GATE 2025 EC Q.39: solution and answer

GATE 2025 · ECQ.39 · 2 marks · MCQ

In the circuit below, $M_{1}$ is an ideal AC voltmeter and $M_{2}$ is an ideal AC ammeter. The source voltage (in Volts) is $v_{s} (t) = 100 cos (200 t)$ .

What should be the value of the variable capacitor $C$ such that the RMS readings on $M_{1}$ and $M_{2}$ are 25 V and 5 A, respectively?

GATE 2025 EC question 39: A sinusoidal voltage source v_s(t) (+ terminal at top, - at bottom) is connected to the left-hand input terminals of a… — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
25 μF
(B)
4 μF
(C)
0.25 μF
(D)
Insufficient information to find $C$

Answer and solution

ANSWER

$C = 25 μ F$ (option A): the readings force $∣ Z ∣ = 25/5 = 5 Ω$ , which equals $R$ , so the branch must be at resonance at $ω = 200 rad/s$ , giving $C = 1/ (ω^{2} L)$ .

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

The box is linear and time-invariant, so in sinusoidal steady state every voltage and current at its output is at the source frequency, $ω = 200 rad/s$ . We never need to know what is inside the box: the two meter readings describe the load branch completely.
The ideal ammeter $M_{2}$ has zero impedance and drops no voltage, so the ideal voltmeter $M_{1}$ reads the voltage across the whole series branch of $5 Ω$ , $C$ and $1 H$ . $M_{2}$ reads the current through that same branch.
Ohm's law in rms magnitudes: $∣ Z ∣ = \frac{∣ V ∣}{∣ I ∣} = \frac{25 V}{5 A} = 5 Ω$ .
Series branch impedance: $Z = 5 + j (ω L - \frac{1}{ω C}) Ω$ , so $∣ Z ∣ = 5^{2} + X^{2}$ with $X = ω L - \frac{1}{ω C}$ .
$25 + X^{2} = 5 \Rightarrow X = 0$ . The branch must be at series resonance: $ω L = \frac{1}{ω C}$ .
$C = \frac{1}{ω ^{2} L} = \frac{1}{( 200 ) ^{2} \times 1} F = \frac{1}{40000} F = 25 μ F$
Check: $X_{L} = 200 \times 1 = 200 Ω$ and $X_{C} = \frac{1}{200 \times 25 \times 1 0 ^{- 6}} = 200 Ω$ . They cancel, so $∣ Z ∣ = 5 Ω$ , and $25 V /5 Ω = 5 A$ matches both readings. Answer: (A).

KEY IDEA

The meters measure the load branch directly, so the unknown LTI box does not matter. The only thing the box sets is the frequency, which stays at $ω = 200 rad/s$ . The readings give $∣ Z ∣ = V_{rms} / I_{rms} = 5 Ω$ , which equals the branch resistance, so the net reactance must be zero (series resonance) and $C = 1/ (ω^{2} L)$ .

WATCH OUT FOR

Picking (D) because the contents of the box are unknown. The box only has to be LTI, which fixes the frequency at 200 rad/s, and the voltmeter and ammeter already give the branch's voltage and current. A second mistake is to use the 100 V source amplitude, or to mix peak and rms values, instead of the ratio of the two meter readings. A third is to use $f = 200 Hz$ instead of $ω = 200 rad/s$ , which gives the wrong capacitance.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2025 EC Q.39: solution and answer

Answer and solution

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