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In steady state, find energy stored in the capacitor -
image
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

1 Answer

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Best answer
Correct Answer - B
image
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)`

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