Sumpner's Back-to-Back Test on Transformers

i Learning Objectives

By the end of this chapter you should be able to:

State what the OC and SC tests cannot determine, and why it matters.
Explain the principle of the back-to-back connection.
Describe how the two supplies divide the work between core and windings.
Explain why the mains provides only the losses.
Carry out the polarity check before closing the secondary loop.
Compute each transformer's iron and copper loss from the two wattmeters.
Predetermine the efficiency from the readings.
Compare the test with the OC and SC tests and with direct loading.

Section 47-1

What the Two Tests Cannot Do

Chapter 46 obtained every parameter of the equivalent circuit from two cheap tests. But look at what each test actually subjected the machine to.

Table 47.1 — What each test does and does not apply.
Test	Rated voltage?	Rated current?	Reveals temperature rise?
Open circuit	Yes	No — about 5 %	No
Short circuit	No — about 5 %	Yes	No
Sumpner	Yes	Yes	Yes

🌡

Why This Matters

Rating is a thermal limit, and neither test creates it

In the open-circuit test the iron is fully heated but the copper is not; in the short-circuit test the copper is fully heated but the iron is not. Neither test ever produces both losses at once, so neither produces the real temperature rise.

Since a transformer's kVA rating is set by the temperature its insulation can tolerate, a heat run is required to prove the design — and that means both losses simultaneously, for hours.

This is exactly Chapter 39's argument again. Swinburne's test on a DC machine predicted efficiency but proved nothing about temperature rise or commutation, and Hopkinson's back-to-back test supplied the remedy by coupling two identical machines so the supply provided only the losses. Sumpner's test does the same for transformers — and it is more elegant, because with nothing rotating, no mechanical coupling is needed at all.

Section 47-2

The Principle

Two identical transformers are connected so that each supplies the other. Because the power circulates between them, the mains need only make good what is lost on each pass.

♻

Two Supplies, Two Jobs

Flux from one, current from the other

The primaries are connected in parallel across a supply at rated voltage. This establishes rated flux in both cores, so both develop their full iron loss.
The secondaries are connected in series opposition, and a small voltage is injected into that loop. This circulates rated current through all four windings, so all develop their full copper loss.

Rated voltage and rated current are applied at the same time, yet by two independent sources that each need supply only one loss.

The secondaries oppose one another, so with no injected voltage their EMFs cancel and no current flows in the loop. That is what allows a very small injected voltage to control the current entirely.

Section 47-3

The Connection

Sumpner's back-to-back connection. The two wattmeters read the two losses separately.

Connect the two primaries in parallel across the main supply, observing polarity so that both are excited in the same sense.
Connect the two secondaries in series opposition — that is, so that their EMFs cancel round the loop.
Verify the opposition with a voltmeter across the open end of the loop before closing it. It should read zero.
Close the loop through a low-voltage supply — commonly a small auto-transformer or an induction regulator — with its own wattmeter, ammeter and voltmeter.
Raise the injected voltage until the ammeter shows rated secondary current.

Section 47-4

The First Supply — Iron Loss

With the primaries across rated voltage, each core carries its rated flux, and the whole of the iron loss of both machines appears.

🔩

What W₁ Reads

The total iron loss of both cores

\[W_1 = 2 \times \left(\text{iron loss per transformer}\right)\]

\[P_i = \frac{W_1}{2}\]

The primaries also carry the two exciting currents, so \(A_1\) reads \(2I_0\). No load current flows in the primaries, because the secondary loop's current is supplied entirely by the second source.

? Why the Load Current Does Not Come From Supply 1

Because the two secondaries are in opposition. The current circulating in the secondary loop produces an MMF in each core, but those MMFs are in opposite senses relative to the two primaries.

The counterbalancing primary current that one transformer would draw is therefore supplied by the other — the two primaries exchange it between themselves through their parallel connection. The mains sees only the exciting current.

