Transformer on No Load

Chapter 41 assumed a loss-free core of infinite permeability. Both assumptions now go.

With the secondary open, the primary draws a small current \(I_0\), called the no-load or exciting current. It does two jobs at once.

Typical magnitudes: \(I_0\) is 2 to 6 % of full-load current in a power transformer, and rather more in small units. Chapter 46 will use exactly this measurement as the open-circuit test.

If \(\phi_0\) is the angle by which \(I_0\) lags \(V_1\), the two components are obtained by resolving along and perpendicular to \(V_1\).

The whole of the no-load calculation follows from two meter readings — \(I_0\) from an ammeter and \(W_0\) from a wattmeter — together with the applied voltage.

The diagram is built in a fixed order, and it is worth learning that order rather than the picture.

1. Draw the flux \(\Phi\) as the horizontal reference. Everything in the magnetic circuit refers to it.
2. The induced EMFs lag the flux by 90°, from \(e = -N\,\mathrm{d}\Phi/\mathrm{d}t\). Draw \(E_1\) and \(E_2\) pointing downward.
3. \(V_1\) must balance \(E_1\), so \(V_1 = -E_1\) points upward, leading the flux by 90°.
4. Set out \(I_\mu\) in phase with \(\Phi\) and \(I_w\) in phase with \(V_1\).
5. Their sum is \(I_0\), lagging \(V_1\) by \(\phi_0\).

For a step-down transformer \(E_2 \lt E_1\), as drawn. Both point the same way, since both windings link the same flux — their relative direction is a matter of winding sense, which is what the dot convention of Chapter 41 records.

This is not a defect. The no-load current is small in absolute terms, so its poor power factor scarcely matters — and at any appreciable load it is swamped by the load component. It matters only for a transformer left energised but unloaded, which is why distribution transformers on lightly loaded feeders are a recognised source of reactive burden on a network.

The two components can be represented by two circuit elements in parallel across the supply. This is the exciting or no-load branch, and it is the first piece of the equivalent circuit that Chapter 44 completes.

The wattmeter reads the total iron loss. The two components can be separated by testing at two frequencies, exactly as the retardation test of Chapter 39 separated iron from friction.

One assumption has been made silently throughout: that \(I_\mu\) is sinusoidal. It is not.

1. The applied voltage is sinusoidal, so — from \(V_1 \approx E_1 = -N_1\,\mathrm{d}\Phi/\mathrm{d}t\) — the flux must be sinusoidal.
2. But the relation between flux and current is the B-H curve of Chapter 5, which is non-linear and hysteretic.
3. To produce a sinusoidal flux through a non-linear characteristic, the current must be non-sinusoidal.

For single-phase analysis the effect is handled by using the equivalent sinusoidal current — a sine wave of the same RMS value — which is what \(I_\mu\) has meant throughout this chapter. The approximation is good because the harmonics contribute little to the RMS total.

Everything so far describes the steady state. The instant of switching on is quite different, and produces the largest current a healthy transformer ever draws.

• Relaxing the loss-free core and infinite permeability gives a real no-load current \(I_0\), typically 2 to 6 % of full-load current.

• \(I_0\) resolves into \(I_w = I_0\cos\phi_0\), supplying the iron loss, and \(I_\mu = I_0\sin\phi_0\), producing the flux.

• In the phasor diagram \(\Phi\) is the reference, \(E_1\) and \(E_2\) lag it by 90°, \(V_1 = -E_1\) leads it by 90°, and \(I_0\) lies between \(V_1\) and \(\Phi\).

• The no-load power factor is low, 0.1 to 0.2, because \(I_\mu \gg I_w\). Better core steel makes it worse.

• The exciting branch is \(R_0 = V_1^{2}/W_0\) in parallel with \(X_0 = V_1/I_\mu\). Neither is truly constant.

• The iron loss is constant at all loads, since the flux is fixed by the supply — which is what makes an open-circuit test valid.

• Testing at two frequencies at constant \(V/f\) separates the losses, since \(W_0/f = A + Bf\).

• The magnetising current is non-sinusoidal and peaky, rich in third harmonic, because the flux must be sinusoidal through a non-linear B-H curve.

• Switching inrush can reach 8 to 12 times full-load current, because switching at a voltage zero with residual flux demands up to \(2.5\Phi_m\).

• Swapping \(\cos\) and \(\sin\). \(I_w = I_0\cos\phi_0\) is the power component; \(I_\mu = I_0\sin\phi_0\) is the larger one.

• Computing \(R_0\) as \(V_1/I_0\). It is \(V_1/I_w\), equivalently \(V_1^{2}/W_0\).

• Putting \(R_0\) and \(X_0\) in series. They are in parallel, both across \(V_1\).

• Treating \(W_0\) as including copper loss. At 5 % of full-load current the copper loss is 0.25 % of its full-load value — negligible.

• Expecting the iron loss to change with load. The flux is fixed by the supply, so it does not.

• Running a two-frequency test at constant voltage. The voltage must scale with frequency to hold \(B_m\) fixed.

• Assuming \(X_0\) is a constant. It depends on permeability and so on the voltage.

• Drawing \(I_0\) close to \(V_1\). With a power factor of 0.19 it lies close to \(\Phi\), nearly 80° from \(V_1\).

• Believing the magnetising current is sinusoidal. It is peaked and rich in third harmonic.

• Confusing inrush with a fault. Inrush carries a strong second harmonic; a fault does not.

Find \(\cos\phi_0\) first from \(W_0/V_1I_0\); both components and both parameters follow at once.

