Fig Q2 (a) shows an equivalent circuit of a transformer of turns ratio N1/N2 = 4

Question

Fig Q2 (a) shows an equivalent circuit of a transformer of turns ratio N1/N2 = 4. primary is connected to a 240 v , 50 v supply and the secondary to a load RL = (actual real value). In fig Q2a, the transformer parameters are referred to the primary. magnetizing inductance has been moved to the input terminals in order to simp analysis. The core losses are neglected. Determine the numerical value of RL' referred to the primary side. Determine the total impedance Z' in parallel with the inductance Lm Give your answer in the form Z' = x + jY. With the supply voltage represented as 240

Explanation / Answer

i)  RL not given If RL = 0.5 RL' = RL / a2 = 0.5 / 42 = 0.03125 ii) XL1 = L1 = (2)(3.14)(50)(0.015) = 4.71 XL2 = L2 = (2)(3.14)(50)(0.010) = 3.14 XL2 = L2 = (2)(3.14)(50)(0.010) = 3.14 Z' = R1 + jXL1 + jXL2 + R2 + RL' Z' = 1.5 + j4.71 + j3.14 + 1 + 0.03125 Z' = 2.53 + j 7.85 iii) XLm = 2 (3.14)(50)(0.8) = 251.2 Im = V/jXm = 240 / j251.2 = 0.955/_-90o A I2 = V / Z' = 240 / (2.53+j7.85) = 240 / 8.25/_72o = 29/_-72o A I1 = I2 + Im I1 = 0.955/_-90 + 29/_-72 I1 = -j0.944 + 8.96 - j27.58 I1= 8.96 - j 26.63 A I1 = 28 /_-71.4o iv) Input power factor = cos (-71.4) = 0.32 Output power factor = cos (-72) = 0.309 I1 = 0.955/_-90 + 29/_-72 I1 = -j0.944 + 8.96 - j27.58 I1= 8.96 - j 26.63 A I1 = 28 /_-71.4o iv) Input power factor = cos (-71.4) = 0.32 Output power factor = cos (-72) = 0.309

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.