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An AC voltage source V=V0siomegatis conn...

An `AC` voltage source `V=V_0siomegat`is connected across resistance `R` and capacitance `C` as shown in figureure. It is given that `R=1/omegaC`. The peak current is `I_0`. If the angular frequency of the voltage source is changed to `omega/sqrt3,` then the new peak current in the circuit is
.

A

`I_0/2`

B

`I_0/sqrt(2)`

C

`I_0/sqrt(3)`

D

`I_0/3`

Text Solution

Verified by Experts

The correct Answer is:
B
`R=1/(omegaC)=X_C`
`:. Z=sqrt(R^2+X_C^2)=sqrt(2)R` (as `X_C=R`)
`I_0=V_0/z=V_0/sqrt2R`....(i)
When omega becomes `1/sqrt(3)` times `X_C` will become `sqrt(3)` times or `sqrt(3)R`.
`Z^'=sqrt((R^2)+(sqrt3R)^2)=2R`
`I_0'=V_0/Z'=V_0/(2R)=I_0/sqrt(2)`
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