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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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Knowledge Check

  • In the circuit as shown in figure the switch is closed at t = 0 . At the instant of closing the switch

    A
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  • In the circuit shows in Fig the switch is closed at t = 0 .

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    At `t = 0,I_(1) = I_(2) = 0`
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    At any time `t, (I_(0))/(I_(2)) = (L_(2))/(L_(1))`
    C
    At any time t, `I_(1) + I_(2) = (epsilon)/(R )`
    D
    At `t = oo, I_(1)` and `I_(2)` are independent of `L_(1)` and `L_(2)`

  • In the circuit shown above, when then switch S is closed,

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    B
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    C
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