For the LCR circuit shown here,the current is | vedclass

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(C) In the given $LCR$ circuit,the current leads the voltage,which implies that the circuit is capacitive,i.e.,$X_C > X_L$ or $\frac{1}{\omega C} > \o

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parallel with $C$ and has a magnitude $\frac{(1 - \omega^2 LC)}{\omega^2 L}$

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(C) In the given $LCR$ circuit,the current leads the voltage,which implies that the circuit is capacitive,i.e.,$X_C > X_L$ or $\frac{1}{\omega C} > \omega L$. This means $\omega^2 LC To make the power factor unity,the circuit must be at resonance,where $X_L = X_{eq}$,where $X_{eq}$ is the equivalent capacitive reactance.Since the current leads the voltage,we need to decrease the total capacitive reactance to reach resonance. This is achieved by increasing the total capacitance of the circuit.When a capacitor $C'$ is connected in parallel with $C$,the equivalent capacitance becomes $C_{eq} = C + C'$.At resonance,the inductive reactance equals the capacitive reactance:$\omega L = \frac{1}{\omega (C + C')}$$\omega^2 L (C + C') = 1$$C + C' = \frac{1}{\omega^2 L}$$C' = \frac{1}{\omega^2 L} - C = \frac{1 - \omega^2 LC}{\omega^2 L}$Since $\omega^2 LC < 1$,the value of $C'$ is positive. Thus,the capacitor $C'$ must be connected in parallel with $C$.

For the $LCR$ circuit shown here,the current is observed to lead the applied voltage. An additional capacitor $C'$,when joined with the capacitor $C$ present in the circuit,makes the power factor of the circuit unity. The capacitor $C'$ must have been connected in

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