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In inductance of (4//pi) H and the resis...

In inductance of `(4//pi)` H and the resistor R, are connected in series and an alternating emf of frequency 50 Hz is applied across combination. If phase difference between applied emf and current is `45^(@)` then the value of R is

A

`200Omega`

B

`400Omega`

C

`600Omega`

D

`800Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`tanphi=(X_(L))/(R)`
`therefore R=(2pifL)/(tanphi)`
`=(2pixx50xx4)/(tan45xxpi)=400Omega`
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