question_answer

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 : [JEE Main Online Paper (Held On 11 April 2015)]  

A)  series with C and has a magnitude\[\frac{1-{{\omega }^{2}}LC}{{{\omega }^{2}}L}.\]

B)  series with C and has a magnitude\[\frac{C}{({{\omega }^{2}}LC-1)}.\]

C)  parallel with C and has a magnitude\[\frac{C}{({{\omega }^{2}}LC-1)}.\]

D)   parallel with C and has a magnitude\[\frac{1-{{\omega }^{2}}LC}{{{\omega }^{2}}L}.\]

Correct Answer: D

Solution :

As current leads voltage thus Since power factor has to be made ?I? \[\therefore \]Effective capacitance has to be increased thus connecting in parallel. \[\because \]\[\cos \phi =1\] \[\therefore \]\[\phi =0\] \[i\omega L=\frac{i}{\omega \left( C+C' \right)}\] \[\therefore C+C'=\frac{1}{{{\omega }^{2}}L}\]AO I\[\therefore \]\[C'=\frac{1}{{{\omega }^{2}}L}-C\]