Question

For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor $C^{\prime }$, when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor $C^{\prime }$, must have been connected in parallel with C and has a magnitued

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

Answer 1

Answer: (A)

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