Section 47-5

The Second Supply — Copper Loss

With no injected voltage the secondary EMFs cancel exactly, so no current flows. Injecting a small voltage drives a current round the loop, limited only by the two transformers' impedances in series.

🔥

What W₂ Reads

The total copper loss of all four windings

\[V_2 = I_2\left(2Z_{02}\right) \quad\Longrightarrow\quad Z_{02} = \frac{V_2}{2I_2}\]

\[W_2 = 2 \times \left(\text{full-load copper loss per transformer}\right), \qquad P_{\text{Cu}} = \frac{W_2}{2}\]

\[R_{02} = \frac{W_2}{2I_2^{2}}, \qquad X_{02} = \sqrt{Z_{02}^{2} - R_{02}^{2}}\]

The injected voltage is about twice the short-circuit voltage of one transformer, since two impedances are in series — around 10 % of rated rather than 5 %.

Note that the second supply is doing precisely what the short-circuit test did, but on two machines at once and while they are also fully fluxed. The core loss it causes is negligible for the same reason as before: the injected voltage produces almost no additional flux, and what flux there is opposes in one core what it aids in the other.

1 Worked Example 47.1 — The Two Supplies

Problem. Two identical 5 kVA, 500/250 V transformers — those of Example 46.5, with 60 W iron loss and 100 W full-load copper loss each, and \(Z_{01} = 2.5~\Omega\) — are tested by Sumpner's method. Find what each supply must provide and what each wattmeter reads.

Rated currents.

\[I_{HV} = \frac{5000}{500} = 10~\mathrm{A}, \qquad I_{LV} = \frac{5000}{250} = 20~\mathrm{A}\]

First supply, across the paralleled primaries at 500 V.

\[W_1 = 2 \times 60 = 120~\mathrm{W}\]

Each transformer's no-load current, referred to the HV side, is \((1.0)(0.5) = 0.5\) A, so \(A_1\) reads 1.0 A — against a rated HV current of 10 A.

Second supply, injected into the secondary loop. Referring the impedance to the LV side:

\[Z_{02} = K^{2}Z_{01} = (0.25)(2.5) = 0.625~\Omega\]

\[V_2 = I_{LV}\left(2Z_{02}\right) = (20)(1.250) = 25.0~\mathrm{V}\]

\[\frac{25.0}{250}\times100 = 10.0\,\%\ \text{of rated LV voltage}\]

\[W_2 = 2 \times 100 = 200~\mathrm{W}\]

Total drawn from the mains.

\[120 + 200 = 320~\mathrm{W}\]

\[\frac{320}{10\,000}\times100 = 3.20\,\%\ \text{of the combined 10 kVA rating}\]

Comment. Two 5 kVA transformers are held at full flux and full current simultaneously on a supply of 320 W — a ratio of 31.25 : 1. That is what makes an all-day heat run affordable.

Note the injected voltage: 10.0 % of rated, exactly twice the 5.0 % percentage impedance found for one machine in Example 46.5. Two impedances in series, so twice the volts. This is a useful check that the connection is correct — if the required voltage comes out at 5 % rather than 10 %, only one transformer is in the loop.

Section 47-6

Procedure and the Polarity Check

⚠ The Polarity Check Is Not Optional

Before closing the secondary loop, a voltmeter is placed across its open ends with the primaries energised at rated voltage.

Reads zero

The secondaries are in opposition. Correct. The loop may be closed through the injecting supply.

Reads twice rated

They are in addition. Reverse one secondary. Closing the loop in this state would short-circuit both transformers through their own EMFs — a fault current of some twenty times rated.

In Example 47.1 the two readings would be 0 V and 500 V — impossible to confuse, provided the check is actually made.

Energise the primaries at rated voltage with the secondary loop open. Record \(W_1\) — this is the iron loss, and it will not change afterwards.
Measure across the open ends of the secondary loop. Confirm zero.
Close the loop through the low-voltage supply, set to zero output.
Raise the injected voltage slowly until \(A_2\) shows rated secondary current. Record \(V_2\) and \(W_2\).
Hold both supplies steady for the duration of the heat run, monitoring winding and oil temperatures until they stabilise.

