Set 19 — Voltage Regulation
A transformer that delivered its rated secondary voltage under every load would need no regulation calculation at all. It does not, because the current has to pass through the winding resistance and the leakage reactance measured in the previous two sets, and the drop across them is subtracted from the induced e.m.f. This set works that drop out on both sides of the machine, first exactly with phasors and then with the approximate formula that every examination expects.
The thread running through all five problems is that the two windings must be collapsed onto one side before anything can be added. Once \(R_{02}\) and \(X_{02}\) exist, the drop is one line of arithmetic, the regulation is that drop as a percentage, and the same information expressed per unit lets the last problem answer the question without a single ohm being quoted.
Everything is referred to one side first. With \(a = N_1/N_2\), an impedance moves across the transformer through \(a^2\), so the two windings collapse into a single series branch:
\[ R_{01} = R_1 + a^2R_2, \qquad X_{01} = X_1 + a^2X_2, \qquad Z_{01} = \sqrt{R_{01}^2 + X_{01}^2} \]\[ R_{02} = \frac{R_{01}}{a^2}, \qquad X_{02} = \frac{X_{01}}{a^2}, \qquad Z_{02} = \frac{Z_{01}}{a^2} \]The secondary drop is a phasor difference, not an arithmetic one. Exactly,
\[ \mathbf{E}_2 = \mathbf{V}_2 + \mathbf{I}_2\left(R_{02} + jX_{02}\right) \]The approximate drop is the projection of \(\mathbf{I}_2Z_{02}\) on \(\mathbf{V}_2\), and is what almost every examination question wants:
\[ \Delta V = I_2\left(R_{02}\cos\phi \pm X_{02}\sin\phi\right) \]Plus for a lagging power factor, minus for a leading one. The error against the exact expression is under a tenth of a percent at ordinary regulations.
Regulation is that drop as a fraction of the no-load secondary voltage, the load being removed at constant primary voltage:
\[ \text{Regulation} = \frac{E_2 - V_2}{E_2}\times100\% \approx \frac{I_2\left(R_{02}\cos\phi \pm X_{02}\sin\phi\right)}{E_2}\times100\% \]Some texts divide by \(V_2\) instead, giving a marginally larger figure. This page uses \(E_2\) throughout, which is also the base on which \(\%R\) and \(\%X\) are defined.
Percentage resistance and reactance are the same drops expressed on the rated voltage, and they are properties of the machine alone:
\[ \%R = \frac{I_2R_{02}}{E_2}\times100 = \frac{P_{cu,\text{FL}}}{S}\times100, \qquad \%X = \frac{I_2X_{02}}{E_2}\times100 \]At a fraction \(x\) of full load the drop simply scales, because the current does:
\[ \%\Delta V = x\left(\%R\cos\phi \pm \%X\sin\phi\right) \]
A 30 kVA, 2400/120 V, 50 Hz transformer has a high-voltage winding resistance of 0.1 Ω and a leakage reactance of 0.22 Ω. The low-voltage winding resistance is 0.035 Ω and its leakage reactance is 0.012 Ω. Find the equivalent winding resistance, reactance and impedance referred to
- the high-voltage side
- the low-voltage side.
Fix the notation before touching a number. The high-voltage winding is winding 1, the low-voltage winding is winding 2, and the turns ratio follows the voltages:
Everything that follows is one multiplication or one division by \(a^2 = 400\). Getting \(a\) the right way up is the whole problem.
Referred to the high-voltage side, the small secondary values are multiplied by 400 and become the dominant terms:
Referred to the low-voltage side, the primary values are divided by 400 and become almost negligible:
Check the two sets against each other. Referring the whole equivalent impedance in one step must reproduce the second calculation:
The same test applies term by term: \(14.1/400 = 0.03525\) and \(5.02/400 = 0.01255\). If the two routes disagree, one of the two windings has been referred the wrong way.
What the numbers are worth. The full-load currents are
That is 7.3% of the rating, which is far heavier than a real 30 kVA unit would be built with — the winding data here is illustrative rather than typical. The arithmetic identity, however, holds for any transformer: copper loss is the same number whichever side it is computed on.
