| 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 11-04-2015] |
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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 :
| [d] As current leads voltage thus |
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| Since power factor has to be made I |
| \[\therefore \]Effective capacitance has to be increased thus connecting in parallel. |
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| \[\because \]\[\cos \phi =1\] \[\therefore \]\[\phi =0\] |
| \[i\omega L=\frac{i}{\omega \left( C+C' \right)}\] \[\therefore C+C'=\frac{1}{{{\omega }^{2}}L}\]AO I\[\therefore \]\[C'=\frac{1}{{{\omega }^{2}}L}-C\] |
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