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In the circuit shown in figure : R = 1...

In the circuit shown in figure :
`R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega` and `C = (sqrt(3))/(2) mF`. Current in `L - R_(1)` circuit is `I_(1)` in `C - R_(1)` circuit is `I_(2)` and the main current is `I`

At some instant current in `L - R_(1)` circuit is `10 A`. At the same instant current in `C - R_(2)` branch will be

A

`5 A`

B

`5 sqrt(2) A`

C

`5 sqrt(6) A`

D

`5 sqrt(3) A`

Text Solution

Verified by Experts

The correct Answer is:
D
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