In an LCR series circuit, an inductor

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Question:

In an LCR series circuit, an inductor $30 \mathrm{mH}$ and a resistor $1 \Omega$ are connected to an AC source of angular frequency $300 \mathrm{rad} / \mathrm{s}$. The value of capacitance for which, the current leads the voltage

by $45^{\circ}$ is $\frac{1}{x} \times 10^{-3} \mathrm{~F}$. Then the value of $x$ is______.

Solution:

$\tan \phi=\frac{\mathrm{x}_{\mathrm{C}}-\mathrm{x}_{\mathrm{L}}}{\mathrm{R}}$

$\tan 45=\frac{x_{C}-x_{L}}{R}$

$x_{C}-x_{L}=R$

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

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

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

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

$\mathrm{C}=\frac{1}{3} \times 10^{-3}$

$X=3$

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