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Average power from given sinusoidal voltage and current with a phase difference

This type of question, asked 6 times: 2 May 2023 S1, 3 May 2023 S1, 16 May 2023 S1, 9 May 2024 S1, 11 Apr 2026 S1, 17 Apr 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    2 May 2023 S1

    23
    In an A.C. circuit, the potential difference VV and current II are given respectively by V=100sin⁡(100t)V = 100\sin(100t) V, I=100sin⁡(100t+π3)I = 100\sin\left(100t + \frac{\pi}{3}\right) mA. The power dissipated in the circuit will be [Given cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}]

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  2. Question 2

    3 May 2023 S1

    26
    In an AC circuit E=200sin⁡(50t)E=200\sin(50t) volt and I=100sin⁡ ⁣(50t+π3)I=100\sin\!\left(50t+\frac{\pi}{3}\right) mA. The power dissipated in the circuit is [sin⁡30∘=0.5, sin⁡60∘=32\sin30^\circ=0.5,\ \sin60^\circ=\frac{\sqrt{3}}{2}]

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  3. Question 3

    16 May 2023 S1

    31
    In LCR series circuit, an alternating e.m.f. 'e' and current 'i' are given by equations e=160sin⁡(100t)e = 160\sin(100t) V and i=250sin⁡ ⁣(100t+π3)i = 250\sin\!\left(100t + \frac{\pi}{3}\right) mA. The average power dissipated in the circuit is

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  4. Question 4

    9 May 2024 S1

    47
    In an AC circuit, the current is i=5sin⁡(100t−π2) Ai=5\sin\left(100t-\frac{\pi}{2}\right)\,A & voltage is e=200sin⁡(100t)e=200\sin(100t) volt. Power consumption in the circuit is (cos⁡90∘=0\cos90^\circ=0)

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  5. Question 5

    11 Apr 2026 S1

    137
    In an AC circuit E=50sin⁡(500t),I=600sin⁡(500t+π3)mAE = 50\sin(500t), I = 600\sin\left(500t + \dfrac{\pi}{3}\right)mA. What is the power dissipated in the circuit? [cos⁡60∘=0.5][\cos 60^\circ = 0.5]

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  6. Question 6

    17 Apr 2026 S2

    139
    In a series LCR circuit alternating e.m.f. and current are given by the equations V=V0sin⁡(ωt)V = V_0\sin(\omega t) and I=I0sin⁡(ωt+π3)I = I_0\sin\left(\omega t + \dfrac{\pi}{3}\right) respectively. The average power dissipated in the circuit over one cycle of a.c. is (cos⁡60∘=0.5)(\cos 60^\circ = 0.5)

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