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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

`0.8`

B

`2 sqrt(2) A`

C

`0.5 A`

D

`sqrt(0.4) A`

Text Solution

Verified by Experts

The correct Answer is:
B

`IR=(V)/(R)=(200)/(100)=2A`
`I=(V)/(X_(L)-X_(C))=(200)/(100)=2A`
`I=sqrt(I_(R)^(2)+I'^(2))=2 sqrt(2)`.
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