PYQs / GATE EE / 2019 / Set 1 / Q53 GATE EE 2019 Set 1 — Question 53 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 Medium Sinusoidal Steady-State Transient & AC Steady-State Electric Circuits Complex Power AC Power & Three-Phase
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Last updated 5 September 2026
Question The voltage across and the current through a load are expressed as follows
v ( t ) = − 170 sin ( 377 t − π 6 ) V v(t) = -170 \sin \left( 377t - \frac{\pi}{6} \right) \text{ V} v ( t ) = − 170 sin ( 377 t − 6 π ) V i ( t ) = 8 cos ( 377 t + π 6 ) A i(t) = 8 \cos \left( 377t + \frac{\pi}{6} \right) \text{ A} i ( t ) = 8 cos ( 377 t + 6 π ) A The average power in watts (round off to one decimal place) consumed by the load is
_____ .
Solution Given:
v ( t ) = − 170 sin ( 377 t − π 6 ) = 170 sin ( 377 t − π 6 + π ) = 170 sin ( 377 t + 5 π 6 ) v(t) = -170 \sin \left( 377t - \frac{\pi}{6} \right) = 170 \sin \left( 377t - \frac{\pi}{6} + \pi \right) = 170 \sin \left( 377t + \frac{5\pi}{6} \right) v ( t ) = − 170 sin ( 377 t − 6 π ) = 170 sin ( 377 t − 6 π + π ) = 170 sin ( 377 t + 6 5 π ) Converting to cosine form using
sin ( θ ) = cos ( θ − π / 2 ) \sin(\theta) = \cos(\theta - \pi/2) sin ( θ ) = cos ( θ − π /2 ) :
v ( t ) = 170 cos ( 377 t + 5 π 6 − π 2 ) = 170 cos ( 377 t + π 3 ) v(t) = 170 \cos \left( 377t + \frac{5\pi}{6} - \frac{\pi}{2} \right) = 170 \cos \left( 377t + \frac{\pi}{3} \right) v ( t ) = 170 cos ( 377 t + 6 5 π − 2 π ) = 170 cos ( 377 t + 3 π ) i ( t ) = 8 cos ( 377 t + π 6 ) i(t) = 8 \cos \left( 377t + \frac{\pi}{6} \right) i ( t ) = 8 cos ( 377 t + 6 π ) The average power
P P P is given by:
P = V r m s I r m s cos ( ϕ v − ϕ i ) P = V_{rms} I_{rms} \cos(\phi_v - \phi_i) P = V r m s I r m s cos ( ϕ v − ϕ i ) Where
V r m s = 170 2 V_{rms} = \frac{170}{\sqrt{2}} V r m s = 2 170 ,
I r m s = 8 2 I_{rms} = \frac{8}{\sqrt{2}} I r m s = 2 8 ,
ϕ v = π 3 = 60 ∘ \phi_v = \frac{\pi}{3} = 60^\circ ϕ v = 3 π = 6 0 ∘ , and
ϕ i = π 6 = 30 ∘ \phi_i = \frac{\pi}{6} = 30^\circ ϕ i = 6 π = 3 0 ∘ .
P = ( 170 2 ) ( 8 2 ) cos ( 60 ∘ − 30 ∘ ) = 170 × 8 2 cos ( 30 ∘ ) P = \left( \frac{170}{\sqrt{2}} \right) \left( \frac{8}{\sqrt{2}} \right) \cos(60^\circ - 30^\circ) = \frac{170 \times 8}{2} \cos(30^\circ) P = ( 2 170 ) ( 2 8 ) cos ( 6 0 ∘ − 3 0 ∘ ) = 2 170 × 8 cos ( 3 0 ∘ ) P = 680 × 3 2 = 340 3 ≈ 588.897 W P = 680 \times \frac{\sqrt{3}}{2} = 340\sqrt{3} \approx 588.897 \text{ W} P = 680 × 2 3 = 340 3 ≈ 588.897 W Rounding off to one decimal place, the average power is
588.9 588.9 588.9 W.
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Solution Given:
v ( t ) = − 170 sin ( 377 t − π 6 ) = 170 sin ( 377 t − π 6 + π ) = 170 sin ( 377 t + 5 π 6 ) v(t) = -170 \sin \left( 377t - \frac{\pi}{6} \right) = 170 \sin \left( 377t - \frac{\pi}{6} + \pi \right) = 170 \sin \left( 377t + \frac{5\pi}{6} \right) v ( t ) = − 170 sin ( 377 t − 6 π ) = 170 sin ( 377 t − 6 π + π ) = 170 sin ( 377 t + 6 5 π ) Converting to cosine form using
sin ( θ ) = cos ( θ − π / 2 ) \sin(\theta) = \cos(\theta - \pi/2) sin ( θ ) = cos ( θ − π /2 ) :
v ( t ) = 170 cos ( 377 t + 5 π 6 − π 2 ) = 170 cos ( 377 t + π 3 ) v(t) = 170 \cos \left( 377t + \frac{5\pi}{6} - \frac{\pi}{2} \right) = 170 \cos \left( 377t + \frac{\pi}{3} \right) v ( t ) = 170 cos ( 377 t + 6 5 π − 2 π ) = 170 cos ( 377 t + 3 π ) i ( t ) = 8 cos ( 377 t + π 6 ) i(t) = 8 \cos \left( 377t + \frac{\pi}{6} \right) i ( t ) = 8 cos ( 377 t + 6 π ) The average power
P P P is given by:
P = V r m s I r m s cos ( ϕ v − ϕ i ) P = V_{rms} I_{rms} \cos(\phi_v - \phi_i) P = V r m s I r m s cos ( ϕ v − ϕ i ) Where
V r m s = 170 2 V_{rms} = \frac{170}{\sqrt{2}} V r m s = 2 170 ,
I r m s = 8 2 I_{rms} = \frac{8}{\sqrt{2}} I r m s = 2 8 ,
ϕ v = π 3 = 60 ∘ \phi_v = \frac{\pi}{3} = 60^\circ ϕ v = 3 π = 6 0 ∘ , and
ϕ i = π 6 = 30 ∘ \phi_i = \frac{\pi}{6} = 30^\circ ϕ i = 6 π = 3 0 ∘ .
P = ( 170 2 ) ( 8 2 ) cos ( 60 ∘ − 30 ∘ ) = 170 × 8 2 cos ( 30 ∘ ) P = \left( \frac{170}{\sqrt{2}} \right) \left( \frac{8}{\sqrt{2}} \right) \cos(60^\circ - 30^\circ) = \frac{170 \times 8}{2} \cos(30^\circ) P = ( 2 170 ) ( 2 8 ) cos ( 6 0 ∘ − 3 0 ∘ ) = 2 170 × 8 cos ( 3 0 ∘ ) P = 680 × 3 2 = 340 3 ≈ 588.897 W P = 680 \times \frac{\sqrt{3}}{2} = 340\sqrt{3} \approx 588.897 \text{ W} P = 680 × 2 3 = 340 3 ≈ 588.897 W Rounding off to one decimal place, the average power is
588.9 588.9 588.9 W.
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