In an LCR series circuit,an inductor 30 , {mH | vedclass

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Question

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

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$3$

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(A) The phase angle $\phi$ in an $LCR$ 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 $\phi = -45^{\circ}$.Thus,$	an(-45^{\circ}) = \frac{X_C - X_L}{R} \implies -1 = \frac{X_C - X_L}{R}$.This gives $X_L - X_C = R$.Given $L = 30 \, {mH} = 0.03 \, {H}$,$R = 1 \, \Omega$,and $\omega = 300 \, {rad/s}$.$X_L = \omega L = 300 	imes 0.03 = 9 \, \Omega$.Substituting the values: $9 - X_C = 1 \implies X_C = 8 \, \Omega$.Since $X_C = \frac{1}{\omega C}$,we have $\frac{1}{300 	imes C} = 8$.$C = \frac{1}{300 	imes 8} = \frac{1}{2400} = \frac{1}{2.4} 	imes 10^{-3} \, {F}$.Wait,re-evaluating the phase condition: If current leads voltage,$\phi$ is negative,so $	an \phi = \frac{X_C - X_L}{R} = 	an(-45^{\circ}) = -1$.$X_L - X_C = R \implies 9 - X_C = 1 \implies X_C = 8$. This leads to $C = \frac{1}{2400} \approx 0.416 	imes 10^{-3} \, {F}$.If the question implies the magnitude of the phase difference is $45^{\circ}$ and current leads,then $X_C > X_L$ is incorrect; it should be $X_L - X_C = R$ if current lags. For current to lead,$X_C - X_L = R$ is not possible if $X_C $X_C - 9 = 1 \implies X_C = 10 \, \Omega$.$\frac{1}{300 	imes C} = 10 \implies C = \frac{1}{3000} = \frac{1}{3} 	imes 10^{-3} \, {F}$.Thus,$x = 3$.

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