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(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current leads the voltage,the circuit is

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$\frac{1}{2\pi f(2\pi fL + R)}$

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(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current leads the voltage,the circuit is capacitive,and the phase angle is $\phi = -45^o$ (or we use the magnitude of the lead as $	an(45^o) = \frac{X_C - X_L}{R}$).Substituting the values: $	an 45^o = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.Since $	an 45^o = 1$,we have $1 = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.$R = \frac{1}{2\pi fC} - 2\pi fL$.$\frac{1}{2\pi fC} = 2\pi fL + R$.Therefore,$C = \frac{1}{2\pi f(2\pi fL + R)}$.

In a circuit,$L, C$ and $R$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^o$. The value of $C$ is

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