Solved Problems · Set 15

Transformer Fundamentals and the EMF Equation

Part 3 · Transformers — one equation, E = 4.44 f N Φm, fixes the turns, the core area and the flux density of every transformer in the book, and this set drills it from both ends.

Prof. Mithun Mondal 8 solved problems GATE · ESE · University

Set 15 — Transformer Fundamentals and the EMF Equation

A transformer has no moving part and no commutator, so almost everything about its design follows from a single relation between the applied voltage, the number of turns and the flux that has to alternate in the core. This set establishes that relation and then uses it in both directions: given the voltages, find the turns and the core area; given the turns and the core, find the flux density and check it against what the iron will stand.

The second thread is the turns ratio. Voltage scales with it, current scales inversely with it, and the volt-ampere rating is the same on both sides — which is all that is needed to size the windings of an ideal transformer at full load. The last problem adds a power factor, and Problem 5 shows how the same per-turn arithmetic handles a three-phase core once every quantity is taken per phase.

Part 3 · Single-Phase Transformers · 8 solved problems

i Method Recap

  • The e.m.f. equation. A sinusoidal flux \(\Phi = \Phi_m\sin\omega t\) linking \(N\) turns induces an average of \(4f\Phi_mN\), and multiplying by the form factor 1.11 gives the r.m.s. value:

    \[ E = 4.44\,f\,N\,\Phi_m = 4.44\,f\,N\,B_mA \]

  • E.m.f. per turn is the same in both windings, because both link the same core flux. It is often the quickest way into a problem:

    \[ \frac{E_1}{N_1} = \frac{E_2}{N_2} = 4.44\,f\,\Phi_m \]

  • The turns ratio \(a = N_1/N_2\) ties voltages and currents in opposite senses, since an ideal transformer conserves volt-amperes:

    \[ \frac{E_1}{E_2} = \frac{N_1}{N_2} = a, \qquad \frac{I_1}{I_2} = \frac{1}{a} \]

  • Net iron area, not gross area. The core is a stack of insulated laminations, so the area carrying flux is the gross cross-section times the stacking factor:

    \[ A_{\text{net}} = k_s A_{\text{gross}}, \qquad \Phi_m = B_mA_{\text{net}} \]

  • Rated currents come from the kVA, which is the same on both sides:

    \[ I_1 = \frac{S}{V_1}, \qquad I_2 = \frac{S}{V_2} \]

  • Three-phase cores are worked per phase. Each limb carries one phase winding, so use the phase voltage, the phase current and one third of the rated kVA — then convert to line quantities only at the end.

VideoWalkthrough
Problem 1CoreEMF per Turn and Core Area

The maximum flux density in the core of a 250/3000 V, 50 Hz single-phase transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 V, determine

  1. the primary and secondary turns
  2. the area of the core.
Solution

The e.m.f. per turn is common to both windings, since both are wound on the same limb and link the same flux. Dividing each winding voltage by it gives the turns:

\[ N_1 = \frac{E_1}{E/\text{turn}} = \frac{250}{8} = 31.25 \approx 32, \qquad N_2 = \frac{3000}{8} = 375 \]

A winding can only have a whole number of turns, so the low-voltage side is wound with 32. Rounding up is the safe direction: more turns mean less flux for the same applied voltage, and therefore a lower flux density than designed.

The area follows from the e.m.f. equation applied to one turn, which avoids depending on either turn count:

\[ \frac{E}{\text{turn}} = 4.44\,f\,B_mA \;\Longrightarrow\; 8 = 4.44 \times 50 \times 1.2 \times A \]
\[ A = \frac{8}{266.4} = 0.0300\ \text{m}^2 = 300\ \text{cm}^2 \]

The same answer from the whole secondary winding, as a check:

\[ 3000 = 4.44 \times 50 \times 375 \times 1.2 \times A \;\Longrightarrow\; A = \frac{3000}{99{,}900} = 0.0300\ \text{m}^2\ \checkmark \]

A note on the rounding. The voltage ratio is exactly 12 : 1, so a design that keeps the ratio exact would use 32 and 384 turns, giving 7.81 V per turn and a flux density of 1.17 T. No whole-number pair delivers exactly 8 V per turn and a 12 : 1 ratio, which is why the 8 V figure should be read as a design target rather than an exact constraint.

