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A call is connected across 250 V, 50 Hz ...

A call is connected across 250 V, 50 Hz supply and it draws a current of 2.5 A and consumes power of 400 W. Find the self inductance and power factor.
Data : `E_(rms) = 250 V, v = 50 Hz, I_(rms) = 2.5 A, P = 400 W, L = ?, cos phi = ?`

Text Solution

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Power `P = R_(rms)I_(rms) cos phi`
`:. Cos phi = (P)/(E_(rms)I_(rms))`
`= (400)/(250 xx 2.5)`
`cos phi = 0.64`
Impedance `Z = (E_(rms))/(I_(rms)) = (250)/(2.5) = 100 Omega`
also, `sin phi = (X_(L))/(Z)`
`:. X_(L) = Z. sin phi = Z sqrt(1- cos^(2) phi)`
`=100 sqrt(1- (0.64)^(2)]`
`:. X_(L) = 76.8 Omega`
But `X_(L) = L omega = L2pi V`
`:. L = (X_(L))/(2pi v) = (76.8)/(2pi xx 50)`
`:. L = 0.244 H`
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