Voltage Regulation of a Transformer

Each winding has both a resistance and a leakage reactance, which combine into a winding impedance.

Leakage reactance can be transferred from one winding to the other in the same way as resistance, and for the same reason — the criterion is equal reactive volt-amperes, so the factor is again \(K^{2}\).

At no load the secondary terminal voltage equals the induced EMF, since no current flows to cause a drop. Write it \({}_0V_2\). On load, the drop across \(Z_{02}\) reduces it to \(V_2\).

Resolving the impedance drop along the direction of \(V_2\) — and neglecting the quadrature component, which Example 44.3 showed contributes under a volt to a phasor of over two thousand — gives the approximate drop.

Dividing the approximate drop by the no-load secondary voltage gives the regulation directly.

Since regulation is proportional to \(I_2\), it is proportional to the load fraction. A transformer with 4 % regulation at full load has 2 % at half load — a fact worth remembering, since it means the figure need only be quoted once.

It is convenient to express the two drops as percentages of the rated voltage once and for all, since they are properties of the transformer alone and do not depend on the load's power factor.

The regulation then takes a particularly compact form:

The approximate formula discards the quadrature component of the drop. Keeping it gives the exact result.

Treating the regulation \(v_r\cos\phi + v_x\sin\phi\) as a function of the power-factor angle, with \(\phi\) positive when lagging, two special values stand out.

The equivalent circuit is a diagram in which the resistance and leakage reactance of the transformer are imagined to be external to the winding, so that the winding itself becomes ideal. Chapter 44 built it; the sequence of simplification is recalled here because it leads directly to the impedance expression below.

The total impedance between the input terminals is

Omitting \(I_0\) altogether removes \(Z_m\) from the expression and leaves the single series impedance \(Z_{01} + Z_L'\) — which is the circuit every regulation calculation in this chapter has silently used.

• Leakage reactance transfers by \(K^{2}\), like resistance: \(X_2' = X_2/K^{2}\), \(X_{01} = X_1 + X_2'\).

• Voltage regulation is \(\left({}_0V_2 - V_2\right)/{}_0V_2 \times 100\) at constant supply voltage and frequency.

• The approximate drop is \(I_2R_{02}\cos\phi \pm I_2X_{02}\sin\phi\), with + for lagging and − for leading.

• The sign rule follows from geometry: \(I_2R_{02}\) lies along the current, \(jI_2X_{02}\) at 90° ahead, so only the reactive term changes sign.

• In percentage terms, \(v_r = I_2R_{02}/{}_0V_2\) and \(v_x = I_2X_{02}/{}_0V_2\), and regulation is \(v_r\cos\phi \pm v_x\sin\phi\).

• The exact value is \(V_2 = \sqrt{{}_0V_2^{2} - b^{2}} - a\); the approximation sets \(b = 0\) and errs by about 1 % of the regulation.

• Zero regulation at \(\tan\phi = -R_{02}/X_{02}\) — a leading power factor.

• Maximum regulation at \(\tan\phi = X_{02}/R_{02}\) — a lagging power factor, and equal to \(\sqrt{v_r^{2}+v_x^{2}}\), the percentage impedance.

• The two angles are exactly 90° apart, and the maximum-regulation angle is the impedance angle.

• Regulation is proportional to load, unlike efficiency.

• Reversing the sign rule. Plus for lagging, minus for leading. Lagging loads make the voltage fall further.

• Dividing by \(V_2\) instead of \({}_0V_2\). Both conventions exist; state which is used.

• Using \(I_1\) with \(R_{02}\), or \(I_2\) with \(R_{01}\). Match the current to the side.

• Expecting regulation to vary as load². It is proportional to load; copper loss varies as load².

• Thinking negative regulation is an error. At leading power factor the voltage genuinely rises.

• Looking for zero regulation at a lagging power factor. It is always leading.

• Assuming maximum regulation is at unity power factor. It is at the impedance angle, typically 70–80° lagging.

• Forgetting that \(\sqrt{v_r^{2}+v_x^{2}}\) is the nameplate percentage impedance. The worst-case regulation is already printed on the machine.

• Using the approximate formula on a high-impedance transformer. At 15 % impedance the error becomes significant.

• Believing lower impedance is always better. It improves regulation but raises the fault level.

Compute \(v_r\) and \(v_x\) first — they are properties of the transformer, and every question then follows from them.

1. P45.1 A 400/100 V transformer has \(R_{02} = 0.03~\Omega\) and \(X_{02} = 0.08~\Omega\), delivering 50 A. Find the regulation at 0.8 lagging.

