PhysicsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2023-24: question 33(b)

From the official CBSE sample question paper for 2023-24, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerAlternating Current

a. A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit.
b. Plot a graph to show the variation of current with frequency of the ac source, explaining the nature of its variation for two different resistances $R_{1}$ and $R_{2}$ ( $R_{1} < R_{2}$ ) at resonance.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

Use current as the reference phasor, since it is common to all three series elements.

Hint 2

$V_{L}$ and $V_{C}$ point in opposite directions, so combine them first, then add $V_{R}$ at right angles.

Hint 3

At resonance $X_{L} = X_{C}$ , so the peak current depends only on $R$ .

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.

a. Let $v = v_{m} sin ω t$ be applied to L, C and R in series. The same current $i = i_{m} sin (ω t + ϕ)$ flows through all three elements. The voltage amplitudes are $V_{R} = i_{m} R$ , $V_{L} = i_{m} X_{L}$ and $V_{C} = i_{m} X_{C}$ .
1 mark
Phasor diagram (described): $I$ and $V_{R}$ lie along the same line. $V_{L}$ is perpendicular to $I$ and leads it by $9 0^{\circ}$ . $V_{C}$ is opposite to $V_{L}$ and lags $I$ by $9 0^{\circ}$ . Taking $V_{L} > V_{C}$ , the net reactive voltage is $(V_{L} - V_{C})$ , perpendicular to $V_{R}$ .
1 mark
The source voltage is the phasor sum of these, so by Pythagoras:
$v_{m}^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2} = i_{m}^{2} [R^{2} + (X_{L} - X_{C})^{2}]$
1 mark
Hence $i_{m} = \frac{v _{m}}{R ^{2} + ( X _{L} - X _{C} ) ^{2}} = \frac{v _{m}}{Z}$ .
The impedance is $Z = R^{2} + (X_{L} - X_{C})^{2}$ . The phase angle is $tan ϕ = \frac{X _{L} - X _{C}}{R}$ .
1 mark
b. Graph of $i_{m}$ against $ω$ for both resistances:
- Each curve rises to a peak at the resonant frequency $ω_{0} = 1/ L C$ , where $X_{L} = X_{C}$ and $Z = R$ , so $i_{ma x} = v_{m} / R$ . On either side of $ω_{0}$ the current falls.
- Curve (i), for $R_{1}$ , has the higher and sharper peak, because $R_{1} < R_{2}$ .
- Curve (ii), for $R_{2}$ , has a lower and broader peak.
- Both peaks occur at the same $ω_{0}$ , since $ω_{0}$ does not depend on $R$ .
1 mark

Answer: $Z = R^{2} + (X_{L} - X_{C})^{2}$ ; at resonance $ω_{0} = 1/ L C$ , the $R_{1}$ curve has a higher, sharper peak and the $R_{2}$ curve a lower, broader peak.

Where the marks are

States the same current through all elements and the voltage amplitudes $V_{R}$ , $V_{L}$ , $V_{C}$ (1 mark)
Correct phasor diagram with the phase relations (1 mark)
Applies Pythagoras: $v_{m}^{2} = i_{m}^{2} [R^{2} + (X_{L} - X_{C})^{2}]$ (1 mark)
Obtains $Z = R^{2} + (X_{L} - X_{C})^{2}$ (and $tan ϕ$ ) (1 mark)
Resonance curves: same $ω_{0}$ ; $R_{1}$ gives a higher, sharper peak than $R_{2}$ (1 mark)

Revise the chapter

Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)
Topic: Series LCR Circuit and Resonance

Practise this chapterBoard-style questions from Alternating Current, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Physics sample paper 2023-24: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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