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 $220V,50Hz$. 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, $X_L=2\pi /L$
$=2\pi \times 50\times \left[\frac{200}{\pi }\times 10^{-3}\right]=20\Omega$
$X_C=\frac{1}{2\pi fC}=\frac{1\times \pi }{2\pi \times 50\times 10^{-3}}=10\Omega$
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 $\frac{\pi }{4}$.