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Voltage regulation of transformer for lagging load

Short answer

Voltage regulation is the rise in secondary voltage from full load to no load expressed as a percentage of no-load voltage. For lagging load, approximate percentage regulation equals %R cos phi plus %X sin phi, so V2 falls below E2.

Define voltage regulation of a transformer. Deduce the expression for the voltage regulation when it is connected with lagging load.

Voltage regulation

Voltage regulation of a transformer is the change in secondary terminal voltage from no load to full load, expressed as a fraction (or percentage) of the no-load voltage, with the primary voltage held constant:

%Reg=E2−V2E2×100=0V2−V20V2×100\%\text{Reg} = \frac{E_2 - V_2}{E_2} \times 100 = \frac{{}_0V_2 - V_2}{{}_0V_2}\times 100

where E2=0V2E_2 = {}_0V_2 is the no-load secondary voltage and V2V_2 is the full-load secondary voltage. A small regulation is desirable (power transformers: about 2-6%).

Expression for lagging load

Use the equivalent circuit referred to the secondary: E2=V2+I2(R02+jX02)E_2 = V_2 + I_2(R_{02} + jX_{02}). Take V2V_2 as reference; I2I_2 lags V2V_2 by ϕ\phi.

Phasor diagram (approximate):

                                   C
                                  /
                                 /  I2.X02
                                /   (perp. to I2)
  O -------------------- A     /
          V2              \   /
                           \ /
                    I2.R02  B
              (parallel to I2)

  E2 = OC (join O to C);  I2 lags V2 by phi

Construction: from O draw V2V_2 = OA. Draw I2R02I_2R_{02} = AB parallel to I2I_2, and I2X02I_2X_{02} = BC perpendicular to I2I_2 (leading it by 90°). Then OC = E2E_2. Drop a perpendicular from C to the extended OA line, meeting it at D.

OD=V2+I2R02cos⁡ϕ+I2X02sin⁡ϕCD=I2X02cos⁡ϕ−I2R02sin⁡ϕE2=OC=OD2+CD2\begin{aligned} OD &= V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi \\ CD &= I_2X_{02}\cos\phi - I_2R_{02}\sin\phi \\ E_2 &= OC = \sqrt{OD^2 + CD^2} \end{aligned}

Because the angle between E2E_2 and V2V_2 is very small, OC≈ODOC \approx OD:

E2−V2≈I2R02cos⁡ϕ+I2X02sin⁡ϕE_2 - V_2 \approx I_2R_{02}\cos\phi + I_2X_{02}\sin\phi

Therefore, for a lagging power factor:

%Reg=I2R02cos⁡ϕ+I2X02sin⁡ϕE2×100\%\text{Reg} = \frac{I_2R_{02}\cos\phi + I_2X_{02}\sin\phi}{E_2}\times 100

or, in per-unit form,

%Reg=%Rcos⁡ϕ+%Xsin⁡ϕ\%\text{Reg} = \%R\cos\phi + \%X\sin\phi

where %R=I2R02E2×100\%R = \frac{I_2R_{02}}{E_2}\times100 and %X=I2X02E2×100\%X = \frac{I_2X_{02}}{E_2}\times100.

A more exact expression keeps the CD term:

%Reg=%Rcos⁡ϕ+%Xsin⁡ϕ+(%Xcos⁡ϕ−%Rsin⁡ϕ)2200\%\text{Reg} = \%R\cos\phi + \%X\sin\phi + \frac{(\%X\cos\phi - \%R\sin\phi)^2}{200}

Notes

  • For lagging loads regulation is always positive (V2<E2V_2 < E_2).
  • For leading loads the sign of the sin⁡ϕ\sin\phi term becomes negative, so regulation can be zero or negative.
  • Regulation is zero when tan⁡ϕ=−R02/X02\tan\phi = -R_{02}/X_{02} (a leading pf), and maximum when tan⁡ϕ=X02/R02\tan\phi = X_{02}/R_{02} (a lagging pf).

✓ Answer checked by Claude Opus 5.5 (high effort)

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