Question

An inductance of $(\frac{200}{\pi }) mH$, a capacitance of $\frac{10^{-3}}{\pi}F$ and a resistance of and a resistance of $10 \,\Omega$ are connected in series with an $AC$ source of $220 V, 50\, Hz$. The phase angle of thecircuit is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (B)

The phase angle $(\theta)$ between $I$ and $V$ is given by
$\tan \,\theta=\frac{X_{L}-X_{C}}{R}\,\,\,\,\dots(i)$
where, $X_{L}=2 \pi / L$
$=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 $
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 \tan \theta =\tan \frac{\pi}{4} $
$\therefore \theta =\frac{\pi}{4}$
The phase angle of the circuit is $\frac{\pi}{4}$.