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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 AC is
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The average power in the circuit over one cycle of AC is
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Verified Answer
The correct answer is:
$\frac{E_0 I_0}{2} \cos \phi$
Key Idea: The power is defined as the rate at which work is being done in the circuit.
Power $=$ rate of work done in one complete cycle.
$\begin{aligned}
& P_{a \nu}=\frac{W}{T} \\
& P_{a \nu}=\frac{\left(E_0 I_0 \cos \phi\right) T / 2}{T} \\
& P_{a \nu}=\frac{E_0 I_0 \cos \phi}{2}
\end{aligned}$
where $\cos \phi$ is called the power factor of an AC circuit.
Power $=$ rate of work done in one complete cycle.
$\begin{aligned}
& P_{a \nu}=\frac{W}{T} \\
& P_{a \nu}=\frac{\left(E_0 I_0 \cos \phi\right) T / 2}{T} \\
& P_{a \nu}=\frac{E_0 I_0 \cos \phi}{2}
\end{aligned}$
where $\cos \phi$ is called the power factor of an AC circuit.
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