For a series L-C-R circuit with R = 100 Ω, L = 0.5 mH and C = 0.1 pF connected across 220 V-50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is

According to question, there is a series L-C-R circuit.

Resistance, R = 100$\mathrm{\Omega }$

Capacitance, $\mathrm{C}=0.1\mathrm{pF}=0.1\times {10}^{-12}\mathrm{F}$

Inductance, $\mathrm{L}=0.5\mathrm{mH}=0.5\times {10}^{-3}\mathrm{H}$

Frequency, f = 50 Hz

Voltage, V = 220 V

$\begin{array}{c}\because \text{ Inductive reactance, }{\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}\\ =2\mathrm{\pi fL}\\ =2\mathrm{\pi }\times 50\times 0.5\times {10}^{-3}\\ \Rightarrow {\mathrm{X}}_{\mathrm{L}}=50\mathrm{\pi }\times {10}^{-3}\mathrm{\Omega }\end{array}$

$\because \text{ Capacitive reactance, }{\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{2\mathrm{\pi fC}}$

$\begin{array}{l}\Rightarrow {\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{\pi }\times 50\times {10}^{-12}\times 0.1}\\ \Rightarrow {\mathrm{X}}_{\mathrm{C}}=\frac{1}{100\mathrm{\pi }\times \frac{1}{10}\times {10}^{-12}}\\ \Rightarrow {\mathrm{X}}_{\mathrm{C}}=\frac{{10}^{11}}{\mathrm{\pi }}\mathrm{\Omega }\end{array}$

On comparing X_L with X_C, we can write

$\begin{array}{l}{\mathrm{X}}_{\mathrm{C}}>>{\mathrm{X}}_{\mathrm{L}}\\ \text{ and }\left|{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}\right|\text{>>}\mathrm{R}\end{array}$

which means that the circuit is predominantly capacitive circuit.