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 $220V,50Hz$. 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, $f=50Hz$
Phase angle $\varphi$ is given by
$\tan \varphi =\frac{X_L-X_C}{R}$
where, $X_L=$ inductive impedance $=\omega L$
$=2\pi fL(\because \omega =2\pi f)$
$\therefore X_L=2\pi \times 50\times \frac{100}{\pi }\times 10^{-3}$
$=10\Omega$
and $X_C=$ Capactive impedance
$=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
$=\frac{1\times 2\pi }{2\pi \times 50\times 10^{-3}}\Omega =\frac{100}{5}=20\Omega$
$\therefore$ Phase angle,
$\tan \varphi =\frac{X_C-X_L}{R}=\frac{20-10}{10}=1$ or $\varphi =45^{\circ }$