PYQs / GATE EE / 2014 / Set 3 / Q57 GATE EE 2014 Set 3 — Question 57 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium RMS & Average Values AC Power & Three-Phase Electric Circuits PMMC & Moving Iron Electrical Quantity Measurement Electrical & Electronic Measurements
Electrical & Electronic Measurements → AC Power & Three-Phase → PMMC & Moving Iron
Last updated 5 September 2026
Question A periodic waveform observed across a load is represented by
V ( t ) = { 1 + sin ω t 0 ≤ ω t < 6 π − 1 + sin ω t 6 π ≤ ω t < 12 π V(t) = \begin{cases} 1 + \sin \omega t & 0 \leq \omega t < 6\pi \\ -1 + \sin \omega t & 6\pi \leq \omega t < 12\pi \end{cases} V ( t ) = { 1 + sin ω t − 1 + sin ω t 0 ≤ ω t < 6 π 6 π ≤ ω t < 12 π The measured value, using moving iron voltmeter connected across the load, is
Correct answer (A) √((3)/(2))
Solution A moving iron voltmeter measures the RMS value of the waveform. The period of the waveform is
T = 12 π / ω T = 12\pi/\omega T = 12 π / ω . Let
θ = ω t \theta = \omega t θ = ω t .
V r m s 2 = 1 12 π ∫ 0 12 π V 2 ( θ ) d θ = 1 12 π [ ∫ 0 6 π ( 1 + sin θ ) 2 d θ + ∫ 6 π 12 π ( − 1 + sin θ ) 2 d θ ] V_{rms}^2 = \frac{1}{12\pi} \int_0^{12\pi} V^2(\theta) d\theta = \frac{1}{12\pi} \left[ \int_0^{6\pi} (1 + \sin \theta)^2 d\theta + \int_{6\pi}^{12\pi} (-1 + \sin \theta)^2 d\theta \right] V r m s 2 = 12 π 1 ∫ 0 12 π V 2 ( θ ) d θ = 12 π 1 [ ∫ 0 6 π ( 1 + sin θ ) 2 d θ + ∫ 6 π 12 π ( − 1 + sin θ ) 2 d θ ] Evaluating the integrals:
∫ 0 6 π ( 1 + 2 sin θ + sin 2 θ ) d θ = [ θ − 2 cos θ + θ 2 − sin 2 θ 4 ] 0 6 π = ( 6 π − 2 + 3 π ) − ( 0 − 2 ) = 9 π \int_0^{6\pi} (1 + 2\sin \theta + \sin^2 \theta) d\theta = [\theta - 2\cos \theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}]_0^{6\pi} = (6\pi - 2 + 3\pi) - (0 - 2) = 9\pi ∫ 0 6 π ( 1 + 2 sin θ + sin 2 θ ) d θ = [ θ − 2 cos θ + 2 θ − 4 s i n 2 θ ] 0 6 π = ( 6 π − 2 + 3 π ) − ( 0 − 2 ) = 9 π ∫ 6 π 12 π ( 1 − 2 sin θ + sin 2 θ ) d θ = [ θ + 2 cos θ + θ 2 − sin 2 θ 4 ] 6 π 12 π = ( 12 π + 2 + 6 π ) − ( 6 π + 2 + 3 π ) = 9 π \int_{6\pi}^{12\pi} (1 - 2\sin \theta + \sin^2 \theta) d\theta = [\theta + 2\cos \theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}]_{6\pi}^{12\pi} = (12\pi + 2 + 6\pi) - (6\pi + 2 + 3\pi) = 9\pi ∫ 6 π 12 π ( 1 − 2 sin θ + sin 2 θ ) d θ = [ θ + 2 cos θ + 2 θ − 4 s i n 2 θ ] 6 π 12 π = ( 12 π + 2 + 6 π ) − ( 6 π + 2 + 3 π ) = 9 π V r m s 2 = 9 π + 9 π 12 π = 18 π 12 π = 1.5 = 3 2 V_{rms}^2 = \frac{9\pi + 9\pi}{12\pi} = \frac{18\pi}{12\pi} = 1.5 = \frac{3}{2} V r m s 2 = 12 π 9 π + 9 π = 12 π 18 π = 1.5 = 2 3 V r m s = 3 2 V_{rms} = \sqrt{\frac{3}{2}} V r m s = 2 3 Thus, the correct option is (A).
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Correct answer (A) √((3)/(2))
Solution A moving iron voltmeter measures the RMS value of the waveform. The period of the waveform is
T = 12 π / ω T = 12\pi/\omega T = 12 π / ω . Let
θ = ω t \theta = \omega t θ = ω t .
V r m s 2 = 1 12 π ∫ 0 12 π V 2 ( θ ) d θ = 1 12 π [ ∫ 0 6 π ( 1 + sin θ ) 2 d θ + ∫ 6 π 12 π ( − 1 + sin θ ) 2 d θ ] V_{rms}^2 = \frac{1}{12\pi} \int_0^{12\pi} V^2(\theta) d\theta = \frac{1}{12\pi} \left[ \int_0^{6\pi} (1 + \sin \theta)^2 d\theta + \int_{6\pi}^{12\pi} (-1 + \sin \theta)^2 d\theta \right] V r m s 2 = 12 π 1 ∫ 0 12 π V 2 ( θ ) d θ = 12 π 1 [ ∫ 0 6 π ( 1 + sin θ ) 2 d θ + ∫ 6 π 12 π ( − 1 + sin θ ) 2 d θ ] Evaluating the integrals:
∫ 0 6 π ( 1 + 2 sin θ + sin 2 θ ) d θ = [ θ − 2 cos θ + θ 2 − sin 2 θ 4 ] 0 6 π = ( 6 π − 2 + 3 π ) − ( 0 − 2 ) = 9 π \int_0^{6\pi} (1 + 2\sin \theta + \sin^2 \theta) d\theta = [\theta - 2\cos \theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}]_0^{6\pi} = (6\pi - 2 + 3\pi) - (0 - 2) = 9\pi ∫ 0 6 π ( 1 + 2 sin θ + sin 2 θ ) d θ = [ θ − 2 cos θ + 2 θ − 4 s i n 2 θ ] 0 6 π = ( 6 π − 2 + 3 π ) − ( 0 − 2 ) = 9 π ∫ 6 π 12 π ( 1 − 2 sin θ + sin 2 θ ) d θ = [ θ + 2 cos θ + θ 2 − sin 2 θ 4 ] 6 π 12 π = ( 12 π + 2 + 6 π ) − ( 6 π + 2 + 3 π ) = 9 π \int_{6\pi}^{12\pi} (1 - 2\sin \theta + \sin^2 \theta) d\theta = [\theta + 2\cos \theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}]_{6\pi}^{12\pi} = (12\pi + 2 + 6\pi) - (6\pi + 2 + 3\pi) = 9\pi ∫ 6 π 12 π ( 1 − 2 sin θ + sin 2 θ ) d θ = [ θ + 2 cos θ + 2 θ − 4 s i n 2 θ ] 6 π 12 π = ( 12 π + 2 + 6 π ) − ( 6 π + 2 + 3 π ) = 9 π V r m s 2 = 9 π + 9 π 12 π = 18 π 12 π = 1.5 = 3 2 V_{rms}^2 = \frac{9\pi + 9\pi}{12\pi} = \frac{18\pi}{12\pi} = 1.5 = \frac{3}{2} V r m s 2 = 12 π 9 π + 9 π = 12 π 18 π = 1.5 = 2 3 V r m s = 3 2 V_{rms} = \sqrt{\frac{3}{2}} V r m s = 2 3 Thus, the correct option is (A).
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