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In an A.C. circuit the emf ($e$) and the current ($i$) at any instant are given respectively by :
$e = E_0 \sin\omega t$
$i = I_0 \sin(\omega t - \phi)$
The average power in the circuit over one cycle of A.C. is -
$e = E_0 \sin\omega t$
$i = I_0 \sin(\omega t - \phi)$
The average power in the circuit over one cycle of A.C. is -
A
$\dfrac{E_0 I_0}{2}\cos\phi$
B
$E_0 I_0$
C
$\dfrac{E_0 I_0}{2}$
D
$\dfrac{E_0 I_0}{2}\sin\phi$
Detailed Solution
$e = E_0\sin\omega t$ and $i = I_0\sin(\omega t - \phi)$; the phase difference between emf and current is $\phi$.
Average power over one cycle: $P_{av} = E_{rms}\,I_{rms}\cos\phi$
$E_{rms} = \dfrac{E_0}{\sqrt{2}}$ and $I_{rms} = \dfrac{I_0}{\sqrt{2}}$
$P_{av} = \dfrac{E_0}{\sqrt{2}} \cdot \dfrac{I_0}{\sqrt{2}}\cos\phi$
$P_{av} = \dfrac{E_0 I_0}{2}\cos\phi$
Here $\cos\phi$ is the power factor of the circuit.
Average power over one cycle: $P_{av} = E_{rms}\,I_{rms}\cos\phi$
$E_{rms} = \dfrac{E_0}{\sqrt{2}}$ and $I_{rms} = \dfrac{I_0}{\sqrt{2}}$
$P_{av} = \dfrac{E_0}{\sqrt{2}} \cdot \dfrac{I_0}{\sqrt{2}}\cos\phi$
$P_{av} = \dfrac{E_0 I_0}{2}\cos\phi$
Here $\cos\phi$ is the power factor of the circuit.