Ideal Transformer and the EMF Equation

• An ideal transformer is one which has no losses: its windings have no ohmic resistance, there is no magnetic leakage, and hence there are no \(I^{2}R\) losses and no core losses.

• In other words, it consists of two purely inductive coils wound on a loss-free core.

• It is impossible to realise such a transformer in practice.

Consider an ideal transformer whose secondary is open and whose primary is connected to a sinusoidal alternating voltage \(V_1\).

1. The potential difference causes an alternating current to flow in the primary.
2. Since the primary coil is purely inductive and the secondary is open — no output — the primary draws a magnetising current \(I_\mu\) only.
3. \(I_\mu\) is very small in magnitude, magnetises the core, and lags \(V_1\) by 90°.
4. The alternating current \(I_\mu\) produces an alternating flux \(\Phi\), which at all times is proportional to the current and in phase with it.

This changing flux is linked with both \(N_1\) and \(N_2\), and therefore induces an EMF in each:

• A self-induced EMF in the primary, \(E_1\), which at every instant opposes \(V_1\) — the counter EMF or back EMF of the primary.

• A mutually induced EMF in the secondary, \(E_2\), in antiphase with \(V_1\).

• The magnitude of \(E_2\) is proportional to the rate of change of flux and to the number of secondary turns.

Let

Multiplying by the number of turns in each winding gives the two EMFs:

This single fact is what makes transformer design tractable. The designer fixes \(E_t\) from the core — as Example 40.3 did — and every winding on that core then follows by simple division.

For an ideal transformer, input VA equals output VA, since there are no losses:

Combining \(I_2/I_1 = 1/K\) with \(K = N_2/N_1\) gives an equivalent and often more useful statement:

This is the transformer's version of the self-regulation seen in Chapter 34. The primary does not "know" the load; it responds automatically because any imbalance in ampere-turns would disturb the flux, and disturbing the flux immediately alters \(E_1\), which changes the primary current until balance is restored.

Because the transformer scales voltage by \(K\) and current by \(1/K\), it scales impedance by the square of the ratio. This is the most useful single idea in transformer analysis.

The consequence is that a transformer with a load can be replaced by a single impedance on the primary side, and the whole two-circuit problem collapses to one circuit. Chapter 44 uses this to fold the secondary resistance and leakage reactance into the primary, producing the equivalent circuit that carries the rest of Part 3.

Rearranging the EMF equation exposes a fact that governs transformer behaviour everywhere in Part 3.

The same expression shows what happens if the supply changes. Since \(B_m \propto V_1/f\), it is the volts per hertz that matters, not the voltage alone — a rule that recurs throughout AC machine practice.

The EMF equation gives magnitudes. The relative sense of the two windings is shown by dots marked on the terminals.

Two conventions are used in practice: additive polarity, where the induced voltages add around the loop when adjacent terminals are joined, and subtractive polarity, where they oppose. Subtractive polarity is standard for large transformers, because it keeps the voltage between adjacent bushings low.

• An ideal transformer has no winding resistance, no leakage flux, no core loss and infinite permeability. It cannot be realised, but a real transformer is close to it.

• On no load the primary draws only the magnetising current \(I_\mu\), which lags \(V_1\) by 90° and is in phase with the flux it produces.

• The flux covers \(\Phi_m\) in a quarter cycle, giving an average EMF per turn of \(4f\Phi_m\); multiplying by the form factor 1.11 gives \(4.44f\Phi_m\).

• Volts per turn is the same on both windings, because both link the same flux at the same frequency.

• The transformation ratio is \(K = N_2/N_1 = E_2/E_1 = V_2/V_1\), and \(I_2/I_1 = 1/K\).

• Ampere-turns balance: \(N_1I_1 = N_2I_2\). The secondary MMF is cancelled as fast as it appears, so the flux is undisturbed.

• Impedance refers as \(K^{2}\): \(Z' = Z/K^{2}\), with \(V' = V/K\) and \(I' = KI\). Power is unchanged.

• Flux is fixed by the supply, \(\Phi_m = V_1/4.44fN_1\), so iron loss and magnetising current are constant at all loads. What matters is the volts per hertz.

• Writing 4.44 as 4. The 4 is the quarter-cycle factor; the form factor 1.11 must multiply it.

• Using peak instead of RMS. \(4.44fN\Phi_m\) gives the RMS EMF directly.

• Using \(B_m\) with the gross core area. Use the net iron area, \(k_sA_{\text{gross}}\).

• Defining \(K\) upside down. Here \(K = N_2/N_1\), so \(K \gt 1\) means step-up. Some books use the reciprocal — check before quoting a formula.

• Referring impedance with \(K\) instead of \(K^{2}\). Voltage and current each scale by \(K\), so impedance scales by its square.

• Thinking the flux depends on the load. It is set by \(V_1\), \(f\) and \(N_1\) alone.

• Expecting the frequency to change. Both sides are always at supply frequency.

• Running a transformer at reduced frequency at rated voltage. \(B_m \propto V/f\), so the core saturates.

• Forgetting the power check after referring. \(I'^{2}Z'\) must equal \(I^{2}Z\).

• Ignoring polarity when paralleling. Opposed polarities produce a short circuit.

Find the volts per turn first; almost everything else follows by multiplication or division.

