In a circuit,L,C,and R are connected in serie | vedclass

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(A) In an $LCR$ series circuit,the phase difference $\phi$ between voltage and current is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current

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

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(A) In an $LCR$ series circuit,the phase difference $\phi$ between voltage and current is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current leads the voltage,the phase angle is $\phi = -45^{\circ}$.Therefore,$	an(-45^{\circ}) = \frac{X_C - X_L}{R} = -1$.This implies $X_L - X_C = R$,or $X_C - X_L = -R$.However,the standard convention for current leading is $	an \phi = \frac{X_L - X_C}{R}$ where $\phi$ is the phase of voltage relative to current. If current leads,$\phi = -45^{\circ}$,so $	an(-45^{\circ}) = -1 = \frac{X_L - X_C}{R}$,which gives $X_C - X_L = R$.Substituting $X_C = \frac{1}{2 \pi f C}$ and $X_L = 2 \pi f L$:$\frac{1}{2 \pi f C} - 2 \pi f L = R$$\frac{1}{2 \pi f C} = R + 2 \pi f L$$C = \frac{1}{2 \pi f (R + 2 \pi f L)}$.

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

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