# In an a.c. circuit the voltage applied is $\dpi{100} E=E_{0}\, sin\, \omega t.$ The resulting current in the circuit is $\dpi{100} I=I_{0}sin\left ( \omega t-\frac{\pi }{2} \right )$ .The power consumption in the circuit is given by Option 1) $P=\sqrt{2}E_{0}I_{0}\; \;$ Option 2) $\; P=\frac{E_{0}I_{0}}{\sqrt{2}}\; \;$ Option 3) $\; P=zero\; \;$ Option 4) $\; P=\frac{E_{0}I_{0}}{2}$

As we learnt in

Phase difference between voltage and current -

$\phi = 90^{\circ}\left ( or -\pi /2 \right )$

Given : $E=E_{0}\; sin\; \omega t$

$I=I_{0}\, sin\left ( \omega t-\frac{\pi }{2} \right )$

Since the phase difference $\dpi{100} (\phi )$  between voltage and current is $\dpi{100} \frac{\pi }{2}$ .

$\dpi{100} \therefore$  Power factor  $\dpi{100} cos\, \phi =cos\frac{\pi }{2}=0$

$\dpi{100} Power\; consumption\, =\, E_{rms}I_{rms}cos\phi =0$.

Option 1)

$P=\sqrt{2}E_{0}I_{0}\; \;$

This option is incorrect

Option 2)

$\; P=\frac{E_{0}I_{0}}{\sqrt{2}}\; \;$

This option is incorrect

Option 3)

$\; P=zero\; \;$

This option is correct

Option 4)

$\; P=\frac{E_{0}I_{0}}{2}$

This option is incorrect

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