Question:
In an AC circuit \(E = 50sin(500t)\), \(I = 600sin(500t+\frac{π}{3})mA\). What is the power dissipated in the circuit? [\(cos60^{\circ} = 0.5\)]
In an AC circuit \(E = 50sin(500t)\), \(I = 600sin(500t+\frac{π}{3})mA\). What is the power dissipated in the circuit? [\(cos60^{\circ} = 0.5\)]
Show Hint
Use $P=V_{rms}I_{rms}\cos\phi$ with phase $\frac\pi3$.
Updated On: Oct 1, 2026
- \(10\) W
- \(75\) W
- \(7.5\) W
- \(50\) W
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Peak values
\(E_0=50\) V and \(I_0=600\) mA \(=0.6\) A. The phase difference is \(\phi=\frac\pi3=60^{\circ}\).
Step 2: Average power
\(P=\frac{E_0I_0}{2}\cos\phi=\frac{50\times0.6}{2}\times0.5=7.5\) W. Option (C).
Final Answer:
The power dissipated is \(7.5\) W, option (C). \[ \boxed{\text{(C) }7.5\text{ W}} \]
\(E_0=50\) V and \(I_0=600\) mA \(=0.6\) A. The phase difference is \(\phi=\frac\pi3=60^{\circ}\).
Step 2: Average power
\(P=\frac{E_0I_0}{2}\cos\phi=\frac{50\times0.6}{2}\times0.5=7.5\) W. Option (C).
Final Answer:
The power dissipated is \(7.5\) W, option (C). \[ \boxed{\text{(C) }7.5\text{ W}} \]
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