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A 0.21-H inductor and a 88-Omega resisto...

A 0.21-H inductor and a `88-Omega` resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. `(Use pi= 22//7)`

A

`2A, tan^(-1)3//4`

B

`14.4A, tan^(-1)7//8`

C

`14.4A, tan^(-1)8//7`

D

`3.28 A, tan^(-1)2//11`

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(rms)=V_(rms)/Z=V_(rms)/sqrt(R^(2)+(omegaL)^(2))=2A`
`tanphi=(omegaL)/R=66/88=3/4`
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