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An AC source producing emf epsilon = e...

An AC source producing `emf`
`epsilon = epsilon_0 [cos(100 pi s^(-1)) t + cos (500 pi s^(-1))t]`
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be
`I = i_1 cos [(100 pi s^(-1) t + varphi_1] + i_2 cos [(500 pi s^(-1))t +phi_2]`.

A

`i_(1)gti_(2)`

B

`i_(1) =i_(2)`

C

`i_(1)lt i_(2)`

D

none

Text Solution

Verified by Experts

The correct Answer is:
C

Impedance of `RC` circuit
`Z = sqrt(R^(2) + ((1)/(omegaC))^(2)`
`i_(1) = (E_(0))/(sqrt(R^(2) + ((1)/(100piC))^(2))) i_(2) = (E_(0))/(sqrt(R^(2) + ((1)/(500 pi C))^(2)))`
`i_(1) lt i_(2)` .
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