The following data refer to a single-phase transformer:
turn ratio 19.5 : 1, \(R_1 = 25\ \Omega\), \(X_1 = 100\ \Omega\), \(R_2 = 0.06\ \Omega\), \(X_2 = 0.25\ \Omega\). The no-load current is 1.25 A, leading the flux by \(30^\circ\). The secondary delivers 200 A at a terminal voltage of 500 V and a power factor of 0.8 lagging.
Determine, with the aid of a phasor diagram,
- the primary applied voltage
- the primary power factor
- the efficiency.
Take the secondary terminal voltage as reference and write the load current as a phasor. A lagging power factor puts the current below the real axis:
Climb through the secondary impedance to reach the induced e.m.f. behind it:
Refer the e.m.f. to the primary through the turns ratio \(a = 19.5\). The primary must supply the reversed phasor \(-\mathbf{E}_1\), which is the back e.m.f. the applied voltage works against:
Place the no-load current using the flux. The flux lags \(-\mathbf{E}_1\) by \(90^\circ\), so it sits at \(183.48^\circ - 90^\circ = 93.48^\circ\), and \(I_0\) is given as leading it by \(30^\circ\):
This is the only place the \(30^\circ\) enters. It fixes both the magnetising and the core-loss components of \(I_0\) at once.
Build the primary current as the exciting current plus the reflected load current:
Add the primary impedance drop to obtain the applied voltage:
So the primary applied voltage is about 11.54 kV — some 2.6% above \(E_1\), the difference being the primary resistance and leakage reactance drop.
The primary power factor is the cosine of the angle between \(\mathbf{V}_1\) and \(\mathbf{I}_1\):
Markedly worse than the load's 0.8, because the magnetising current and the leakage reactance both add lagging reactive volt-amperes that the load never asked for.
Core loss from the no-load current. The flux lags \(V_1\) by \(90^\circ\) and \(I_0\) leads the flux by \(30^\circ\), so \(I_0\) lags \(V_1\) by \(60^\circ\):
Copper loss from the equivalent secondary resistance, which counts both windings at once:
Assemble the efficiency:
Check the input independently from the primary phasors, which used a different route to every number:
Agreement to a fraction of a percent confirms the phasor diagram, the power factor and the loss accounting together.
A 50 kVA, 4400/220 V transformer has \(R_1 = 3.45\ \Omega\) and \(R_2 = 0.009\ \Omega\). The reactances are \(X_1 = 5.2\ \Omega\) and \(X_2 = 0.015\ \Omega\). Calculate for this transformer
- the equivalent resistance referred to the primary
- the equivalent resistance referred to the secondary
- the equivalent reactance referred to both primary and secondary
- the equivalent impedance referred to both primary and secondary
- the total copper loss, first from the individual winding resistances and then from the equivalent resistance referred to each side.
Full-load currents and the turns ratio first, since part (v) needs the currents and everything else needs \(a\):
Equivalent resistances. Referred to the primary the secondary resistance is multiplied by \(a^2 = 400\):
The two windings contribute almost equally here — a sign of a well-balanced design, since the copper is then shared evenly between them.
Equivalent reactances, by exactly the same rule:
Equivalent impedances are the phasor sums, not the arithmetic ones:
Total copper loss, three ways. From the individual windings:
All three must agree, and they do. The equivalent resistance is defined precisely so that this happens — it is the resistance that dissipates the true total loss when carrying the current of the side it is referred to.
What the parameters predict about regulation, which is what they were computed for:
A 230/460 V transformer has a primary resistance of 0.2 Ω and a reactance of 0.5 Ω; the corresponding secondary values are 0.75 Ω and 1.8 Ω. Find the secondary terminal voltage when the transformer supplies 10 A at 0.8 power factor lagging.
