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 $\tan \phi==\frac{ X _{ L }- X _{ C }}{ R }=\frac{\omega L -\frac{1}{ m }}{ R }$
For power factor to be one the current and voltage to be in same phase i.e. $\phi$ to
be zero
adding capacitor of capacitance $C$ ' in series of $C$, the reactance will be
$X _{ L }- X _{ C }=\omega L -\frac{1}{\omega\left(\frac{C C^{\prime}}{ C + C }\right)}$
$\omega L -\frac{1}{\omega\left(\frac{C C}{C+C}\right)}=0$
$\frac{ CC ^{\prime}}{ C + C ^{\prime}}=\omega^{2} L$
$\omega^{2} LCC ^{\prime}= C + C ^{\prime}$
$C ^{\prime}=\frac{ C }{\omega^{2} LC -1}$
adding capacitor of capacitance $C$ ' in parallel of $C$, the reactance will be
$X _{ L }- X _{ C }=\omega L -\frac{1}{\omega\left( C + C ^{\prime}\right)}$
$\omega L -\frac{1}{\omega\left( C + C ^{\prime}\right)}=0$
$C ^{\prime}=\frac{1}{\omega^{2} L }- C =\frac{1-\omega^{2} LC }{\omega^{2} L }$