In steady state, find energy stored in the capacitor -

Options

A. `(1)/(2)C[(ER_(1))/(r+R_(1)+R_(2))]^(2)`
B. `(1)/(2)C[E_(0)+((ER_(1))/(r+R_(1)+R_(2))).R_(1)]^(2)`
C. `(1)/(2)CE_(0)^(2)`
D. none of the above

Answer

B

Explanation

In steady state current in capacitor branch is zero hence `r_(0)` can be removed
Voltage across `R_(1):` `IR_(1)=(ER_(1))/(r+R_(1)+R_(2))`
`:.` Voltage across capacitor `=(ER_(1))/(r+R_(1)+R_(2))+E_(0)`
`:.` Energy stored in capacitor `=(1)/(2)C(E_(0)+(ER_(1))/(r+R_(1)+R_(2)))^(2)`