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250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω. Determine its regulation while supplying full load on 0.8 leading p.f. The secondary terminal voltage on full load 0.8 and leading p.f.?
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250/125 V, 5 KVA single phase transformer has primary resistance of 0....
Calculation of Regulation


Regulation refers to the percentage drop in voltage from no-load to full-load conditions. It is calculated using the formula:


% Regulation = (Vno-load - Vfull-load) / Vfull-load x 100%


Where Vno-load is the secondary voltage at no-load and Vfull-load is the secondary voltage at full-load.


Given data:



  • Primary voltage (V1) = 250 V

  • Secondary voltage (V2) = 125 V

  • Power rating (S) = 5 KVA

  • Primary resistance (R1) = 0.2 Ω

  • Primary reactance (X1) = 0.75 Ω

  • Secondary resistance (R2) = 0.05 Ω

  • Secondary reactance (X2) = 0.2 Ω

  • Power factor (pf) = 0.8 leading



First, we need to calculate the equivalent impedance of the transformer referred to the primary side:


Zeq = (R1 + jX1) + [(R2 + jX2) / (k^2)]


Where k is the turns ratio = V1 / V2


Substituting the given values:


k = V1 / V2 = 250 / 125 = 2


Zeq = (0.2 + j0.75) + [(0.05 + j0.2) / (2^2)] = 0.2125 + j0.8875 Ω


Next, we can calculate the current in the secondary winding:


I2 = S / V2 = 5 x 10^3 / 125 = 40 A


Since the power factor is leading, the angle of the current (θ) is negative:


θ = cos^-1(pf) = cos^-1(0.8) = -36.87°


Hence, the current can be represented as:


I2 = 40 ∠-36.87° A


Now, we can find the voltage drop in the equivalent impedance:


Vdrop = Zeq x I2 = (0.2125 + j0.8875) x 40 ∠-36.87° = 8.5 ∠53.13° V


The secondary voltage under full-load conditions is given by:


V2(full-load) = V2 - Vdrop = 125 - 8.5 = 116.5 V


Therefore, the percentage regulation is:


% Regulation = (125 - 116.5) / 116.5 x 100% = 7.27%


Calculation of Secondary Terminal
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250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω. Determine its regulation while supplying full load on 0.8 leading p.f. The secondary terminal voltage on full load 0.8 and leading p.f.?
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250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω. Determine its regulation while supplying full load on 0.8 leading p.f. The secondary terminal voltage on full load 0.8 and leading p.f.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω. Determine its regulation while supplying full load on 0.8 leading p.f. The secondary terminal voltage on full load 0.8 and leading p.f.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω. Determine its regulation while supplying full load on 0.8 leading p.f. The secondary terminal voltage on full load 0.8 and leading p.f.?.
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