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

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

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

Q33(a) · 5 marks · Long answerAlternating Current

a. Draw graphs showing the variations of inductive reactance and capacitive reactance with frequency of applied ac source.
b. Draw the phasor diagram for a series LRC circuit connected to an AC source.
c. When an alternating voltage of 220V is applied across a device X, a current of 0.25A flows which lags behind the applied voltage in phase by π/2 radian. If the same voltage is applied across another device Y, the same current flows but now it is in phase with the applied voltage.
(i) Name the devices X and Y.
(ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y.

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

Write the formulas $X_{L} = 2 π f L$ and $X_{C} = 1/ (2 π f C)$ and ask how each depends on $f$ .

Hint 2

Recall which element makes the current lag by $9 0^{\circ}$ and which keeps it in phase.

Hint 3

For the series combination, the reactance and resistance combine as $R^{2} + X_{L}^{2}$ , not by simple addition.

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. $X_{L} = 2 π f L$ , so the graph of $X_{L}$ against $f$ is a straight line through the origin. $X_{C} = \frac{1}{2 π f C}$ , so the graph of $X_{C}$ against $f$ is a rectangular hyperbola: it is very large at low $f$ and approaches the $f$ -axis as $f$ increases. Both are drawn on the same $f$ -axis, with the curves labelled $X_{L}$ and $X_{C}$ .
1 mark
b. Phasor diagram:
- Draw $I$ along the reference (x) axis.
- Draw $V_{R} = I R$ along $I$ .
- Draw $V_{L} = I X_{L}$ vertically up, leading $I$ by $9 0^{\circ}$ .
- Draw $V_{C} = I X_{C}$ vertically down, lagging $I$ by $9 0^{\circ}$ .
The resultant $V$ is the hypotenuse of a triangle with sides $V_{R}$ and $(V_{L} - V_{C})$ . It makes an angle $ϕ$ with $I$ , where $tan ϕ = \frac{X _{L} - X _{C}}{R}$ .
1 mark
c(i). In X, the current lags the voltage by $π /2$ , so X is a pure inductor. In Y, the current is in phase with the voltage, so Y is a pure resistor.
1 mark
c(ii). $X_{L} = \frac{V}{I} = \frac{220}{0.25} = 880 Ω$ and $R = \frac{220}{0.25} = 880 Ω$ .
1 mark
For the series combination, $Z = R^{2} + X_{L}^{2} = 8802 \approx 1244.5 Ω$ .
So $I = \frac{V}{Z} = \frac{220}{880 2} = \frac{1}{4 2} \approx 0.177 A$ .
1 mark

Answer: (c)(i) X = inductor, Y = resistor; (ii) $Z = 8802 Ω$ , $I \approx 0.177$ A

Where the marks are

Graphs: $X_{L}$ straight line through origin, $X_{C}$ hyperbola (½ + ½ mark)
Correct phasor diagram for series LCR with $V_{L}$ leading and $V_{C}$ lagging $I$ (1 mark)
Identifies X as inductor and Y as resistor (½ + ½ mark)
Finds $X_{L} = R = 880 Ω$ (1 mark)
Finds $Z = 8802 Ω$ and $I = 0.177$ A (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 paperSample paper 2023-24: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

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

Hints

Step-by-step solution

Revise the chapter

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