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For the circuit shown in the figure, fin...

For the circuit shown in the figure, find the charge stored on capacitor in steady state.

A

`(RC)/(R+R_0)E`

B

`(RC)/(R_0)E-E_(0)`

C

zero

D

`(RC)/(R+R_0)(E-E_0)`

Text Solution

Verified by Experts

The correct Answer is:
D

`i=([E-E-0])/(R+R_0)`
Now,`V_a-V_b=(E-E_0)-iR_0`
`=(E-E_0)[-R_0/(R+R_0)]`
`(R(E-E))/(R+R_0)`
`:. q=C(V_a-V_b)=(CR(E-E_0))/(R+R_0)`
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