Question

In an AC circuit, E and I are given by \(E=150\,\sin (150t)\) V and \(I=150\,\sin \left(150t+\frac{\pi }{3}\right)\) A. The power dissipated in the circuit is
\(\{\cos 60^{\circ }=1/2\}\)

• A. \(106\ \text{W}\)
• B. \(150\ \text{W}\)
• C. \(5625\ \text{W}\)
• D. zero

Hint:

Average power is the product of the rms voltage and rms current times the cosine of the phase angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Average power in an AC circuit is \(P=E_{rms}I_{rms}\cos\phi\).

Step 2: Calculation
Peak values are \(E_{0}=150\ \text{V}\) and \(I_{0}=150\ \text{A}\), and the phase difference is \(\phi=\frac{\pi}{3}\). So \(P=\frac{E_{0}I_{0}}{2}\cos 60^{\circ}=\frac{150\times 150}{2}\times\frac{1}{2}\).

Step 3: Compare the options
\(P=5625\ \text{W}\), which is the third option.

Final Answer:
The power dissipated is \(5625\ \text{W}\). \[ \boxed{\text{(C) }5625\ \text{W}} \]