PYQs / GATE EE / 2021 / Set 1 / Q58 GATE EE 2021 Set 1 — Question 58 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Hard RMS & Average Values AC Power & Three-Phase Electric Circuits RMS & Average Value Signal Representation Signals & Systems
Electric Circuits → AC Power & Three-Phase → RMS & Average Value
Last updated 5 September 2026
Question The waveform shown in solid line is obtained by clipping a full-wave rectified sinusoid (shown dashed). The ratio of the RMS value of the full-wave rectified waveform to the RMS value of the clipped waveform is
__________ . (Round off to 2 decimal places.)
Correct answer 1.2 to 1.23
Solution Let the full-wave rectified sinusoid be
v ( t ) = V m ∣ sin ( ω t ) ∣ v(t) = V_m |\sin(\omega t)| v ( t ) = V m ∣ sin ( ω t ) ∣ . Its RMS value is
V r m s 1 = V m 2 V_{rms1} = \frac{V_m}{\sqrt{2}} V r m s 1 = 2 V m .
The clipped waveform
v c ( t ) v_c(t) v c ( t ) is clipped at
0.707 V m = V m 2 0.707 V_m = \frac{V_m}{\sqrt{2}} 0.707 V m = 2 V m . This corresponds to an angle
θ = π 4 \theta = \frac{\pi}{4} θ = 4 π .
Over one period
[ 0 , π ] [0, \pi] [ 0 , π ] , the clipped waveform is:
v c ( θ ) = V m sin θ v_c(\theta) = V_m \sin \theta v c ( θ ) = V m sin θ for
0 ≤ θ ≤ π 4 0 \leq \theta \leq \frac{\pi}{4} 0 ≤ θ ≤ 4 π and
3 π 4 ≤ θ ≤ π \frac{3\pi}{4} \leq \theta \leq \pi 4 3 π ≤ θ ≤ π v c ( θ ) = V m 2 v_c(\theta) = \frac{V_m}{\sqrt{2}} v c ( θ ) = 2 V m for
π 4 ≤ θ ≤ 3 π 4 \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4} 4 π ≤ θ ≤ 4 3 π The mean square value of the clipped waveform is:
V r m s 2 2 = 1 π [ 2 ∫ 0 π / 4 ( V m sin θ ) 2 d θ + ∫ π / 4 3 π / 4 ( V m 2 ) 2 d θ ] V_{rms2}^2 = \frac{1}{\pi} \left[ 2 \int_0^{\pi/4} (V_m \sin \theta)^2 d\theta + \int_{\pi/4}^{3\pi/4} \left(\frac{V_m}{\sqrt{2}}\right)^2 d\theta \right] V r m s 2 2 = π 1 [ 2 ∫ 0 π /4 ( V m sin θ ) 2 d θ + ∫ π /4 3 π /4 ( 2 V m ) 2 d θ ] V r m s 2 2 = V m 2 π [ 2 ∫ 0 π / 4 1 − cos 2 θ 2 d θ + 1 2 ( 3 π 4 − π 4 ) ] V_{rms2}^2 = \frac{V_m^2}{\pi} \left[ 2 \int_0^{\pi/4} \frac{1 - \cos 2\theta}{2} d\theta + \frac{1}{2} \left(\frac{3\pi}{4} - \frac{\pi}{4}\right) \right] V r m s 2 2 = π V m 2 [ 2 ∫ 0 π /4 2 1 − cos 2 θ d θ + 2 1 ( 4 3 π − 4 π ) ] V r m s 2 2 = V m 2 π [ ( θ − sin 2 θ 2 ) 0 π / 4 + π 4 ] = V m 2 π [ π 4 − 1 2 + π 4 ] = V m 2 π ( π 2 − 1 2 ) = V m 2 ( 1 2 − 1 2 π ) V_{rms2}^2 = \frac{V_m^2}{\pi} \left[ \left( \theta - \frac{\sin 2\theta}{2} \right)_0^{\pi/4} + \frac{\pi}{4} \right] = \frac{V_m^2}{\pi} \left[ \frac{\pi}{4} - \frac{1}{2} + \frac{\pi}{4} \right] = \frac{V_m^2}{\pi} \left( \frac{\pi}{2} - \frac{1}{2} \right) = V_m^2 \left( \frac{1}{2} - \frac{1}{2\pi} \right) V r m s 2 2 = π V m 2 [ ( θ − 2 sin 2 θ ) 0 π /4 + 4 π ] = π V m 2 [ 4 π − 2 1 + 4 π ] = π V m 2 ( 2 π − 2 1 ) = V m 2 ( 2 1 − 2 π 1 ) The ratio is:
Ratio = V r m s 1 V r m s 2 = V m / 2 V m 2 ( 0.5 − 0.5 / π ) = 1 2 0.5 ( 1 − 1 / π ) = 1 1 − 1 / π = π π − 1 ≈ 1.211 \text{Ratio} = \frac{V_{rms1}}{V_{rms2}} = \frac{V_m/\sqrt{2}}{\sqrt{V_m^2 (0.5 - 0.5/\pi)}} = \frac{1}{\sqrt{2} \sqrt{0.5(1 - 1/\pi)}} = \frac{1}{\sqrt{1 - 1/\pi}} = \sqrt{\frac{\pi}{\pi - 1}} \approx 1.211 Ratio = V r m s 2 V r m s 1 = V m 2 ( 0.5 − 0.5/ π ) V m / 2 = 2 0.5 ( 1 − 1/ π ) 1 = 1 − 1/ π 1 = π − 1 π ≈ 1.211 Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer 1.2 to 1.23
Solution Let the full-wave rectified sinusoid be
v ( t ) = V m ∣ sin ( ω t ) ∣ v(t) = V_m |\sin(\omega t)| v ( t ) = V m ∣ sin ( ω t ) ∣ . Its RMS value is
V r m s 1 = V m 2 V_{rms1} = \frac{V_m}{\sqrt{2}} V r m s 1 = 2 V m .