Because the iron loss is established in step 1 and is unaffected by the loop current, the two wattmeters read the two losses independently and simultaneously — which no other transformer test achieves.

Section 47-7

Working Up the Results

\[P_i = \frac{W_1}{2}, \qquad P_{\text{Cu}} = \frac{W_2}{2}\]

and thereafter every result of Chapters 44 to 46 follows, for a single transformer:

\[\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^{2}P_{\text{Cu}}}, \qquad x_{\max} = \sqrt{\frac{P_i}{P_{\text{Cu}}}}\]

Note the one thing that must be halved and the one thing that must not. The wattmeter readings are for two machines and must be halved. But the current read by \(A_2\) is the current in each transformer, since the two secondaries are in series — it is not halved. Confusing the two is the commonest error in these calculations.

2 Worked Example 47.2 — Efficiency From the Readings

Problem. A Sumpner test on two identical 5 kVA transformers gives \(W_1 = 120\) W and \(W_2 = 200\) W at rated current. Find the losses of one transformer, its full-load efficiency at 0.8 power factor, and the load for maximum efficiency.

Losses per transformer.

\[P_i = \frac{120}{2} = 60~\mathrm{W}, \qquad P_{\text{Cu}} = \frac{200}{2} = 100~\mathrm{W}\]

Full-load efficiency at 0.8 power factor.

\[\eta = \frac{(5000)(0.8)}{4000 + 60 + 100} = \frac{4000}{4160} = 96.15\,\%\]

Maximum efficiency.

\[x = \sqrt{\frac{60}{100}} = 0.7746 \quad\Longrightarrow\quad 3.873~\mathrm{kVA}\]

\[\eta_{\max} = \frac{(3873)(0.8)}{3098 + 120} = 96.27\,\%\]

Comment. These are exactly the figures obtained in Example 46.5 from the separate open-circuit and short-circuit tests. Sumpner's test gives the same answers — and in addition proves the machine can survive the heat.

3 Worked Example 47.3 — Recovering the Impedance

Problem. In the same test, the injected voltage needed to circulate the rated 20 A in the secondary loop is 25.0 V. Find \(Z_{02}\), \(R_{02}\) and \(X_{02}\) for one transformer, and the percentage impedance.

Impedance of one transformer. Two are in series, so

\[Z_{02} = \frac{V_2}{2I_2} = \frac{25.0}{(2)(20)} = \frac{25.0}{40} = 0.6250~\Omega\]

Resistance of one transformer. \(W_2\) is the loss of two, so

\[R_{02} = \frac{W_2}{2I_2^{2}} = \frac{200}{(2)(400)} = 0.2500~\Omega\]

Reactance.

\[X_{02} = \sqrt{(0.6250)^{2} - (0.2500)^{2}} = \sqrt{0.3906 - 0.0625} = 0.5728~\Omega\]

Percentage impedance.

\[\frac{I_2Z_{02}}{V_{LV}}\times100 = \frac{(20)(0.6250)}{250}\times100 = 5.00\,\%\]

Comment. The values agree with Example 46.5 referred to the LV side: \(R_{02} = K^{2}R_{01} = (0.25)(1.000) = 0.250~\Omega\) and \(X_{02} = (0.25)(2.291) = 0.573~\Omega\) \(\checkmark\)

Notice the two different divisors. The voltage is divided by 2 because two impedances share it in series; the power is divided by 2 because two machines dissipate it. But the current is not divided, because the same current passes through both. Getting one of these wrong changes the answer by a factor of two or four.

Section 47-8

Merits and Limitations

Merits

Both transformers run at rated voltage and rated current simultaneously, so a genuine heat run is possible.
The mains supplies only the losses — a few percent of rating.
Iron and copper losses are read on separate wattmeters, so no subtraction is needed.
The load can be varied continuously by the injected voltage, and overload tests cost no more.
No load bank, no coupling and no mechanical arrangement of any kind.

