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The circuit shown in the figure consists...

The circuit shown in the figure consists of a battery of emf `epsilon = 10 V` , a capacitor of capacitance `C = 1.0 mu F` and three resistors of value `R_(1) = 2 Omega, R_(2) = 2 Omega` and `R_(3) = 1 Omega`, Initially the capacitor is completely uncharged and the switch `S` is open. The switch `S` is closed at `t = 0`. Then

A

the current through resistor `R_(3)` at the moment the switched closed is zero

B

the current through resistro `R_(3)` a long time after the switched closed is `5 A`

C

the ratio of current through `R_(1)` and `R_(2)` is always constant

D

all of these

Text Solution

Verified by Experts

The correct Answer is:
D
(d) At `t = `, the equivalent circuit is
`I = (epsilon)/(R_(3) + (R_(1) R_(2))/(R_(1) + R_(2))) = (10)/(1 + ((2)(2))/(2 + 2)) = (10)/(1 + 1) = 5 A`
Also, `I_(R_(1)) R_(1) = I_(R_(2)) R_(2) implies (I_(R_(1)))/(I_(R_(2))) = (R_(2))/(R_(1))`
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