For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional…

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:

1. Parallel with C and has a magnitude $\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}$
2. Series with C and has a magnitude $\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}$
3. Series with C and has a magnitude $\frac{C}{\left({\omega }^{2}LC-1\right)}$
4. Parallel with C and has a magnitude $\frac{C}{\left({\omega }^{2}LC-1\right)}$

Solution

Current is leading,  it means inductive reactance is more than capacitive reactance

For making power factor to be unity the value of C is to be increased.

$\Rightarrow$ another capacitor $C^{\prime }$ should be connected in parallel. So in that condition $X_L=X_C$

$\Rightarrow \omega L=\frac{1}{\omega C_{eq}}\Rightarrow C_{eq}=\frac{1}{{\omega }^{2}L}$

$\Rightarrow C+C^{\prime }=\frac{1}{{\omega }^{2}L}\Rightarrow C^{\prime }=\frac{1}{{\omega }^{2}L}-C$

$\Rightarrow C^{\prime }=\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}$

Asked in: JEE Main 2015 (11 Apr Online)

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