Question

An LCR series circuit with $100\Omega$ resistance is connected to an AC source of $200 \,V$ and angular frequency $300\, rad/s$. When only the capacitance is removed, the current lags behind the voltage by $60^{\circ}$ . When only the inductance is removed, the current leads the voltage by $60^{\circ}$. Calculate the current (in ampere) in the LCR circuit.

Answer 1

If first case, $tan \,60^{\circ} = \frac{X_L}{R} = \frac{\omega L}{R}$
or $\sqrt{3} = \frac{300 L}{100}$
$\therefore L = 0.58 \,H$
In second case, $tan \,60^{\circ} = \frac{X_C}{R} = \frac{1}{\omega CR}$
or $\sqrt{3} = \frac{1}{300 C \times 100}$
$\therefore C = 19.2\,\mu F$
The impedance of the circuit is
$Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C)^2}}$
$ = \sqrt{100^2 + ( 300 \times 0.58 - \frac{1}{300 \times 19.2 \times 10^{-6}})^2}$
$ = 100\,\Omega$
Current, $ i = \frac{V}{Z} = \frac{200}{100} = 2\,A$