Single Choice

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 parallel with C and has a magnitued

A$$\frac {1-\omega^2LC}{\omega^2L}$$ parallel with C
Correct Answer
B$$\frac {1-\omega^2LC}{\omega^2L}$$ series with C
C$$\frac {C}{(\omega^2LC-1)}$$ parallel with C
D$$\frac {C}{(\omega^2LC-1)}$$

Solution

The impedance of circuit is given by
$$Z= \sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}$$
and the current lag voltage by $$\tan\phi==\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$$
For power factor to be one the current and voltage to be in same phase i.e. $$\phi$$ to be zero
adding capacitor of capacitance C' in series of C, the reactance will be
$$X_L-X_C=\omega L-\frac{1}{\omega (\frac{CC'}{C+C'})}$$
$$\omega L-\frac{1}{\omega (\frac{CC'}{C+C'})}=0$$
$$\frac{CC'}{C+C'}=\omega ^2L$$
$$\omega ^2LCC'=C+C'$$
$$C'=\frac{C}{\omega ^2LC-1}$$
adding capacitor of capacitance C' in parallel of C, the reactance will be
$$X_L-X_C=\omega L-\frac{1}{\omega (C+C')}$$
$$\omega L-\frac{1}{\omega (C+C')}=0$$
$$C'=\frac{1}{\omega^2 L}-C=\frac{1-\omega^2 LC}{\omega^2 L}$$
Hence correct option is A.

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