1. P42.1 A 440 V transformer takes 1.0 A at 150 W on no load. Find the power factor and both components.

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   \[\cos\phi_0 = \frac{150}{(440)(1.0)} = 0.3409, \qquad \sin\phi_0 = 0.9401\]

   \[I_w = 0.3409~\mathrm{A}, \qquad I_\mu = 0.9401~\mathrm{A}\]

2. P42.2 For P42.1, find \(R_0\) and \(X_0\).

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   \[R_0 = \frac{V_1^{2}}{W_0} = \frac{(440)^{2}}{150} = 1291~\Omega\]

   \[X_0 = \frac{440}{0.9401} = 468.0~\Omega\]

3. P42.3 A 3300 V transformer has \(I_w = 0.15\) A and \(I_\mu = 0.90\) A. Find \(I_0\), the power factor and the iron loss.

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   \[I_0 = \sqrt{(0.15)^{2} + (0.90)^{2}} = \sqrt{0.8325} = 0.9124~\mathrm{A}\]

   \[\cos\phi_0 = \frac{0.15}{0.9124} = 0.1644, \qquad W_0 = (3300)(0.15) = 495~\mathrm{W}\]

4. P42.4 A transformer gives 400 W at 500 V, 50 Hz and 168 W at 250 V, 25 Hz. Separate the losses at 50 Hz.

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   \(V/f\) is 10 in both cases, so \(B_m\) is unchanged.

   \[\frac{400}{50} = 8.0, \qquad \frac{168}{25} = 6.72\]

   \[B = \frac{8.0 - 6.72}{25} = 0.0512, \qquad A = 8.0 - 2.56 = 5.44\]

   \[P_h = (5.44)(50) = 272~\mathrm{W}, \qquad P_e = (0.0512)(2500) = 128~\mathrm{W}\]

5. P42.5 A 100 kVA transformer has an iron loss of 900 W and a full-load copper loss of 1600 W. Find the load for maximum efficiency and that efficiency at unity power factor.

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   \[x = \sqrt{\frac{900}{1600}} = 0.75 \quad\Longrightarrow\quad 75~\mathrm{kVA}\]

   \[\eta_{\max} = \frac{75\,000}{75\,000 + 1800} = 97.66\,\%\]

6. P42.6 A transformer draws 2.5 A on no load. Its full-load current is 50 A. Express \(I_0\) as a percentage, and find the no-load copper loss as a fraction of the full-load value.

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   \[\frac{2.5}{50} = 5\,\%\]

   \[\frac{I_0^{2}R_1}{I_{1}^{2}R_1} = (0.05)^{2} = 0.0025 = 0.25\,\%\]

   This is why the no-load copper loss is neglected and \(W_0\) taken entirely as iron loss.

7. P42.7 Explain why the no-load power factor of a transformer is so low, and why improving the core steel makes it lower still.

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   Because \(\cos\phi_0 = I_w/I_0\) and the magnetising component \(I_\mu\) is far larger than the working component \(I_w\). The core must carry a substantial flux — needing real MMF — while the iron loss is deliberately kept small.

   Better steel makes it worse because it reduces the hysteresis loop area, and hence \(I_w\), without much reducing the MMF needed to drive the flux. \(I_w\) falls, \(I_\mu\) does not, so the ratio falls.

   This is not a defect. The absolute value of \(I_0\) is what matters, and it is small. At any real load the current is dominated by the load component and the overall power factor is that of the load.

8. P42.8 Why is the magnetising current non-sinusoidal, and why does this matter in three-phase systems?

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   The applied voltage is sinusoidal, so from \(v \approx -N\,\mathrm{d}\Phi/\mathrm{d}t\) the flux must be sinusoidal. But flux and current are related by the non-linear, hysteretic B-H curve. Producing a sinusoidal flux through a non-linear characteristic requires a non-sinusoidal current — peaked, because near the flux peak the core enters saturation and disproportionately more current is needed.

   Fourier analysis shows a strong third harmonic, typically 10–40 % of the fundamental.

   In three-phase systems the third harmonics of the three phases are all in phase with one another. They therefore cannot circulate in a star connection without a neutral, and their suppression distorts the flux and the phase voltages. Providing a delta winding gives them a path in which to circulate, which is a principal reason for the delta connection.

9. P42.9 Explain the cause of switching inrush and why it is worst at a voltage zero.

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   Flux is the integral of the applied voltage: \(\Phi = \frac{1}{N_1}\int v\,\mathrm{d}t\). The steady-state flux wave must lag the voltage by 90°, so it is at its negative peak when the voltage passes through zero going positive.

   If the transformer is switched on at a voltage zero, the flux must start from whatever is in the core (zero, or the residual) and integrate over a full half cycle, reaching \(2\Phi_m\) — twice its normal peak. With residual flux of, say, \(0.5\Phi_m\) the demand becomes \(2.5\Phi_m\).

   The core cannot carry this; it saturates deeply, its permeability collapses towards that of air, and the current needed rises to 8–12 times full load.

   Switching at the voltage peak is the benign case: the flux then starts from zero exactly as the steady state requires, and there is no transient at all.

10. P42.10 How can a protective relay distinguish inrush from an internal fault, given that both give a large current?

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    By the harmonic content. Inrush current is the result of one-sided saturation, so its waveform is strongly asymmetrical and contains a large second harmonic — commonly 15–25 % of the fundamental. A genuine internal fault produces a substantially sinusoidal current with very little second harmonic.

    Differential relays therefore use second-harmonic restraint: the trip is blocked while the second-harmonic content exceeds a set fraction.

    The distinction matters because the two demand opposite responses. An inrush is harmless and decays in a few cycles, so tripping would be a nuisance; a fault must be cleared at once.