Question 141 of 190

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:

AParallel with C and has a magnitude $\dfrac{1-{\omega }^{2}LC}{{\omega }^{2}L}$BSeries with C and has a magnitude $\dfrac{1-{\omega }^{2}LC}{{\omega }^{2}L}$CSeries with C and has a magnitude $\dfrac{C}{\left({\omega }^{2}LC-1\right)}$DParallel with C and has a magnitude $\dfrac{C}{\left({\omega }^{2}LC-1\right)}$

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Answer: (A) Parallel with C and has a magnitude $\dfrac{1-{\omega }^{2}LC}{{\omega }^{2}L}$

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 ${C}^{'}$ should be connected in parallel. So in that condition ${X}_{L}={X}_{C}$

$\Rightarrow \omega L=\dfrac{1}{\omega {C}_{eq}}\Rightarrow {C}_{eq}=\dfrac{1}{{\omega }^{2}L}$

$\Rightarrow C+{C}^{'}=\dfrac{1}{{\omega }^{2}L}\Rightarrow {C}^{'}=\dfrac{1}{{\omega }^{2}L}-C$

$\Rightarrow {C}^{'}=\dfrac{1-{\omega }^{2}LC}{{\omega }^{2}L}$