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   \[\Delta V = (50)\left[(0.03)(0.8) + (0.08)(0.6)\right] = (50)(0.072) = 3.60~\mathrm{V}\]

   \[\text{Regulation} = \frac{3.60}{100}\times100 = 3.60\,\%\]

2. P45.2 For P45.1, find the regulation at unity and at 0.8 leading.

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   \[\text{unity: } \Delta V = (50)(0.03) = 1.50~\mathrm{V} \quad\Longrightarrow\quad 1.50\,\%\]

   \[\text{0.8 lead: } \Delta V = (50)\left[0.024 - 0.048\right] = -1.20~\mathrm{V} \quad\Longrightarrow\quad -1.20\,\%\]

3. P45.3 For P45.1, find \(v_r\), \(v_x\) and the percentage impedance.

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   \[v_r = \frac{(50)(0.03)}{100}\times100 = 1.50\,\%, \qquad v_x = \frac{(50)(0.08)}{100}\times100 = 4.00\,\%\]

   \[\sqrt{v_r^{2} + v_x^{2}} = \sqrt{2.25 + 16.00} = 4.272\,\%\]

4. P45.4 For P45.3, find the power factors of zero and maximum regulation, and the maximum value.

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   \[\text{zero: } \tan\phi = -\frac{1.50}{4.00} = -0.375 \quad\Longrightarrow\quad 20.56^{\circ}\ \text{lead}, \quad \cos\phi = 0.9363\]

   \[\text{max: } \tan\phi = \frac{4.00}{1.50} = 2.6667 \quad\Longrightarrow\quad 69.44^{\circ}\ \text{lag}, \quad \cos\phi = 0.3511\]

   \[\text{Regulation}_{\max} = 4.272\,\%\]

   Check: \(20.56 + 69.44 = 90.00^{\circ}\) \(\checkmark\)

5. P45.5 A transformer has 2 % resistive and 5 % reactive drop. Find the regulation at 0.9 lagging and at 0.9 leading.

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   \[\sin\phi = 0.4359\]

   \[\text{lagging: } (2)(0.9) + (5)(0.4359) = 1.80 + 2.179 = 3.979\,\%\]

   \[\text{leading: } 1.80 - 2.179 = -0.379\,\%\]

6. P45.6 A transformer with \({}_0V_2 = 240\) V shows \(V_2 = 231\) V at full load. Find the regulation on both bases.

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   \[\text{down: } \frac{240 - 231}{240}\times100 = 3.750\,\%\]

   \[\text{up: } \frac{240 - 231}{231}\times100 = 3.896\,\%\]

   The two differ by 0.146 points — small, but the base must be stated.

7. P45.7 Explain the rule governing the \(\pm\) sign in the approximate drop.

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   Plus for lagging, minus for leading. It follows from the phasor geometry rather than convention.

   The resistive drop \(\mathbf{I}_2R_{02}\) lies along the current, so its component along \(V_2\) is \(I_2R_{02}\cos\phi\) regardless of which way the current leans — always positive.

   The reactive drop \(j\mathbf{I}_2X_{02}\) lies 90° ahead of the current. When the current lags \(V_2\), this places the reactive drop partly along \(V_2\), adding to the fall. When the current leads, it points partly against \(V_2\) and subtracts.

   Only the reactive term changes sign, and since \(X_{02}\) is typically three or four times \(R_{02}\), it usually decides the result — including whether the regulation is positive at all.

8. P45.8 Why does maximum regulation occur at the impedance angle?

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   Write the regulation as \(v_r\cos\phi + v_x\sin\phi\). This is the dot product of the load's unit phasor with the vector \((v_r, v_x)\) — that is, with the impedance triangle itself.

   A dot product is greatest when the two directions coincide. So the regulation is greatest when the load's power-factor angle equals the angle of the impedance triangle, \(\theta = \arctan(X_{02}/R_{02})\), and its value is then the length of that vector, \(\sqrt{v_r^{2} + v_x^{2}}\).

   Formally: \(\mathrm{d}/\mathrm{d}\phi\left(v_r\cos\phi + v_x\sin\phi\right) = -v_r\sin\phi + v_x\cos\phi = 0\) gives \(\tan\phi = v_x/v_r\).

   Zero regulation occurs when the two are perpendicular, which is why the two special angles are exactly 90° apart.

9. P45.9 Why is the percentage impedance the most useful single figure on a transformer nameplate?

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   Because it fixes three things at once.

   Worst-case regulation. \(\sqrt{v_r^{2}+v_x^{2}}\) is exactly the maximum possible regulation, so the voltage drop can never exceed it whatever the load.

   Short-circuit current. A secondary fault is limited only by this impedance, giving \(100/Z_{\%}\) times full-load current — which sets the switchgear rating and the winding bracing.

   Load sharing in parallel. Transformers in parallel divide the load in inverse proportion to their percentage impedances, so matching them is a condition of satisfactory parallel operation.

   The first two pull in opposite directions: low impedance gives good regulation but a high fault level. Choosing it is the central compromise of transformer specification.

10. P45.10 A consumer's voltage must stay within \(\pm6\,\%\). The transformer has 4 % regulation at full load and the cable a further 3 %. What can be done?

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    The total drop at full load is \(4 + 3 = 7\,\%\), exceeding the 6 % allowance — while at no load the drop is zero, so the voltage sits at the top of the band. The swing, not the drop, is the problem.

    Remedies, in rough order of cost:

    Tap changing. Set the transformer taps to raise the no-load voltage by about 3.5 %, centring the swing in the band. An off-load tap changer suffices if the loading pattern is stable; an on-load tap changer follows it continuously.

    Power-factor correction. Since \(v_x \gg v_r\), improving the load power factor from say 0.8 to 0.95 cuts the transformer's regulation by roughly a third at no capital cost to the network.

    Reinforce the cable to reduce its 3 %, or relocate the transformer closer to the load centre.

    Specifying a lower-impedance transformer would help the regulation but raise the fault level, and is rarely the right answer on its own.