Question

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)`

Answer 1

Correct Answer - B
The peak value of the 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))`.