In the steady state, the energy stored in the capacitor is:

Options

A. `(1)/(2)C(E_(1)+E_(2))^(2)`
B. `(1)/(2)C(E_(1)-E_(2))^(2)`
C. `(1)/(2)C((E_(1)R_(1)+E_(1)R_(2))/(r_(1)+r_(2)+R_(1)+R_(2)))^(2)`
D. `(1)/(2)C(E_(2)+(E_(1)R_(1))/(r_(1)+R_(1)+R_(2)))^(2)`

Answer

D

Explanation

When the capacitor plate acquire full charge `q_(0)` there will be no current in the capacitor arm. Applying kirchhoff's second law to the current carrying circuit
`i(R_(1)_R_(2))=E_(1)-ir_(1)` or `i=(E_(1))/(r_(1)+R_(1)+R_(2))`
Now `V_(a)-V_(b)=-iR_(1)=(E_(1)R_(2))/(r_(1)+R_(1)+R_(2))`
and `V_(C)=(q_(0))/(C)=E_(2)+iR_(1)=E_(2)+(E_(1)R_(1))/(r_(1)+R_(1)+R_(2))`
Now energy stored in the capacitor
`U=(1)/(2)CV_(c)^(2),=(1)/(2)C[E_(2)+(E_(1)R_(1))/(r_(1)+R_(1)+R_(2))]^(2)`