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In the circuit as shown in figure the sw...

In the circuit as shown in figure the switch is closed at `t = 0`.

A long time after closing the switch

A

voltage drop across the capacitor is `E`

B

curent through the battery is `E//R_(1) +R_(2)`

C

enegru stores in the capacitor is
`(1)/(2)C((R_(2)E)/(R_(1)+R_(2)))^(2)`

D

current through the capacitor becomes zero

Text Solution

Verified by Experts

The correct Answer is:
B, C, D
Finally capacitor acts as open circuit
(A) Voltage drop across capacitor are
`V_(C ) = (R_(2)E)/(R_(1)+R_(2))`
(B) `I =(E)/(R_(1)+R_(2)) (C ) U = (1)/(2) C ((R_(2)E)/(R_(1)+R_(2)))^(2)`
(D) `I_(C ) = 0`
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