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In the circuit shows in Fig. kay `K` is closed at `t = 0`, the current through the key at the instant `t = 10^(-3) In 2 s` is
image
A. `2 A`
B. `3.5 A`
C. `2.5 A`
D. `0`

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Correct Answer - C
Current in branches containing `L` and `R` will flow independently
`I_(1) = (20)/(10)(1 - e^(-(t)/(5xx10^(-4)))) = (3)/(2) = 1.5 A`
image
`I_(2) = (20)/(10) e^(-(t)/(10^(-3))) = 1.0 A`
`I = I_(1) + I_(2) = 2.5 A`

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