This is a step-up transformer, so the turns ratio is less than one and the primary impedances are divided by \(a^2\) when they are carried to the secondary:
Gather the whole transformer on the secondary side, where the load lives:
Referring a 0.2 Ω primary resistance to the high-voltage side makes it 0.8 Ω. On a step-up transformer the primary contribution grows, and it is the term most often left out.
The no-load secondary voltage is what the terminal voltage is measured against. With 230 V applied,
Apply the approximate drop with the lagging sign, since a lagging current makes both terms subtract from the terminal voltage:
Confirm with the exact expression, which keeps the quadrature component instead of discarding it:
A difference of 0.5 V in 460, or 0.11%. The approximate formula is entirely adequate, and this is why it is the one used everywhere.
As a regulation the answer reads:
Change the power factor to 0.8 leading and the reactive term reverses:
The terminal voltage now rises on load, a regulation of −2.26%. Nothing has changed in the transformer; only the phase of the current has.
Calculate the percentage voltage drop of a transformer whose percentage resistance is 2.5% and percentage reactance 5%, rated at 500 kVA, when it is delivering 400 kVA at 0.8 power factor lagging.
Read what the percentages mean. They are the resistive and reactive drops at full-load current, as a percentage of rated voltage:
Equivalently, the full-load copper loss is 2.5% of 500 kVA, that is 12.5 kW. No voltage, current or ohmic value is needed anywhere in this problem.
Scale to the actual current. The drop is proportional to current, so with \(I\) the actual and \(I_f\) the full-load current:
The second form is the first one multiplied top and bottom by the rated voltage: \(I\cos\phi\) becomes the load kW and \(I\sin\phi\) the load kVAR.
Resolve the load into its active and reactive parts:
Substitute:
The same result from the load-fraction form: \(x = 400/500 = 0.8\) and \(\%\Delta V = 0.8(2.5\times0.8 + 5\times0.6) = 0.8\times5 = 4.0\%\).
Two contrasting cases, from the same two numbers:
| Load condition | Calculation | % drop |
|---|---|---|
| 400 kVA, 0.8 lagging | \(0.8(2.5\times0.8 + 5\times0.6)\) | 4.00% |
| 400 kVA, unity p.f. | \(0.8(2.5\times1)\) | 2.00% |
| 400 kVA, 0.8 leading | \(0.8(2.5\times0.8 - 5\times0.6)\) | −0.80% |
| Full load, worst case | \(\cos\phi = R/Z = 2.5/5.59\) | 5.59% |
The maximum possible drop is \(\sqrt{(\%R)^2 + (\%X)^2} = 5.59\%\), reached when the load power factor equals the transformer's own, \(\cos\phi = 0.447\) lagging. No load can make it worse than that.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Turns ratio | \(a = N_1/N_2 = V_1/V_2\) | Every problem; \(a < 1\) for the step-up of Problem 4 |
| Referred to primary | \(R_{01} = R_1 + a^2R_2,\ X_{01} = X_1 + a^2X_2\) | Problems 1, 3 |
| Referred to secondary | \(R_{02} = R_2 + R_1/a^2,\ X_{02} = X_2 + X_1/a^2\) | Problems 1, 2, 3, 4 |
| Cross-check | \(Z_{02} = Z_{01}/a^2\) | Two independent routes must agree — Problems 1, 3 |
| Equivalent impedance | \(Z_{01} = \sqrt{R_{01}^2 + X_{01}^2}\) | Phasor sum, never \(R + X\) — Problems 1, 3 |
| Exact secondary e.m.f. | \(\mathbf{E}_2 = \mathbf{V}_2 + \mathbf{I}_2(R_{02} + jX_{02})\) | Problems 2, 4 |
| Approximate drop | \(\Delta V = I_2(R_{02}\cos\phi \pm X_{02}\sin\phi)\) | + lagging, − leading — Problems 4, 5 |
| Exact terminal voltage | \(V_2 = \sqrt{E_2^2 - q^2} - \Delta V\), \(q = I_2(X_{02}\cos\phi - R_{02}\sin\phi)\) | 0.11% below the approximation in Problem 4 |
| Voltage regulation | \(\dfrac{E_2 - V_2}{E_2}\times100\%\) | Negative for leading loads — Problems 4, 5 |
