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4.2 kV/120 V transformer voltage regulation

Short answer

Referred to secondary, equivalent resistance and reactance are 0.04114 Ω and 0.10286 Ω. No-load voltage E2 = √((120+500×0.04114)² + (500×0.10286)²) = 149.684 V. Voltage regulation = (149.684-120)/120×100 = 24.74%.

A 4.2 kV/120 V, 50 Hz, 1-phase transformer has following series parameters: R1=1.4 ΩR_1 = 1.4\ \Omega, X1=3.5 ΩX_1 = 3.5\ \Omega, R2=0.04 ΩR_2 = 0.04\ \Omega, X2=0.1 ΩX_2 = 0.1\ \Omega. If the transformer draws 500 A current on the secondary at rated terminal voltage, calculate voltage regulation at unity power factor.

Voltage regulation is the change in secondary terminal voltage from no load to load, as a fraction of the rated terminal voltage:

%VR=E2−V2V2×100\%\text{VR} = \frac{E_2 - V_2}{V_2}\times 100

where E2E_2 is the no-load secondary voltage and V2V_2 is the load terminal voltage.

Equivalent parameters referred to secondary

a=4200120=35,a2=1225a = \frac{4200}{120} = 35, \qquad a^2 = 1225 R02=R2+R1a2=0.04+1.41225=0.04114 ΩX02=X2+X1a2=0.1+3.51225=0.10286 Ω\begin{aligned} R_{02} &= R_2 + \frac{R_1}{a^2} = 0.04 + \frac{1.4}{1225} = 0.04114\ \Omega \\ X_{02} &= X_2 + \frac{X_1}{a^2} = 0.1 + \frac{3.5}{1225} = 0.10286\ \Omega \end{aligned}

No-load voltage at unity p.f.

With V2=120V_2 = 120 V, I2=500I_2 = 500 A and cos⁡ϕ=1\cos\phi = 1:

E2=(V2+I2R02)2+(I2X02)2=(120+20.571)2+(51.429)2=149.684 V\begin{aligned} E_2 &= \sqrt{(V_2 + I_2R_{02})^2 + (I_2X_{02})^2} \\ &= \sqrt{(120 + 20.571)^2 + (51.429)^2} \\ &= 149.684\ \text{V} \end{aligned}

Regulation

%VR=149.684−120120×100=24.74%\%\text{VR} = \frac{149.684 - 120}{120}\times 100 = 24.74\%

Answer: voltage regulation at unity power factor =24.74%= 24.74\% (no-load secondary voltage =149.684= 149.684 V).

✓ Answer checked by Claude Opus 5.5 (high effort)

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