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In the circuit shown in fig. X(C ) = 100...

In the circuit shown in fig. `X_(C ) = 100 Omega,(X_L)=200 Omega and R=100 Omega`. The effective current through the source is

A

`2 A`

B

`2 sqrt(2) A`

C

`0.5 A`

D

`sqrt(0.4) A`

Text Solution

Verified by Experts

The correct Answer is:
2
`I_(R) = (V)/(R ) = (200)/(100) = 2A`, `I_(LC) = (200)/(X_(L) - X_(C)) = 2A`
`I = sqrt(2^(2) + 2^(2)) = 2 sqrt(2) A` as `I_(R )` inphase with `V`
`I_(LC)` lages behind `V` by `pi//2`
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