A transformer has a turns-ratio of 4 and Its primary Is connected to a 240V, 50H

Question

A transformer has a turns-ratio of 4 and Its primary Is connected to a 240V, 50Hz supply. Figure Q2 shows the simplified equivalent circuit in which the magnetizing branch has been moved to the primary terminals. The transformer parameters (all referred to the primary) are as follows: Lm = 0.5H, lr1 = Ir2 = 12mH R1 = 1 ohm, R2 = 0.8 ohm Determine the rms magnitude of the magnetizing current Im A 0.5 ohm load resistance (actual value) is now connected to the secondary. Determine the value of the load resistance referred to the primary, RL', and the rms magnitude of the primary current I1. Determine the efficiency of the transformer under this operating condition.

Explanation / Answer

Magnetising current XLm = 2fLm = (2)(3.14)(50)(0.5) = 157 Im = V / jXLm Im = 240 / j157 Im = 240 / j157 Im = -j 1.53A or Im = 1.53/_-90o A Magnitude of Im = 1.53A XL1 = 2fL1 = (2)(3.14)(50)(12*10-3) = 3.768 Since L1 = L2, XL2 = 3.768 The efffective impedance in the secondary is ZL = R1 +jXL1 + jXL2 + R2 + RL ZL = 1 + j3.768 + j3.768 + 0.8 + 0.5 ZL = 2.3 + j7.536 ZL = 7.88/_73o Magnitude of ZL = 7.88 Imepedance referred to the primary will be Zin = ZL / a2 Given a = 4 Zin = 7.88/_73o / 42 Zin = 0.5 /_73o Magnitude of Zin = 0.5 Current I2 = V / ZL = 240 / 7.88 = 30.45A I1 = Im + I2 I1 = 1.53 + 30.45 I1 = 31.98A Pin = I12 Zin = (31.98)2 (0.5) = 1023W Pout = I22 RL = (30.45)2 (0.5) = 464 Efficiency = Pout / Pin = 464 / 1023 = 0.4535 Effeciency = 45.35% Magnitude of Zin = 0.5 Current I2 = V / ZL = 240 / 7.88 = 30.45A I1 = Im + I2 I1 = 1.53 + 30.45 I1 = 31.98A Pin = I12 Zin = (31.98)2 (0.5) = 1023W Pout = I22 RL = (30.45)2 (0.5) = 464 Efficiency = Pout / Pin = 464 / 1023 = 0.4535 Effeciency = 45.35%

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