In an L-C-R series circuit, an inductor 30 mH and a resistor 1Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45° is $\frac{1}{\mathrm{x}}\times {10}^{-3}\mathrm{F}$. Then, the value of x is …… .

Given,

Inductance, L = 30 mH

Resistance, R= 1$\mathrm{\Omega }$

Angular frequency, $\mathrm{\omega }$ = 300 rad/s

We know that in L-C-R circuit, $\tan \mathrm{\varphi }=\frac{{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}$

where, $\mathrm{\varphi }$ = phase angle = 45°

${\mathrm{X}}_{\mathrm{C}}=\text{ capacitive reactance }=\frac{1}{\mathrm{\omega C}}$

${\mathrm{X}}_{\mathrm{L}}=\text{ inductive reactance }=\mathrm{\omega L}$

$\begin{array}{l}\Rightarrow \tan {45}^{\circ }=\frac{{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\\ \Rightarrow {\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}=\mathrm{R} \left[\because \tan {45}^{\circ }=1\right]\\ \Rightarrow \frac{1}{\mathrm{\omega C}}-\mathrm{\omega L}=\mathrm{R}\Rightarrow \frac{1}{\mathrm{\omega C}}-300\times 30\times {10}^{-3}=1\\ \Rightarrow \frac{1}{\mathrm{\omega C}}=10\Rightarrow \mathrm{\omega C}=\frac{1}{10}\\ \Rightarrow \mathrm{C}=\frac{1}{10\mathrm{\omega }}\Rightarrow \mathrm{C}=\frac{1}{10\times 300}\\ \Rightarrow \mathrm{C}=\frac{1}{3}\times {10}^{-3}\mathrm{F} ...\left(\mathrm{i}\right)\end{array}$

According to question, the value of capacitance is $\frac{1}{\mathrm{x}}\times {10}^{-3}$. So, on comparing it with Eq. (i), we can say x = 3.