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An inductor 200 mH, capacitor 500mu F an...

An inductor 200 mH, capacitor `500mu F` and resistor `10 Omega` are connected in series with a 100 V variable frequency ac source. What is the frequency at which the power factor of the circuit is unity?

A

`10.22 Hz`

B

`12.4 Hz`

C

`19.2 Hz`

D

`15.9 Hz`

Text Solution

Verified by Experts

The correct Answer is:
D
Here `L = 200 mH = 200xx10^(-3)H=0.2H`
`C=500 mu F=500xx10^(-6)=5xx10^(-4)F`
`R=10 Omega` and `V_("rms")=100 V`
Power factor `=cos phi =(R )/(Z)=1` (Given)
`therefore Z=R rArr sqrt(R^(2)+(X_(L)-X_(C ))^(2))=R`
`R^(2)+(X_(L)-X_(C ))^(2)=R^(2) rArr X_(L)-X_(C )=0`
or `X_(L)=X_(C )` (This is resonance condition)
`2pi upsilon L=(1)/(2pi upsilon C) rArr 4pi^(2)upsilon^(2)LC=1`
`therefore upsilon=(1)/(2pi sqrt(LC))=(1)/(2xx3.14sqrt(0.2xx5xx10^(-4))`
`=(1)/(2xx3.14xx10^(-2))=(100)/(6.28)=15.9 Hz`
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