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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 = ? |
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Answer» Solution :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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