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Question

Consider L,C,R circuit as shown in figure, with a.c. source of peak value V and angular frequency ω. Then the peak value of current through the ac source.
786198_5429a99f3e2547c487329d4d42bb785a.jpg

A
V1R2+(ωL1ωC)2
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B
V1R2+(ωC1ωL)2
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C
VR2+(ωL1ωC)2
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D
VRωCω2C2+R(ω2C21)2
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Solution

The correct option is B V1R2+(ωC1ωL)2
Xc=1jωc
XL=jωL
And R are in parallel,
Therefore, equivalent admittance of the circuit is
Y=1Xc+1XL+1R.
Y=1R+j(cω1Lω)
|Y|=1R2+(ωc1ωL)2
Current I=VY=V1R2+(ωc1ωL)2

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