Question

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

• A. 3
• B. 3.0
• C. 3.00

Answer 1

Answer: (A), (B), (C)

We know that,

$\tan \varphi =\frac{X_C-X_L}{R}$

$\Rightarrow \tan {45}^{\circ }=\frac{X_C-X_L}{R}$

$\Rightarrow X_C-X_L=R$

$\Rightarrow \frac{1}{\omega C}-\omega L=R$

$\Rightarrow \frac{1}{\omega C}-300\times 0.03=1$

$\Rightarrow \frac{1}{\omega C}=10$

$\Rightarrow C=\frac{1}{10\omega }=\frac{1}{10\times 300}$

$\Rightarrow C=\frac{1}{3}\times {10}^{-3}\text{F}=\frac{1}{x}\times {10}^{-3}\text{F}$

$\therefore x=3$