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Question:
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In an AC circuit the emf $(e)$ and the current (i) at any instant are given respectively by
$$
\begin{array}{l}
e=E_{0} \sin \omega t \\
i=I_{0} \sin (\omega t-\phi)
\end{array}
$$
The average power in the circuit over one cycle of AC is
Options:
$$
\begin{array}{l}
e=E_{0} \sin \omega t \\
i=I_{0} \sin (\omega t-\phi)
\end{array}
$$
The average power in the circuit over one cycle of AC is
Solution:
1805 Upvotes
Verified Answer
The correct answer is:
$\frac{E_{0} I_{0}}{2} \cos \phi$
Power $=$ rate of work done in one complete cycle.
Or
$$
P_{\mathrm{av}}=\frac{W}{T}
$$
Or
$$
P_{\mathrm{av}}=\frac{\left(E_{0} I_{0} \cos \phi\right) T / 2}{T}
$$
or
$$
P_{\mathrm{av}}=\frac{E_{0} I_{0} \cos \phi}{2}
$$
where $\cos \phi$ is called the power factor of an $\mathrm{AC}$ circuit.
Or
$$
P_{\mathrm{av}}=\frac{W}{T}
$$
Or
$$
P_{\mathrm{av}}=\frac{\left(E_{0} I_{0} \cos \phi\right) T / 2}{T}
$$
or
$$
P_{\mathrm{av}}=\frac{E_{0} I_{0} \cos \phi}{2}
$$
where $\cos \phi$ is called the power factor of an $\mathrm{AC}$ circuit.
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