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

  • 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}

 

Answers (1)

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 (\phi )  between voltage and current is \frac{\pi }{2} .

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

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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