Question

A capacitance of $\left(\frac{10^{-3}}{2 \pi}\right) F$ and an inductance of $\left(\frac{100}{\pi}\right) mH$ and a resistance of $10 \,\Omega$ are connected in series with an $AC$ voltage source of $220 \,V , \,50 \,Hz$. The phase angle of the circuit is

• A. $60^{\circ}$
• B. $30^{\circ}$
• C. $45^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (C)

Capacitance, $C=\frac{10^{-3}}{2 \,\pi} F$
Inductance, $L=\left(\frac{100}{\pi}\right) mH =\frac{100}{\pi} \times 10^{-3} H$
Resistance, $R=10 \,\Omega$
Voltage of source $=220$ volt
Frequency of source, $f=50 \,Hz$
Phase angle $\phi$ is given by
$\tan \phi=\frac{X_{L}-X_{C}}{R}$
where, $X_{L}=$ inductive impedance $=\omega L$
$=2 \pi f L \,\,\,(\because \omega=2 \pi f)$
$ \therefore X_{L} =2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} $
$=10 \,\Omega $
and $X_{C}=$ Capactive impedance
$=\frac{1}{\omega C}=\frac{1}{2 \,\pi f C}$
$=\frac{1 \times 2 \pi}{2 \pi \times 50 \times 10^{-3}} \Omega=\frac{100}{5}=20 \,\Omega$
$\therefore $ Phase angle,
$\tan \phi=\frac{X_{C}-X_{L}}{R}=\frac{20-10}{10}=1$ or $\phi=45^{\circ}$