Question

In the LCR series circuit, an alternating e.m.f. \(e\) and current \(i\) are given by the equation \(e=100\,\sin (100t)V\) and \(i=100\,\sin \left(100t+\frac{\pi }{3}\right)mA\).
The average power dissipated in the circuit will be
\(\{\sin (\pi /3)=3^{0.5}/2,\ \cos (\pi /3)=1/2\}\)

• A. \(7\) W
• B. \(3.5\) W
• C. \(5\) W
• D. \(2.5\) W

Hint:

Average power is half of the peak emf times peak current times the cosine of the phase angle.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Average power \(P=\dfrac{1}{2}E_0I_0\cos\phi\), where \(\phi\) is the phase difference.

Step 2: Read the values
\(E_0=100\) V, \(I_0=100\) mA \(=0.1\) A, and \(\phi=\dfrac{\pi}{3}\) (the current leads the emf).

Step 3: Calculation
\(P=\dfrac{1}{2}\times100\times0.1\times\cos\dfrac{\pi}{3}=5\times\dfrac{1}{2}\).

Step 4: Result
\(P=2.5\) W.

Final Answer:
The average power dissipated is 2.5 W. \[ \boxed{\text{(D) }2.5\text{ W}} \]