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A resistor of $500 \Omega$, an inductance of $0.5 \mathrm{H}$ are in series with an AC which is given by $V=100 \sqrt{2} \sin (1000 t)$. The power factor of the combination is
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The correct answer is:
$\frac{1}{\sqrt{2}}$
Voltage $V=100 \sqrt{2} \sin (1000 t)$
and $\quad \mathrm{V}=\mathrm{V}_{0} \sin \omega t$
On comparing $\omega=1000$
Inductive reactance
$$
\begin{aligned}
X_{\mathrm{L}} &=\omega \mathrm{L}=1000 \times 0.5=500 \Omega \\
\therefore \quad \cos \phi &=\frac{500}{\sqrt{(500)^{2}+(500)^{2}}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$
and $\quad \mathrm{V}=\mathrm{V}_{0} \sin \omega t$
On comparing $\omega=1000$
Inductive reactance
$$
\begin{aligned}
X_{\mathrm{L}} &=\omega \mathrm{L}=1000 \times 0.5=500 \Omega \\
\therefore \quad \cos \phi &=\frac{500}{\sqrt{(500)^{2}+(500)^{2}}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$
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