Question

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

Answer 1

The correct option is A $\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}$ parallel with C

power factor $\varphi$ is defined by $\tan \varphi =\frac{\omega L-\frac{1}{\omega C}}{R}$
when a capacitance C' is added in parallel,
$\tan {\varphi }^{\prime }=\frac{\omega L-\frac{1}{\omega (C+C^{\prime })}}{R}$
$\varphi =0$
$\Rightarrow \omega L-\frac{1}{\omega (C+C^{\prime })}=0$
$\Rightarrow \omega L=\frac{1}{\omega (C+C^{\prime })}$
$\Rightarrow \omega C^{\prime }=\frac{1}{\omega L}-\omega C$
$C^{\prime }=\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}$