Question

An LCR series circuit with $100\Omega$ resistance is connected to an AC source of $200V$ and angular frequency $300rad/s$. 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

Answer: 2

If first case, $tan60^{\circ }=\frac{X_L}{R}=\frac{\omega L}{R}$
or $\sqrt{3}=\frac{300L}{100}$
$\therefore L=0.58H$
In second case, $tan60^{\circ }=\frac{X_C}{R}=\frac{1}{\omega CR}$
or $\sqrt{3}=\frac{1}{300C\times 100}$
$\therefore C=19.2\mu F$
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, $i=\frac{V}{Z}=\frac{200}{100}=2A$