuit.
(ii) When does an inductor act as a conductor in a circuit? Give reason for it.
(iii) An electric lamp is designed to operate at 110 V dc and 11 A current. If the lamp is operated on 220 V, 50 Hz ac source with a coil in series, then find the inductance of the coil.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

In an RC circuit, the voltages across R and C are 90° out of phase, so add them as phasors, not arithmetically.

Hint 2

What is the reactance $2 π ν L$ when the frequency is zero?

Hint 3

Find the lamp's resistance from its dc rating. The current must stay 11 A on 220 V ac, so find Z and then $X_{L}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Let the current be $i = i_{m} sin ω t$ . Across the resistor, $v_{R}$ is in phase with $i$ and its peak value is $v_{R m} = i_{m} R$ . Across the capacitor, $v_{C}$ lags $i$ by $π /2$ and its peak value is $v_{C m} = i_{m} X_{C}$ , where $X_{C} = \frac{1}{ω C}$ .
Phasor diagram (in words): the phasor $i_{m}$ and $V_{R m}$ are drawn along the same direction. $V_{C m}$ is drawn at $9 0^{\circ}$ behind them (clockwise). The source phasor $v_{m}$ is the resultant of these two perpendicular phasors.
1 mark
Since the two phasors are perpendicular:
$v_{m}^{2} = v_{R m}^{2} + v_{C m}^{2} = (i_{m} R)^{2} + (i_{m} X_{C})^{2}$
So $i_{m} = \frac{v _{m}}{R ^{2} + X _{C}^{2}}$ .
Impedance $Z = \frac{v _{m}}{i _{m}} = R^{2} + X_{C}^{2} = R^{2} + \frac{1}{ω ^{2} C ^{2}}$ .
The phase angle is given by $tan ϕ = X_{C} / R$ , with the current leading the voltage.
1 mark
(ii) An inductor acts as a conductor (only its small wire resistance remains) in a dc circuit.
Reason: $X_{L} = ω L = 2 π ν L$ . For dc, $ν = 0$ , so $X_{L} = 0$ .
Equivalently, a steady current has $d I / d t = 0$ , so the induced emf $ε = - L d I / d t = 0$ and nothing opposes the current.
1 mark
(iii) Resistance of the lamp:
$R = \frac{V}{I} = \frac{110}{11} = 10 Ω$
On ac, the same current of 11 A must flow. So the impedance is:
$Z = \frac{V _{r m s}}{I _{r m s}} = \frac{220}{11} = 20 Ω$
1 mark
Using $Z^{2} = R^{2} + X_{L}^{2}$ :
$400 = 100 + X_{L}^{2} \Rightarrow X_{L} = 300 = 103 Ω$
Then $X_{L} = 2 π f L$ gives:
$L = \frac{10 3}{2 π \times 50} = \frac{3}{10 π} H \approx 0.055 H$
1 mark

Answer: (i) $Z = R^{2} + (1/ ω C)^{2}$ ; (ii) in a dc circuit, since $X_{L} = 2 π ν L = 0$ for $ν = 0$ ; (iii) $L = \frac{3}{10 π}$ H ≈ 0.055 H

Where the marks are

Phasor relations $v_{R m} = i_{m} R$ (in phase) and $v_{C m} = i_{m} X_{C}$ (lagging by 90°) (1 mark)
Obtains $Z = R^{2} + X_{C}^{2}$ (1 mark)
States dc and gives the reason $X_{L} = 0$ for $ν = 0$ (1 mark)
Finds $R = 10 Ω$ and $Z = 20 Ω$ (1 mark)
Finds $X_{L} = 103 Ω$ and $L = 3 / (10 π)$ H (1 mark)

Revise the chapter

Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Alternating Current, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2024: question 32(a)

Hints

Step-by-step solution

Revise the chapter

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