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In a LR circuit of 3 mH dinductance and ...

In a LR circuit of 3 mH dinductance and `4 Omega` resistance, emf `E=4cos 1000t` volt is applied. The amplitude of current is

A

`(4)/(sqrt(7))A`

B

`1.0A`

C

`(4)/(7)A`

D

`0.8A`

Text Solution

Verified by Experts

The correct Answer is:
D
Impedence, `Z=sqrt(R^(2)+X_(L)^(2))`
Here, `R=4Omega,X_(L)=L_(omega)`
`=3xx10^(-3)xx1000Omega`
`=3Omega`
Then, `Z=sqrt((4)^(2)+(3)^(2))`
or `Z=sqrt(16+9)=sqrt(25)`
or `Z=5Omega`
Hence, current `I_(0)=(E_(0))/(Z)`
`=(4)/(5)=0.8A`
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