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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 i

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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, meaning $X_C > X_L$. The phase angle is $\phi = 45^{\circ}$.Substituting the values: $	an 45^{\circ} = \frac{X_C - X_L}{R}$.Since $	an 45^{\circ} = 1$, we have $1 = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.Rearranging the terms: $R = \frac{1}{2\pi fC} - 2\pi fL$.$\frac{1}{2\pi fC} = 2\pi fL + R$.Therefore, $C = \frac{1}{2\pi f(2\pi fL + R)}$.

When $L$, $C$, and $R$ are connected in series with an $AC$ source of frequency $f$, the current leads the voltage by a phase angle of $45^{\circ}$. What is the value of $C$?

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