Limitations

Two identical transformers are required — often impossible for a one-off machine.
Two supplies are needed, one of them variable.
The setup is more elaborate than the OC and SC tests.
The load power factor cannot be varied, so the test is at the machine's own impedance angle.
An error in polarity is dangerous if the loop is closed unchecked.

Table 47.2 — The three approaches compared, for the 5 kVA machines of Example 47.1.
	OC + SC	Sumpner	Direct loading
Machines needed	One	Two identical	One
Supply required	250 VA + 250 VA	320 W	5 kVA
Load bank	None	None	4 kW
Rated voltage applied	OC only	Yes	Yes
Rated current applied	SC only	Yes	Yes
Heat run possible	No	Yes	Yes
Losses read separately	Yes	Yes	No
Power factor variable	n/a	No	Yes
Typical use	Routine test	Type test, heat run	Small machines only

4 Worked Example 47.4 — The Cost of a Heat Run

Problem. A six-hour heat run is required on the two 5 kVA transformers. Compare the energy consumed by Sumpner's method with that of loading both directly at 0.8 power factor. Take electricity at ₹8 per kWh.

By Sumpner's method.

\[P = 320~\mathrm{W} = 0.320~\mathrm{kW} \quad\Longrightarrow\quad (0.320)(6) = 1.92~\mathrm{kWh}\]

\[\text{cost} = (1.92)(8) = \text{Rs}~15.36\]

By direct loading of both machines.

\[P = (2)(5000)(0.8) = 8000~\mathrm{W} = 8~\mathrm{kW} \quad\Longrightarrow\quad (8)(6) = 48~\mathrm{kWh}\]

\[\text{cost} = (48)(8) = \text{Rs}~384\]

Ratio.

\[\frac{48}{1.92} = 25 \quad\text{times less energy}\]

Comment. The saving is a factor of 25, and it is not merely the money. Direct loading also requires a 4 kW load bank per machine, and somewhere to put 8 kW of heat for six hours. Sumpner's method needs neither.

The advantage grows with size. On a pair of 5 MVA transformers with 1 % total loss the mains would provide 100 kW instead of 8 MW — the difference between a routine works test and an impossibility.

5 Worked Example 47.5 — Running at Other Loads

Problem. The injected voltage is reduced so that the loop current falls to half, three-quarters and then raised to 125 % of rated. Tabulate \(W_2\), the total mains power, and the efficiency each transformer would show at 0.8 power factor.

Reasoning. \(W_1\) is fixed at 120 W by the primary voltage. \(W_2\) varies as the square of the loop current.

Table 47.3 — Sumpner test at several loop currents.
Load \(x\)	\(I_2\) (A)	\(W_1\) (W)	\(W_2\) (W)	Mains total (W)	Efficiency
0.50	10	120	50.0	170.0	95.92 %
0.75	15	120	112.5	232.5	96.27 %
1.00	20	120	200.0	320.0	96.15 %
1.25	25	120	312.5	432.5	95.85 %

Sample working at \(x = 0.75\).

\[W_2 = (200)(0.75)^{2} = 112.5~\mathrm{W}, \qquad P_{\text{Cu}} = 56.25~\mathrm{W}\]

\[\eta = \frac{(5000)(0.75)(0.8)}{3000 + 60 + 56.25} = \frac{3000}{3116.25} = 96.27\,\%\]

Comment. The efficiency peaks near \(x = 0.775\), as Example 47.2 predicted, and the table shows it clearly at 0.75. A single Sumpner setup therefore produces the whole efficiency curve, simply by turning one knob.

Note that the 125 % overload costs only 432.5 W from the mains — 4.3 % of the combined rating. Testing a transformer beyond its rating is essentially free by this method, which is why it is the standard type test.

