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In the circuit shown key is closed at t=...

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

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.

Text Solution

Verified by Experts

The correct Answer is:
B, D
`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`
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Knowledge Check

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