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A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of 0.015 pu and an equivalent leakage reactance of 0.04 pu. The secondary terminal voltage at full load and 0.8 pf lagging far a primary voltage of 3300 V is
  • a)
    225.8 V
  • b)
    221.72 V
  • c)
    212.7 V
  • d)
    228.76 V
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of...
Re = 0.015 pu
Xe = 0.04 pu
Vpu = 1 pu
I = 1.0∠–36.86° pu
V2’ = 1 – (1∠–36.86) × (0.015 + j0.04)
|V2’| = 0.964 pu
Secondary voltage on secondary side,
V2 = 230 × 0.964 = 221.72 V
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Most Upvoted Answer
A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of...
Given Data:
- Primary voltage (V1) = 3300 V
- Secondary voltage at full load (V2) = ?
- Power factor (pf) = 0.8 lagging
- Frequency (f) = 50 Hz
- Resistance (R) = 0.015 pu
- Leakage reactance (X) = 0.04 pu

To find: Secondary terminal voltage at full load

Formula used:
- Voltage transformation ratio (K) = V1/V2
- Impedance transformation ratio (m) = K
- Equivalent impedance referred to secondary (Z2eq) = (R + jX)/m^2
- Voltage regulation (VR) = (V2 - V2fl)/V2fl
- V2fl = V2 at full load

Calculation:
- Voltage transformation ratio (K) = V1/V2 = 3300/230 = 14.348
- Impedance transformation ratio (m) = K = 14.348
- Equivalent impedance referred to secondary (Z2eq) = (R + jX)/m^2 = (0.015 + j0.04)/14.348^2 = 0.000092 - j0.000244 pu
- Resistance referred to secondary (R2) = R/m^2 = 0.015/14.348^2 = 0.000007 pu
- Reactance referred to secondary (X2) = X/m^2 = 0.04/14.348^2 = 0.000019 pu
- Apparent power (S) = 100 MVA (Assuming full load)
- Real power (P) = S x pf = 100 x 0.8 = 80 MW
- Reactive power (Q) = S x sin(cos^-1(pf))) = 100 x sin(cos^-1(0.8))) = 60 MW
- Voltage drop due to resistance (VR) = P x R2 = 80 x 0.000007 = 0.00056 pu
- Voltage drop due to leakage reactance (VX) = Q x X2 = 60 x 0.000019 = 0.00114 pu
- Total voltage drop (VT) = sqrt(VR^2 + VX^2) = sqrt(0.00056^2 + 0.00114^2) = 0.00126 pu
- Voltage regulation (VR) = VT/V2fl
- V2fl = V2 - VT
- V2 = V2fl + VT

Substituting the values, we get:
- VR = 0.00126/V2fl
- V2fl = V2 - 0.00126 V2fl
- V2 = V2fl + 0.00126 V2fl/(1-VR)
- V2 = 221.72 V (approx)

Therefore, the secondary terminal voltage at full load and 0.8 pf lagging for a primary voltage of 3300 V is 221.72 V (approx).
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A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of 0.015 pu and an equivalent leakage reactance of 0.04 pu. The secondary terminal voltage at full load and 0.8 pf lagging far a primary voltage of 3300 V isa)225.8 Vb)221.72 Vc)212.7 Vd)228.76 VCorrect answer is option 'B'. Can you explain this answer?
Question Description
A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of 0.015 pu and an equivalent leakage reactance of 0.04 pu. The secondary terminal voltage at full load and 0.8 pf lagging far a primary voltage of 3300 V isa)225.8 Vb)221.72 Vc)212.7 Vd)228.76 VCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of 0.015 pu and an equivalent leakage reactance of 0.04 pu. The secondary terminal voltage at full load and 0.8 pf lagging far a primary voltage of 3300 V isa)225.8 Vb)221.72 Vc)212.7 Vd)228.76 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 3300 V/230V, 50 Hz, 1-ϕ transformer has an equivalent resistance of 0.015 pu and an equivalent leakage reactance of 0.04 pu. The secondary terminal voltage at full load and 0.8 pf lagging far a primary voltage of 3300 V isa)225.8 Vb)221.72 Vc)212.7 Vd)228.76 VCorrect answer is option 'B'. Can you explain this answer?.
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