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An alternating e.m.f of 100 V is placed ...

An alternating e.m.f of 100 V is placed is applied to a circuit containing a resistance of `40 Omega ` and an inductance L in series . The current is found to lag behind the voltage by an angla a = `tan^(-1) 3//4`
The inductive reactance in this case is :

A

`40 Omega`

B

`30 Omega`

C

`50 Omega`

D

`10sqrt(5) Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`30 Omega `
Explanation : `X_(L) = omegaL = 2piL`
In LR circuit , resistance and inductance are connected in series
` :. " " tan a = 3/4 = (X_(L))/R " "( :. tan a= (V_(L))/(V_(R))=(IX_(L))/(IR))`
`X_(L) = 3/4 R = 3/4 xx 40 = 30 Omega `
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