Question

For a series $LCR$ circuit with $R=100\Omega$, $L=0.5\text{mH}$ and $C=0.1\text{pF}$ connected across $220\text{V}-50\text{Hz AC}$ supply, the phase angle between current and supplied voltage and the nature of the circuit is :

• A. $\approx {90}^{\circ }$ , predominantly capacitive circuit
• B. $\approx {90}^{\circ }$, predominantly inductive circuit
• C. $0^{\circ }$ , resonance circuit
• D. $0^{\circ }$, resistive circuit

Answer 1

The correct option is A $\approx {90}^{\circ }$ , predominantly capacitive circuit

Given:
$R=100\Omega ;L=0.5\text{mH};C=0.1\text{pF}$
$V=220\text{V};f=50\text{Hz}$

Inductive reactance,

$X_L=\omega L=2\pi fL=50\pi \times {10}^{-3}\Omega$

Capacitive reactance,

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{{10}^{11}}{100\pi }=\frac{{10}^9}{\pi }\Omega$

From the given calculation, we get,

$|X_C-X_L|>>R\&X_C>>X_L$

$\Rightarrow X_C>>R$

So, this is a capacitive dominant circuit, and phasor diagram of this circuit is,


So, angle between voltage and current is $\approx {90}^{\circ }$.

Hence, option $\left(\text{D}\right)$ is correct.