In the circuit shown key is closed at `t=0` with capacitor initially uncharged.

Options

A. Rate of increase of capacitors energy, just after the switch is closed is `(E^(2))/R`
B. Maximum power supplied by the cell is `(E^(2))/R`
C. At `t=RC In 2`, capacitor is charged up to half of the energy stored by it in steady state
D. At `t=RC In 2`, potential difference across capacitor is same that across resistor.

Answer

B, D

Explanation

`P=VI=EI=E.(E^(2))/R(1-e^((-t)/(RC)))`
`=P_(max)=(E^(2))/R`
Rate of increase of capacitor's energy `=V_(c)I=q/CI`
As `q=CE(1-e^(1/(RC)))`
`implies` At `t=RC In 2,q=(CE)/2`