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In a circuit $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^{\circ}$. The value of $\mathrm{C}$ is:
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Verified Answer
The correct answer is:
$\frac{1}{2 \pi f(2 \pi f L+R)}$
$$
\begin{aligned}
\tan \phi & =\frac{X_C-X_L}{R} \\
\tan \left(\frac{\pi}{4}\right) & =\frac{\frac{1}{\omega C}-\omega L}{R} \\
\Rightarrow \quad R & =\frac{1}{\omega C}-\omega L \\
(R+2 \pi f L) & =\frac{2}{2 \pi f C} \\
C & =\frac{1}{2 \pi f(R+2 \pi f L)}
\end{aligned}
$$
\begin{aligned}
\tan \phi & =\frac{X_C-X_L}{R} \\
\tan \left(\frac{\pi}{4}\right) & =\frac{\frac{1}{\omega C}-\omega L}{R} \\
\Rightarrow \quad R & =\frac{1}{\omega C}-\omega L \\
(R+2 \pi f L) & =\frac{2}{2 \pi f C} \\
C & =\frac{1}{2 \pi f(R+2 \pi f L)}
\end{aligned}
$$
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