The clipped waveform
v c ( t ) v_c(t) v c ( t ) is clipped at
0.707 V m = V m 2 0.707 V_m = \frac{V_m}{\sqrt{2}} 0.707 V m = 2 V m . This corresponds to an angle
θ = π 4 \theta = \frac{\pi}{4} θ = 4 π .
Over one period
[ 0 , π ] [0, \pi] [ 0 , π ] , the clipped waveform is:
v c ( θ ) = V m sin θ v_c(\theta) = V_m \sin \theta v c ( θ ) = V m sin θ for
0 ≤ θ ≤ π 4 0 \leq \theta \leq \frac{\pi}{4} 0 ≤ θ ≤ 4 π and
3 π 4 ≤ θ ≤ π \frac{3\pi}{4} \leq \theta \leq \pi 4 3 π ≤ θ ≤ π v c ( θ ) = V m 2 v_c(\theta) = \frac{V_m}{\sqrt{2}} v c ( θ ) = 2 V m for
π 4 ≤ θ ≤ 3 π 4 \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4} 4 π ≤ θ ≤ 4 3 π The mean square value of the clipped waveform is:
V r m s 2 2 = 1 π [ 2 ∫ 0 π / 4 ( V m sin θ ) 2 d θ + ∫ π / 4 3 π / 4 ( V m 2 ) 2 d θ ] V_{rms2}^2 = \frac{1}{\pi} \left[ 2 \int_0^{\pi/4} (V_m \sin \theta)^2 d\theta + \int_{\pi/4}^{3\pi/4} \left(\frac{V_m}{\sqrt{2}}\right)^2 d\theta \right] V r m s 2 2 = π 1 [ 2 ∫ 0 π /4 ( V m sin θ ) 2 d θ + ∫ π /4 3 π /4 ( 2 V m ) 2 d θ ] V r m s 2 2 = V m 2 π [ 2 ∫ 0 π / 4 1 − cos 2 θ 2 d θ + 1 2 ( 3 π 4 − π 4 ) ] V_{rms2}^2 = \frac{V_m^2}{\pi} \left[ 2 \int_0^{\pi/4} \frac{1 - \cos 2\theta}{2} d\theta + \frac{1}{2} \left(\frac{3\pi}{4} - \frac{\pi}{4}\right) \right] V r m s 2 2 = π V m 2 [ 2 ∫ 0 π /4 2 1 − cos 2 θ d θ + 2 1 ( 4 3 π − 4 π ) ] V r m s 2 2 = V m 2 π [ ( θ − sin 2 θ 2 ) 0 π / 4 + π 4 ] = V m 2 π [ π 4 − 1 2 + π 4 ] = V m 2 π ( π 2 − 1 2 ) = V m 2 ( 1 2 − 1 2 π ) V_{rms2}^2 = \frac{V_m^2}{\pi} \left[ \left( \theta - \frac{\sin 2\theta}{2} \right)_0^{\pi/4} + \frac{\pi}{4} \right] = \frac{V_m^2}{\pi} \left[ \frac{\pi}{4} - \frac{1}{2} + \frac{\pi}{4} \right] = \frac{V_m^2}{\pi} \left( \frac{\pi}{2} - \frac{1}{2} \right) = V_m^2 \left( \frac{1}{2} - \frac{1}{2\pi} \right) V r m s 2 2 = π V m 2 [ ( θ − 2 sin 2 θ ) 0 π /4 + 4 π ] = π V m 2 [ 4 π − 2 1 + 4 π ] = π V m 2 ( 2 π − 2 1 ) = V m 2 ( 2 1 − 2 π 1 ) The ratio is:
Ratio = V r m s 1 V r m s 2 = V m / 2 V m 2 ( 0.5 − 0.5 / π ) = 1 2 0.5 ( 1 − 1 / π ) = 1 1 − 1 / π = π π − 1 ≈ 1.211 \text{Ratio} = \frac{V_{rms1}}{V_{rms2}} = \frac{V_m/\sqrt{2}}{\sqrt{V_m^2 (0.5 - 0.5/\pi)}} = \frac{1}{\sqrt{2} \sqrt{0.5(1 - 1/\pi)}} = \frac{1}{\sqrt{1 - 1/\pi}} = \sqrt{\frac{\pi}{\pi - 1}} \approx 1.211 Ratio = V r m s 2 V r m s 1 = V m 2 ( 0.5 − 0.5/ π ) V m / 2 = 2 0.5 ( 1 − 1/ π ) 1 = 1 − 1/ π 1 = π − 1 π ≈ 1.211 Understand the concept, then try another question Revisit Electric Circuits with concept notes, common mistakes and an original worked example before your next attempt.
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