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An inductance of $\left(\frac{200}{\pi}\right) \mathrm{mH}$, a capacitance of $\left(\frac{10^{-3}}{\pi}\right) \mathrm{F}$ and a resistance of $10 \Omega$ are connected in series with an AC source $220 \mathrm{~V}, 50 \mathrm{~Hz}$. The phase angle of the circuit is
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
The phase angle $(\theta)$ between $I$ and $V$ is given by
$$
\tan \theta=\frac{X_{L}-X_{C}}{R}
$$
where, $X_{L}=2 \pi f L$
$$
\begin{aligned}
&=2 \pi \times 50 \times\left[\frac{200}{\pi} \times 10^{-3}\right]=20 \Omega \\
X_{C} &=\frac{1}{2 \pi f C}=\frac{1 \times \pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega
\end{aligned}
$$
and $R=10 \Omega$
Substituting values of $X_{L}, X_{C}$ and $R$ is Eq. (i), we get
$\tan \theta=\frac{20-10}{10}=1$ $\Rightarrow \quad \tan \theta=\tan \frac{\pi}{4}$ $\therefore \quad \theta=\frac{\pi}{4}$ The phase angle of the circuit is $\frac{\pi}{4} .$
$$
\tan \theta=\frac{X_{L}-X_{C}}{R}
$$
where, $X_{L}=2 \pi f L$
$$
\begin{aligned}
&=2 \pi \times 50 \times\left[\frac{200}{\pi} \times 10^{-3}\right]=20 \Omega \\
X_{C} &=\frac{1}{2 \pi f C}=\frac{1 \times \pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega
\end{aligned}
$$
and $R=10 \Omega$
Substituting values of $X_{L}, X_{C}$ and $R$ is Eq. (i), we get
$\tan \theta=\frac{20-10}{10}=1$ $\Rightarrow \quad \tan \theta=\tan \frac{\pi}{4}$ $\therefore \quad \theta=\frac{\pi}{4}$ The phase angle of the circuit is $\frac{\pi}{4} .$
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