Question

In series LCR circuit the alternating emf 'e' and the current 'i' are given by the equations \(e=e_{0}\cos \omega t\), \(i=i_{0}\sin \left(\omega t+\frac{\pi }{6}\right)\). The average power dissipated in the circuit over a one complete cycle of A.C is (\(\cos 60=1/2\))

• A. \(\frac{e_{0}i_{0}}{2}\)
• B. \(\frac{e_{0}i_{0}}{4}\)
• C. \(\sqrt{\frac{3}{2}}e_{0}i_{0}\)
• D. Zero

Hint:

Convert the emf to a sine, find the phase difference with the current and use average power equals half of e0 i0 cos of the phase.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Average power over a cycle is \(P=\frac{e_{0}i_{0}}{2}\cos\phi\), where \(\phi\) is the phase difference between emf and current.

Step 2: Phase difference
\(e=e_{0}\cos\omega t=e_{0}\sin\left(\omega t+\frac{\pi}{2}\right)\). Current: \(i=i_{0}\sin\left(\omega t+\frac{\pi}{6}\right)\). So \(\phi=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}\).

Step 3: Calculation
\(P=\frac{e_{0}i_{0}}{2}\cos60^{\circ}=\frac{e_{0}i_{0}}{4}\).

Final Answer:
The average power is e0 i0 divided by 4. \[ \boxed{\text{(B) }\dfrac{e_{0}i_{0}}{4}} \]