| Percentage resistance | \(\%R = \dfrac{I_fR_{02}}{E_2}\times100 = \dfrac{P_{cu,\text{FL}}}{S}\times100\) | The two definitions coincide — Problems 3, 5 |
| Drop at part load | \(\%\Delta V = x(\%R\cos\phi \pm \%X\sin\phi)\) | \(x = S/S_{\text{rated}}\) — Problem 5 |
| Drop from kW and kVAR | \(\%\Delta V = \dfrac{(\%R)\text{kW} + (\%X)\text{kVAR}}{S}\) | Same formula, load resolved — Problem 5 |
| Maximum regulation | \(\%\Delta V_{\max} = \sqrt{(\%R)^2 + (\%X)^2}\) at \(\cos\phi = \%R/\%Z\) | 5.59% in Problem 5 |
| Reflected current | \(\mathbf{I}_2' = -\mathbf{I}_2/a,\quad \mathbf{I}_1 = \mathbf{I}_0 + \mathbf{I}_2'\) | Problem 2 |
| Copper loss | \(P_{cu} = I_1^2R_{01} = I_2^2R_{02}\) | Identical on either side — Problems 1, 2, 3 |
Common Mistakes
Inverting \(a\) when referring an impedance. Multiplying by \(a^2\) moves an impedance towards the primary and dividing moves it towards the secondary; the two answers differ by \(a^4\), a factor of 160 000 in Problem 1.
Adding resistance and reactance arithmetically. \(Z_{01}\) in Problem 3 is 13.23 Ω, not \(7.05 + 11.2 = 18.25\) Ω.
Forgetting the primary term on a step-up transformer. In Problem 4 the 0.2 Ω primary resistance becomes 0.8 Ω on the secondary side and more than doubles \(R_{02}\); using the secondary values alone leaves a drop of 16.8 V instead of 35.2 V.
Using the plus sign for a leading power factor. A leading current subtracts the reactive term, and in Problems 4 and 5 it can make the regulation negative — the terminal voltage rises when load is applied.
Treating the approximate drop as a magnitude subtraction of phasors. It is the projection of \(\mathbf{I}_2Z_{02}\) on \(\mathbf{V}_2\); the exact result in Problem 4 differs by 0.5 V, which is small only because the quadrature term is small.
Neglecting \(I_0\) when it is not negligible. In Problem 2 the exciting current is 12% of the primary current, and dropping it shifts the primary power factor from 0.698 to 0.73.
Putting the load power factor angle where the no-load angle belongs. The core loss in Problem 2 uses \(\cos\phi_0 = \cos60^\circ\), the angle between \(V_1\) and \(I_0\) — not the load's 0.8 and not the primary's 0.698.
Counting a winding resistance twice. Once \(R_{02}\) is used for the copper loss, the individual \(R_1\) and \(R_2\) must not be added again — the three routes in Problem 3 are alternatives, not contributions.
Applying percentage values at the wrong load. The 2.5% and 5% of Problem 5 are full-load figures; at 400 kVA they must be scaled by 0.8 before use.
Quoting regulation without a power factor. The same transformer in Problem 5 gives +4%, +2% or −0.8% at the same kVA. A regulation figure with no power factor attached says nothing.
Regulation answers one of the two questions a user asks of a transformer: how far does my voltage sag when I load the machine? Every number on this page came from the series branch of the equivalent circuit — the resistance and leakage reactance that Sets 17 and 18 measured — and from the phase angle of the load current.
The other question is what the machine costs to run. The same series resistance that produced the voltage drop also produces the copper loss, and it varies as the square of the load; the shunt branch produces the iron loss, and it does not vary at all. Setting those two against the output gives the efficiency, and asking when they are equal gives the load at which the efficiency is greatest.
Next: Set 20 — Efficiency and All-Day Efficiency, where the loss split of the previous sets is turned into an efficiency curve and then into the energy accounting of a distribution transformer over 24 hours.