JEE Main 2015 · 11 Apr · Shift 1 · Physics Q22

← JEE Main 2015 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 22 of 90

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 :

L, C and R in series across an AC source V = V₀ sin ωt.

series with C and has a magnitude

\dfrac{1-\omega^2LC}{\omega^2L}

series with C and has a magnitude

\dfrac C{(\omega^2LC-1)}

parallel with C and has a magnitude

\dfrac C{(\omega^2LC-1)}

parallel with C and has a magnitude

\dfrac{1-\omega^2LC}{\omega^2L}

Answer and worked solution

Official answer: D

The current leads, so $X_C>X_L$ . Unity power factor needs $\omega L=\frac1{\omega C_{eq}}$ , i.e. $C_{eq}=\frac1{\omega^2L}>C$ . So C' goes in parallel:
$C'=\frac1{\omega^2L}-C=\frac{1-\omega^2LC}{\omega^2L}.$

Topics:LCR circuits and resonance Series and parallel combinations

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 22