Question

In an a.c. circuit the voltage applied is $E=E_0\sin \omega t$. The resulting current in the circuit is $I=I_0\sin (\omega t-\frac{\pi }{2})$. The power consumption in the circuit is given by -

• A. $P=\frac{E_0{I}_0}{\sqrt{2}}$
• B. $P=\text{zero}$
• C. $P=\frac{E_0{I}_0}{2}$
• D. $P=\sqrt{2}{E}_0{I}_0$

Answer 1

Answer: (B)

$P=\text{zero}$

Average Power Consumed. $P_{\text{Average}}=VI\cos \varphi$ where $\varphi$ is the phase angle between the current and the voltage and where $V$ and $I$ are understood to be the effective or rms values of the voltage and current. The term $\cos \varphi$ is called the "power factor" for the circuit.
As in our case the phase difference is $\frac{\pi }{2}\&as\cos \left(\frac{\pi }{2}\right)=0$
So power consumed is 0.