Answer the following. (i) You are given three circuit elements X, Y and Z. They are connected one by one across a given ac source. It is found that V and I are in phase for element X. V leads I by ((π)/(4) ) for element Y while I leads V by ((π)/(4) ) for element Z. Identify elements X, Y and Z. (ii) Establish the expression for impedance of circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source. (iii) In a series LCR circuit, obtain the conditions under which (i) impedance is minimum and (ii) wattless current flows in the circuit.

(i) X: resistor, Y: inductor (X_L = R), Z: capacitor (X_C = R); (ii) Z=√(R²+(X_C-X_L)²), current peaks at ω₀ = 1/√(LC); (iii) minimum Z when X_L = X_C; wattless current when φ = π/2 (cos φ = 0)

PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 5 marks · Case-based

CBSE Class 12 Physics 2024: question 32(a)

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, 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 · Case-basedCompetency-basedAlternating Current

Answer the following.

(i)
You are given three circuit elements X, Y and Z. They are connected one by one across a given ac source. It is found that V and I are in phase for element X. V leads I by $(\frac{π}{4})$ for element Y while I leads V by $(\frac{π}{4})$ for element Z. Identify elements X, Y and Z.
0 marks
(ii)
Establish the expression for impedance of circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source.
0 marks
(iii)
In a series LCR circuit, obtain the conditions under which (i) impedance is minimum and (ii) wattless current flows in the circuit.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.

Hints

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

Hint 1

In which element are V and I in phase? Which element makes V lead, and which makes I lead?

Hint 2

In a series circuit the current is common to all elements; draw the voltage phasors relative to it.

Hint 3

The power factor is cos φ; when is the average power zero?

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) X: V and I are in phase, so X is a pure resistor.
½ mark
Y: V leads I by $π /4$ , so Y is an inductor. A lead of exactly $π /4$ means a real inductor with $X_{L} = R$ , since $tan ϕ = X_{L} / R = 1$ .
½ mark
Z: I leads V by $π /4$ , so Z is a capacitor. Again this is a real capacitor (in series with resistance) with $X_{C} = R$ .
½ mark
(ii) Phasor diagram (in words): In a series R, L, C circuit the same current $i_{m}$ flows through all three, so take the current phasor as reference.
- $V_{R m} = i_{m} R$ is in phase with the current.
- $V_{L m} = i_{m} X_{L}$ leads the current by $9 0^{\circ}$ .
- $V_{C m} = i_{m} X_{C}$ lags the current by $9 0^{\circ}$ .
½ mark
$V_{L m}$ and $V_{C m}$ are opposite in direction, so their resultant is $(V_{C m} - V_{L m})$ , at right angles to $V_{R m}$ .
Therefore $V_{m}^{2} = V_{R m}^{2} + (V_{C m} - V_{L m})^{2}$ .
½ mark
Substituting each voltage: $V_{m}^{2} = (i_{m} R)^{2} + (i_{m} X_{C} - i_{m} X_{L})^{2}$ .
½ mark
Impedance:
$Z = \frac{V _{m}}{i _{m}} = R^{2} + (X_{C} - X_{L})^{2}$
where $X_{L} = ω L$ and $X_{C} = \frac{1}{ω C}$ .
½ mark
Variation of current with frequency (graph in words): $i_{m} = \frac{V _{m}}{Z}$ .
- At low frequency the current is small, because $X_{C}$ is large.
- The current rises to a sharp maximum $\frac{V _{m}}{R}$ at the resonant frequency $ω_{0} = \frac{1}{L C}$ .
- At higher frequency it falls again, because $X_{L}$ becomes large.
½ mark
(iii)(i) $Z = R^{2} + (X_{C} - X_{L})^{2}$ is minimum (equal to $R$ ) when $X_{C} = X_{L}$ , i.e. at resonance $ω = \frac{1}{L C}$ .
½ mark
(iii)(ii) Average power is $P = V_{r m s} I_{r m s} cos ϕ$ . A wattless current means $P = 0$ even though $V \neq = 0$ and $I \neq = 0$ .
So $cos ϕ = 0$ , i.e. $ϕ = π /2$ . This happens when the circuit is purely reactive, with $R = 0$ (pure L, pure C, or an LC combination).
½ mark

Answer: (i) X: resistor, Y: inductor (X_L = R), Z: capacitor (X_C = R); (ii) $Z = R^{2} + (X_{C} - X_{L})^{2}$ , current peaks at ω₀ = 1/√(LC); (iii) minimum Z when X_L = X_C; wattless current when φ = π/2 (cos φ = 0)

Where the marks are

Identifies X as a resistor (½ mark)
Identifies Y as an inductor (½ mark)
Identifies Z as a capacitor (½ mark)
Phasor diagram (½ mark)
V_m² = V_R² + (V_C − V_L)² (½ mark)
Substitutes i_mR, i_mX_C, i_mX_L (½ mark)
Z = √(R² + (X_C − X_L)²) (½ mark)
Current vs frequency curve with a peak at resonance (½ mark)
Minimum impedance when X_C = X_L (½ mark)
Wattless current when cos φ = 0, φ = π/2 (½ 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 paper2024 board paper, set 55/5/1: 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 2024: question 32(a)

Hints

Step-by-step solution

Revise the chapter

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