In an a.c. circuit the voltage applied is $E=E_{0} \sin \omega t$. The resulting current in the circuit is $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$. The power consumption in the circuit is given by

(A) $P=\frac{E_{0} I_{0}}{\sqrt{2}}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Alternating Current

In an a.c. circuit the voltage applied is

E = E_{0} sin ω t

. The resulting current in the circuit is

I = I_{0} sin (ω t - \frac{π}{2})

. The power consumption in the circuit is given by

Answer: (B)

Here the current lags the applied voltage by a quarter cycle, so the phase angle is

\frac{π}{2}

. The average power in an a.c. circuit is

P = E_{rms} I_{rms} cos ϕ

, and with

cos \frac{π}{2} = 0

this average power is zero. That is the mark of a purely reactive element, which stores and returns energy each cycle but consumes none on average.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option