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

Given 
Inductance, L = 30mH 
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^o 
X_C = capacitive reactance = $\frac{1}{\mathrm{\omega C}}$
X_L = inductive reactance = $\mathrm{\omega }$L
$\begin{array}{l}\Rightarrow \tan {45}^{\circ }=\frac{{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\\ \Rightarrow {\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}=\mathrm{R}\left[\because \tan {45}^{\circ }=1\right]\\ \Rightarrow \frac{1}{\mathrm{\omega C}}-\mathrm{\omega L}=\mathrm{R}\\ \Rightarrow \frac{1}{\mathrm{\omega C}}-300\times 30\times {10}^{-3}=1\\ \Rightarrow \frac{1}{\mathrm{\omega C}}=10\Rightarrow \mathrm{\omega C}=\frac{1}{10}\\ \Rightarrow \mathrm{C}=\frac{1}{10\mathrm{\omega }}\Rightarrow \mathrm{C}=\frac{1}{10\times 300}\\ \Rightarrow \frac{1}{3}\times {10}^{-3}\mathrm{F}.=0.33\times {10}^{-3}.\end{array}$