If a current $I$ given by $I = I_0 \sin(\omega t - \frac{\pi}{2})$ flows in an $AC$ circuit across which an $AC$ potential of $E = E_0 \sin(\omega t)$ has been applied,then the power consumption $P$ in the circuit will be
- A$P = \frac{E_0 I_0}{\sqrt{2}}$
- B$P = \sqrt{2} E_0 I_0$
- C$P = \frac{E_0 I_0}{2}$
- D$P = 0$
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View SolutionAn e.m.f. $e=E_0 \cos \omega t$ is applied to a circuit containing $L, C$ and $R$ in series where $X_L=3 R$ and $X_C=R$. The average power dissipated in the circuit is
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