Section 47-9

Summary and Key Formulas

The OC test applies rated voltage without rated current; the SC test rated current without rated voltage. Neither reveals the temperature rise, which sets the rating.
Sumpner's test connects two identical transformers with primaries in parallel and secondaries in series opposition.
The first supply, at rated voltage, establishes rated flux in both cores; \(W_1\) reads the total iron loss.
The second supply, at about twice the percentage impedance, circulates rated current; \(W_2\) reads the total copper loss.
The mains supplies only the losses, because the load current is exchanged between the two primaries.
Losses per transformer are \(W_1/2\) and \(W_2/2\). The current is not halved — the secondaries are in series.
A polarity check is essential: the open loop must read zero, not twice rated voltage.
Merits: heat run possible, losses read separately, overload free. Limitations: two identical machines, two supplies, fixed power factor.

Table 47.4 — Formulas of this chapter.
Quantity	Formula	Notes
Iron loss per transformer	\(P_i = W_1/2\)	halve the reading
Copper loss per transformer	\(P_{\text{Cu}} = W_2/2\)	halve the reading
Impedance per transformer	\(Z_{02} = \dfrac{V_2}{2I_2}\)	two in series
Resistance per transformer	\(R_{02} = \dfrac{W_2}{2I_2^{2}}\)	\(I_2\) not halved
Reactance per transformer	\(X_{02} = \sqrt{Z_{02}^{2} - R_{02}^{2}}\)	—
Injected voltage	\(V_2 = 2I_2Z_{02}\)	≈ 2 × % impedance
Primary ammeter	\(A_1 = 2I_0\)	exciting current only
Total mains power	\(W_1 + W_2\)	= all losses of both
Efficiency	\(\eta = \dfrac{xS\cos\phi}{xS\cos\phi + P_i + x^{2}P_{\text{Cu}}}\)	per transformer
Maximum-efficiency load	\(x = \sqrt{P_i/P_{\text{Cu}}}\)	—

Section 47-10

Common Mistakes

Halving the current as well as the readings. The secondaries are in series, so \(A_2\) reads the current in each.
Forgetting to halve the wattmeter readings. Both are for two machines.
Using \(Z_{02} = V_2/I_2\). Two impedances share the injected voltage, so divide by \(2I_2\).
Connecting the secondaries in addition. The open loop must read zero, not twice rated.
Closing the loop without the polarity check. In addition, the two EMFs short-circuit both machines.
Expecting \(A_1\) to read rated current. It reads only \(2I_0\) — the load current is exchanged between the primaries.
Thinking \(W_1\) changes when load is applied. The flux is fixed by the primary voltage.
Applying the test to non-identical transformers. The secondary EMFs would not cancel, and the losses could not be halved.
Expecting to vary the power factor. The loop current is set by the transformers' own impedance angle.
Confusing this with Hopkinson's test. Same idea, but Hopkinson's needs a mechanical coupling; Sumpner's needs none.

Section 47-11

Chapter Review

Practice Problems

Remember the asymmetry: halve the two wattmeter readings and the injected voltage, but never the loop current.

P47.1 A Sumpner test on two identical transformers gives \(W_1 = 300\) W and \(W_2 = 480\) W. Find the iron and copper losses of one transformer.

Show answer

\[P_i = \frac{300}{2} = 150~\mathrm{W}, \qquad P_{\text{Cu}} = \frac{480}{2} = 240~\mathrm{W}\]
P47.2 The machines of P47.1 are rated 20 kVA. Find the full-load efficiency at 0.9 power factor.

Show answer

\[\eta = \frac{(20\,000)(0.9)}{18\,000 + 150 + 240} = \frac{18\,000}{18\,390} = 97.88\,\%\]
P47.3 For P47.2, find the load for maximum efficiency and its value.

Show answer

\[x = \sqrt{\frac{150}{240}} = 0.7906 \quad\Longrightarrow\quad 15.81~\mathrm{kVA}\]

\[\eta_{\max} = \frac{(15\,811)(0.9)}{14\,230 + 300} = \frac{14\,230}{14\,530} = 97.94\,\%\]
P47.4 The 20 kVA transformers are 2000/400 V. If the injected loop voltage is 40 V at the rated LV current, find \(Z_{02}\) and \(R_{02}\) for one machine.