1. P41.1 A 50 Hz transformer has a net core area of 0.012 m² and works at 1.5 T. Find the volts per turn.

   Show answer

   \[\Phi_m = (1.5)(0.012) = 0.0180~\mathrm{Wb}\]

   \[E_t = (4.44)(50)(0.0180) = 3.996~\mathrm{V/turn}\]

2. P41.2 For P41.1, find the turns for a 3300/240 V transformer.

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   \[N_1 = \frac{3300}{3.996} = 825.8 \to 826, \qquad N_2 = \frac{240}{3.996} = 60.1 \to 60\]

   With 826 and 60 turns, \(K = 60/826 = 0.07264\) and the actual secondary voltage is \((0.07264)(3300) = 239.7\) V.

3. P41.3 A 400/100 V transformer supplies 20 A to its secondary. Find \(K\), the primary current and the ampere-turns if \(N_2 = 50\).

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   \[K = \frac{100}{400} = 0.25, \qquad I_1 = KI_2 = (0.25)(20) = 5~\mathrm{A}\]

   \[N_1 = \frac{50}{0.25} = 200, \qquad N_1I_1 = 1000 = N_2I_2 = (50)(20) \quad\checkmark\]

4. P41.4 A 25 \(\Omega\) load is on the secondary of a 1100/220 V transformer. Find the referred impedance and both currents.

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   \[K = 0.2, \qquad Z' = \frac{25}{0.04} = 625~\Omega\]

   \[I_2 = \frac{220}{25} = 8.8~\mathrm{A}, \qquad I_1 = (0.2)(8.8) = 1.76~\mathrm{A}\]

   \[\text{check: } \frac{1100}{1.76} = 625~\Omega \quad\checkmark\]

5. P41.5 An amplifier requires a 4500 \(\Omega\) load and drives a 5 \(\Omega\) loudspeaker. Find the turns ratio needed.

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   \[K^{2} = \frac{Z_2}{Z'} = \frac{5}{4500} = \frac{1}{900} \quad\Longrightarrow\quad K = \frac{1}{30}\]

   A 30 : 1 step-down transformer. For 900 primary turns, the secondary would have 30.

6. P41.6 A transformer designed for 400 V, 50 Hz is connected to 400 V, 40 Hz. Find the change in flux density and comment.

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   \[\frac{B_2}{B_1} = \frac{f_1}{f_2} = \frac{50}{40} = 1.25\]

   The flux density rises 25 %. If the design value was 1.5 T it becomes 1.875 T, approaching saturation — the magnetising current would rise sharply and the transformer would overheat. To use it at 40 Hz the voltage must be reduced to \((400)(40/50) = 320\) V.

7. P41.7 Derive the EMF equation, stating the origin of each factor in 4.44.

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   The flux rises from zero to \(\Phi_m\) in a quarter cycle, that is in \(1/4f\) seconds, so the average rate of change is \(\Phi_m/(1/4f) = 4f\Phi_m\) Wb/s. Rate of change of flux per turn is EMF in volts, so the average EMF per turn is \(4f\Phi_m\).

   For a sinusoid the RMS value is the average multiplied by the form factor \(\pi/2\sqrt2 = 1.11\), giving \(E_t = 4.44f\Phi_m\) and hence \(E = 4.44fN\Phi_m\).

   Origins: the 4 is geometric — the flux traverses its full amplitude in a quarter period. The 1.11 is the sine wave's form factor. A square-wave supply would give 4.00, not 4.44.

8. P41.8 Explain why the volts per turn is the same for both windings.

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   Both windings are wound on the same core and are therefore linked by the same flux, alternating at the same frequency. Faraday's law says the EMF induced in a single turn is the rate of change of the flux linking it — so every turn on that core, in whichever winding, has the same EMF induced in it.

   The windings differ only in how many such turns they contain, which is why \(E_1/N_1 = E_2/N_2 = 4.44f\Phi_m\).

   This is why leakage flux matters: it is precisely the flux that links one winding but not the other, and it is the reason a real transformer's volts per turn differ slightly between windings.

9. P41.9 Explain why the core flux is independent of the load, and give two consequences.

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   From the EMF equation, \(\Phi_m = V_1/(4.44fN_1)\) — and no load current appears. With the supply voltage, frequency and primary turns all fixed, the flux is fixed.

   Mechanism: when the secondary carries current, its MMF \(N_2I_2\) would tend to alter the flux. Any such change alters \(E_1\), which unbalances \(V_1 - E_1\) and immediately draws additional primary current until \(N_1I_1 = N_2I_2\). The secondary MMF is cancelled as fast as it appears.

   Consequences: the iron loss is constant at all loads (so it can be measured at no load, as Chapter 46 does), and the magnetising current is constant at all loads.

10. P41.10 Why may a 50 Hz transformer be used on 60 Hz but not the reverse?

    Show answer

    Because \(B_m \propto V_1/f\). At higher frequency and the same voltage the flux density falls, so a 50 Hz unit on 60 Hz works at \(50/60 = 83\) % of its design flux density — safe, merely under-using the core.

    At lower frequency the flux density rises. A 60 Hz unit on 50 Hz runs at \(60/50 = 120\) % of design flux density, and since the design point is already near the knee of the B-H curve, the core saturates. The magnetising current rises steeply and non-sinusoidally, and the transformer overheats.

    The remedy is to reduce the voltage in proportion, keeping the volts per hertz at its design value.