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An alternating voltage $=200 \sin 100 t$ is applied to a series combination of $R=30 \Omega$ and an inductor of $400 \mathrm{mH}$. The power factor of the circuit is,
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Verified Answer
The correct answer is:
0.6
As we know,
$$
\begin{aligned}
\text { power factor } & =\frac{R}{Z} \\
& =\frac{R}{\sqrt{R^2+\omega L^2}}
\end{aligned}
$$
Given $\omega=100, L=400 \mathrm{mH}$
$$
\begin{aligned}
& =\frac{30}{\sqrt{(30)^2+\left(100 \times 400 \times 10^{-3}\right)^2}} \\
& =\frac{30}{\sqrt{900+160000 \times 10^{-3 \times 2}}} \\
& =\frac{30}{50}=0.6
\end{aligned}
$$
$$
\begin{aligned}
\text { power factor } & =\frac{R}{Z} \\
& =\frac{R}{\sqrt{R^2+\omega L^2}}
\end{aligned}
$$
Given $\omega=100, L=400 \mathrm{mH}$
$$
\begin{aligned}
& =\frac{30}{\sqrt{(30)^2+\left(100 \times 400 \times 10^{-3}\right)^2}} \\
& =\frac{30}{\sqrt{900+160000 \times 10^{-3 \times 2}}} \\
& =\frac{30}{50}=0.6
\end{aligned}
$$
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