Show answer

\[I_{LV} = \frac{20\,000}{400} = 50~\mathrm{A}\]

\[Z_{02} = \frac{40}{(2)(50)} = 0.4000~\Omega, \qquad R_{02} = \frac{480}{(2)(2500)} = 0.0960~\Omega\]

\[X_{02} = \sqrt{0.1600 - 0.00922} = 0.3883~\Omega\]
P47.5 For P47.4, find the percentage impedance of one transformer and the injected voltage as a percentage of rated LV.

Show answer

\[\%Z = \frac{(50)(0.400)}{400}\times100 = 5.00\,\%\]

\[\frac{40}{400}\times100 = 10.0\,\% \quad = 2 \times 5.00\,\% \quad\checkmark\]
P47.6 The loop current in P47.4 is reduced to 40 A. Find the new \(W_2\) and the total mains power.

Show answer

\[W_2 = (480)\left(\frac{40}{50}\right)^{2} = (480)(0.64) = 307.2~\mathrm{W}\]

\[\text{total} = 300 + 307.2 = 607.2~\mathrm{W}\]
P47.7 Why can neither the OC nor the SC test reveal the temperature rise?

Show answer
Because each applies only one of the two rated quantities.
The OC test applies rated voltage, so the core is fully fluxed and the iron loss is fully developed — but the current is only about 5 % of rated, so the copper loss is \((0.05)^{2} = 0.25\,\%\) of its full value. The windings barely warm.

The SC test applies rated current, so the copper loss is fully developed — but the voltage is only about 5 % of rated, so the flux and hence the iron loss are negligible. The core barely warms.

Neither test ever produces both losses at once, so neither produces the real temperature rise. Since the kVA rating is fixed by what the insulation can tolerate thermally, a heat run with both losses present is required to prove the design — and that is precisely what Sumpner's test provides.
P47.8 Explain why the mains supplies only the losses, even though both transformers carry rated current.

Show answer
Because the load current is exchanged between the two primaries rather than drawn from the mains.
The current circulating in the secondary loop produces an MMF in each core. Since the two secondaries are connected in opposition, those MMFs act in opposite senses relative to the two primaries. The counterbalancing primary current that one transformer demands is therefore exactly what the other supplies, and the two primaries pass it between themselves through their parallel connection.

The mains sees only the exciting current of the two machines, so \(A_1\) reads \(2I_0\) — in Example 47.1, 1.0 A against a rated 10 A.

Meanwhile the injecting supply provides only enough voltage to push rated current through two winding impedances in series — about 10 % of rated. Between them the two sources supply the iron loss and the copper loss and nothing else.
P47.9 Describe the polarity check and explain what happens if it is omitted.

Show answer
With the primaries energised at rated voltage and the secondary loop still open, a voltmeter is connected across the open ends.
If it reads zero, the secondaries are in opposition — correct. The loop may be closed through the injecting supply.

If it reads twice the rated secondary voltage, they are in addition. One secondary must be reversed.

If the check is omitted and the loop is closed in the additive state, the two secondary EMFs — some twice rated voltage — are applied to the two winding impedances in series. Since each transformer has about 5 % impedance, the resulting current is roughly \(200/10 = 20\) times rated.

That is a dead short circuit on both machines, with mechanical forces some 400 times normal. The check takes seconds and the consequence of skipping it is destruction of both transformers.
P47.10 Compare Sumpner's test with Hopkinson's test of Chapter 39.

Show answer
The principle is identical: two identical machines are connected so that each supplies the other, the power circulates, and the mains provides only the losses — permitting a full-load heat run at a fraction of the rating.
Hopkinson's test (DC machines) requires a mechanical coupling: one machine motors and drives the other as a generator on a common shaft. The load is adjusted by the two field rheostats, and the stray losses must be found by subtraction and assumed equally divided.

