In an a.c. circuit the voltage applied is $\mathrm{E}=\mathrm{E}_0 \sin \omega \mathrm{t}$. The resulting…

In an a.c. circuit the voltage applied is $\mathrm{E}=\mathrm{E}_0 \sin \omega \mathrm{t}$. The resulting current in the circuit is $\mathrm{I}=\mathrm{I}_0 \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$. The power consumption in the circuit is given by
  1. $\mathrm{P}=\frac{\mathrm{E}_0 \mathrm{I}_0}{\sqrt{2}}$
  2. $\mathrm{P}=$ zero
  3. $\mathrm{P}=\frac{\mathrm{E}_0 \mathrm{I}_0}{2}$
  4. $\mathrm{P} \sqrt{2} \mathrm{E}_0 \mathrm{I}_0$

Solution

$\cos \phi=0$ So power $=0$

Asked in: JEE Main 2007

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