In an AC circuit, E and I are given by E = 150 (150t) V and I = 150 (150t + (π)/(3)) A. The power dissipated in the circuit is \ (60)^ ^ = 1/2\

Correct answer: C. 5625\ W The given equations are E = 150 (150t) and I = 150 (150t + (π)/(3)). The peak voltage E_0 = 150 V and the peak current I_0 = 150 A. The phase difference between voltage and current is φ = (π)/(3) =60^ ^ . The average power dissipated in the circuit is given by P = E_ rms I_ rms φ . P = E_0 2 I_0 2 φ = (E_0 I_0)/(2) φ Substituting the values: P = (150 × 150)/(2) (60^ ^ ) P = (22500)/(2) × (1)/(2) = (22500)/(4) =5625 W. Answer: 5625\ W

MHT CET 202624 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 24 April · Evening Shift · Physics · Alternating Current

In an AC circuit, E and I are given by E = 150 (150t) V and I = 150...

In an AC circuit, E and I are given by

E = 150\sin(150t)

V and

I = 150\sin \left(150t + \dfrac{\pi}{3}\right) A. The power dissipated in the circuit is<br>\{\cos(60)^{}^{\circ} = 1/2\}

Options

A $106\ \text{W}$
B $150\ \text{V}$
C $5625\ \text{W}$
Dzero

Correct answer

C. $5625\ \text{W}$

Step-by-step solution

The given equations are

E = 150\sin(150t)

and

I = 150\sin \left(150t + \dfrac{\pi}{3}\right).<br><br>The peak voltage

E_0 = 150

V and the peak current

I_0 = 150

A.<br><br>The phase difference between voltage and current is

\phi =

\dfrac{\pi}{3} =60^{}^{\circ}

.

The average power dissipated in the circuit is given by

P = E_{rms} I_{rms} \cos\phi

P = \dfrac{E_0}{\sqrt{2}} \dfrac{I_0}{\sqrt{2}} \cos\phi = \dfrac{E_0 I_0}{2} \cos\phi<br><br>Substituting the values:<br><br>

P =

\dfrac{150 \times 150}{2} \cos(60^{}^{\circ})

P = \dfrac{22500}{2} \times \dfrac{1}{2} = \dfrac{22500}{4} =5625

W.

Answer:

5625\ \text{W}

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