GATE 2024 IN Q.30: solution and answer

GATE 2024 · INQ.30 · 1 mark · NAT

The voltage applied and the current drawn by a circuit are

v (t) = 95 + 200 cos (120 π t) + 90 cos (360 π t - 6 0^{\circ}) V

i (t) = 4 cos (120 π t - 6 0^{\circ}) + 1.5 cos (240 π t - 7 5^{\circ}) A

The average power absorbed by the circuit is ________ W (rounded off to nearest integer).

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Answer and solution

ANSWER

The average power absorbed by the circuit is 200 W.

Official answer: 200Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Average power when the waveforms are non-sinusoidal. Every component is a harmonic of 60 Hz ( $120 π$ , $240 π$ and $360 π rad/s$ ), so both waveforms repeat with period $T = \frac{1}{60} s$ . Over that period, the product of a voltage and a current at different frequencies averages to zero. Only a DC voltage with a DC current, or a voltage and a current at the same frequency, give average power: $P = V_{0} I_{0} + n \sum \frac{V _{m, n} I _{m, n}}{2} cos (θ_{v, n} - θ_{i, n})$ Here $V_{m, n}$ and $I_{m, n}$ are peak amplitudes. The same thing in rms form is $V_{n} I_{n} cos (θ_{v, n} - θ_{i, n})$ .
Sort the components by frequency. DC: $V_{0} = 95 V$ , $I_{0} = 0$ . At $120 π rad/s$ (60 Hz): voltage of peak $200 V$ at $0^{\circ}$ , current of peak $4 A$ at $- 6 0^{\circ}$ . As rms phasors these are $\frac{200}{2} ∠ 0^{\circ} = 141.42∠ 0^{\circ} V$ and $\frac{4}{2} ∠ - 6 0^{\circ} = 2.828∠ - 6 0^{\circ} A$ . At $240 π rad/s$ (120 Hz): current only, with peak $1.5 A$ at $- 7 5^{\circ}$ . At $360 π rad/s$ (180 Hz): voltage only, with peak $90 V$ at $- 6 0^{\circ}$ .
DC term. $P_{0} = V_{0} I_{0} = 95 \times 0 = 0 W$ , because the current has no DC component. The 95 V source term only multiplies sinusoidal currents, and a sinusoid averages to zero over a full period.
60 Hz term. Both voltage and current are present at this frequency: $P_{1} = \frac{200 \times 4}{2} cos (0^{\circ} - (- 6 0^{\circ})) = 400 \times 0.5 = 200 W$ The rms phasors give the same result: $141.42 \times 2.828 \times cos 6 0^{\circ} = 400 \times 0.5 = 200 W$ .
120 Hz and 180 Hz terms. Each of these frequencies has only a voltage or only a current, so there is nothing to pair it with. Products such as $cos (120 π t) cos (240 π t - 7 5^{\circ})$ or $cos (360 π t - 6 0^{\circ}) cos (240 π t - 7 5^{\circ})$ average to zero over $T$ . So $P_{2} = P_{3} = 0$ .
Total. $P = P_{0} + P_{1} + P_{2} + P_{3} = 0 + 200 + 0 + 0 = 200 W$

KEY IDEA

Voltage and current components at different frequencies (a DC voltage paired with an AC current counts as different) transfer no average power over a common period. Only same-frequency pairs contribute, each giving $\frac{V _{m} I _{m}}{2} cos (θ_{v} - θ_{i})$ . Here the only such pair is the 60 Hz one: $\frac{200 \times 4}{2} cos 6 0^{\circ} = 200 W$ .

WATCH OUT FOR

Students often multiply the DC voltage by an AC current (writing $95 \times 4$ or similar), or pair the 90 V (180 Hz) voltage with the 1.5 A (120 Hz) current. Both give zero average power, because the frequencies differ. A second common slip is using $V_{m} I_{m} cos ϕ$ with peak values and forgetting the factor $\frac{1}{2}$ , which gives 400 W. Using the wrong sign in the phase difference is harmless here, since $cos (\pm 6 0^{\circ}) = 0.5$ .

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GATE 2024 IN Q.30: solution and answer

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