Sumpner's test (transformers) needs no coupling at all, since nothing rotates. The load is set by one injected voltage, and — a real advantage — the iron and copper losses appear on separate wattmeters, so no subtraction and no assumption of equal division is needed.

Both share the same limitation: two identical machines are required, which restricts them to type testing rather than routine work.

Multiple-Choice Questions

MCQ 1. Sumpner's test requires:
(a) one transformer (b) two identical transformers (c) a load bank (d) a mechanical coupling

Show answer
(b) two identical transformers. No coupling and no load bank are needed.
MCQ 2. In Sumpner's test the primaries are connected:
(a) in series (b) in parallel (c) in opposition (d) open

Show answer
(b) in parallel, across a supply at rated voltage.
MCQ 3. The secondaries are connected:
(a) in parallel (b) in series addition (c) in series opposition (d) short-circuited

Show answer
(c) in series opposition, so their EMFs cancel.
MCQ 4. The first wattmeter \(W_1\) reads:
(a) copper loss of both (b) iron loss of both (c) total loss (d) output

Show answer
(b) iron loss of both cores, since both are at rated flux.
MCQ 5. The injected voltage is about:
(a) rated (b) half rated (c) twice the percentage impedance (d) zero

Show answer
(c) twice the percentage impedance, because two impedances are in series.
MCQ 6. The impedance of one transformer is:
(a) \(V_2/I_2\) (b) \(V_2/2I_2\) (c) \(2V_2/I_2\) (d) \(V_2/4I_2\)

Show answer
(b) \(V_2/2I_2\) — the voltage is shared, the current is not.
MCQ 7. The ammeter in the primary circuit reads:
(a) rated current (b) twice rated (c) \(2I_0\) (d) zero

Show answer
(c) \(2I_0\). The load current is exchanged between the two primaries.
MCQ 8. Before closing the secondary loop, the voltmeter across its open ends should read:
(a) rated voltage (b) twice rated (c) zero (d) half rated

Show answer
(c) zero, confirming opposition. Twice rated means one secondary must be reversed.
MCQ 9. The chief advantage of Sumpner's test over OC and SC tests is that it:
(a) is cheaper (b) permits a heat run (c) needs one machine (d) varies power factor

Show answer
(b) permits a heat run, since rated voltage and rated current are applied together.
MCQ 10. Sumpner's test is the transformer equivalent of:
(a) Swinburne's test (b) Hopkinson's test (c) the retardation test (d) the brake test

Show answer
(b) Hopkinson's test — the back-to-back principle, without the mechanical coupling.

Conceptual Questions

Explain what the OC and SC tests cannot determine and why it matters.
Describe the Sumpner connection and the function of each of the two supplies.
Explain why the mains supplies only the losses.
Describe the polarity check and the consequence of omitting it.
Derive the expressions for the losses and the impedance of one transformer.
Explain why the injected voltage is twice the percentage impedance.
Give the merits and limitations of the test.
Compare Sumpner's test with Hopkinson's test.

Looking Ahead

The single-phase transformer is now complete as a subject: constructed in Chapter 40, modelled in Chapters 41 to 44, its regulation derived in Chapter 45, and every parameter and its temperature rise measured in Chapters 46 and 47.

Chapter 48 returns to efficiency in more detail — including the condition for maximum efficiency in terms of kVA rather than load fraction, and the all-day efficiency of Chapter 44 revisited for realistic duty cycles — and then introduces the autotransformer, in which a single tapped winding serves as both primary and secondary.

The autotransformer's appeal is a striking saving in copper: the weight required falls in the ratio \((1 - K)\) compared with a two-winding transformer of the same rating. At a ratio near unity the saving approaches 100 %, which is why autotransformers dominate where the two voltages are close — starting compensators, interconnecting transformers between transmission voltages, and variable-voltage laboratory supplies. The price is the loss of electrical isolation, and Chapter 48 examines when that price is acceptable.