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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

`2M`

B

`(I_(0))/(sqrt(2))`

C

`(I_(0))/(sqrt(3))`

D

`(I_(0))/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
The peak value of current is
`I_(0)=(V_(0))/(sqrt(R^(2)+1/(omega^(2)C^(2))))=(V_(0))/(sqrt(2)R)`
When the angular frequency is changed to `(omega)/(sqrt(3))` The new peak value is
`I_(0)^(.)=(V_(0))/(sqrt(R^(2)+3/(omega^(2)C^(2))))=(V_(0))/(sqrt(4R^(2)))=(V_(0))/(2R)`
`I_(0)^(.)=(I_(0))/(sqrt(2))`
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