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\[\therefore \]\[C'=\frac{1}{{{\omega }^{2}}L}-C\]You need to login to perform this action.
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