E.m.f. per turn is the design variable a transformer engineer actually chooses. Fix it and the core area is fixed by the permitted flux density, and every winding's turn count is fixed by its voltage. It is the single number that ties the magnetic design to the electrical one.
Answera\(N_1 = 31.25 \approx 32,\ N_2 = 375\) b\(A = 0.03\ \text{m}^2 = 300\ \text{cm}^2\)
Problem 2Exam levelTurns from Core Dimensions

The core of a 100 kVA, 11000/550 V, 50 Hz single-phase core-type transformer has a gross cross-section of 20 cm × 20 cm, with a stacking factor of 0.9. Find

  1. the number of high-voltage and low-voltage turns
  2. the e.m.f. per turn,

if the peak flux density \(B_m\) is not to exceed 1.3 T.

Solution

Take the net iron area first. The laminations are separated by their insulating coating, so only 90% of the measured cross-section carries flux:

\[ A = k_sA_{\text{gross}} = 0.9 \times (0.2 \times 0.2) = 0.036\ \text{m}^2 \]

Using the gross 0.04 m2 would understate the turns by 10% and drive the real flux density to 1.44 T — well into saturation.

Apply the e.m.f. equation to the high-voltage winding with \(B_m\) at its permitted maximum, which gives the minimum acceptable number of turns:

\[ 11{,}000 = 4.44 \times 50 \times N_1 \times 1.3 \times 0.036 = 10.39\,N_1 \]
\[ N_1 = \frac{11{,}000}{10.39} = 1058.8 \]

Now let the low-voltage winding decide the rounding. The ratio is exactly 20 : 1, so

\[ N_2 = \frac{550}{11{,}000}\,N_1 = \frac{1058.8}{20} = 52.9 \approx 53 \]
\[ N_1 = 20 \times 53 = 1060\ \text{turns} \]

Choosing the whole number on the coarse winding first and scaling up keeps the ratio exact; choosing 1059 on the HV side would have forced a fractional LV winding.

The e.m.f. per turn, which must come out the same from either winding:

\[ \frac{E_1}{N_1} = \frac{11{,}000}{1060} = 10.38\ \text{V}, \qquad \frac{E_2}{N_2} = \frac{550}{53} = 10.38\ \text{V}\ \checkmark \]

Check the flux density that these turns actually produce, since the 1.3 T was a limit and not a target:

\[ B_m = \frac{11{,}000}{4.44 \times 50 \times 1060 \times 0.036} = 1.298\ \text{T} < 1.3\ \text{T}\ \checkmark \]

Rounding the turns up has bought a small margin, exactly as it should.

A flux-density limit is a one-sided constraint, so round the turns up and never down. Fewer turns than calculated means more flux per turn, a saturating core, a magnetising current that is no longer sinusoidal, and iron losses far above the design value. The 100 kVA rating plays no part in this calculation at all — it sizes the conductors, not the core.
Answera\(N_1 = 1060,\ N_2 = 53\) b\(E/\text{turn} = 10.38\ \text{V}\) (with \(B_m = 1.298\ \text{T}\))
Problem 3CoreFlux Density and Secondary EMF

A single-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2. If the primary winding is connected to a 50 Hz supply at 520 V, calculate

  1. the peak value of flux density in the core
  2. the voltage induced in the secondary winding.
Solution

The applied voltage fixes the flux, not the other way round. Rearranging the e.m.f. equation for the energised winding:

\[ E_1 = 4.44\,f\,N_1B_mA \;\Longrightarrow\; 520 = 4.44 \times 50 \times 400 \times B_m \times 60\times10^{-4} \]
\[ B_m = \frac{520}{532.8} = 0.976\ \text{Wb/m}^2 \]

Equivalently \(\Phi_m = B_mA = 5.86\ \text{mWb}\), which is the flux both windings link.

The secondary voltage follows from the turns ratio. With \(a = N_1/N_2 = 400/1000 = 0.4\), this is a step-up transformer:

\[ E_2 = \frac{E_1}{a} = \frac{520}{0.4} = 1300\ \text{V} \]

Confirm through the e.m.f. per turn, which is the same 1.3 V on both windings:

\[ \frac{E_1}{N_1} = \frac{520}{400} = 1.3\ \text{V}, \qquad E_2 = 1.3 \times 1000 = 1300\ \text{V}\ \checkmark \]
Flux density is a property of the applied voltage, the frequency and the core — the load never enters it. This is why a transformer's iron loss is essentially constant from no load to full load, and why running a 50 Hz transformer on 60 Hz is harmless while running a 60 Hz one on 50 Hz is not.
Answera\(B_m = 0.976\ \text{T}\) b\(E_2 = 1300\ \text{V}\)
Problem 4CoreFull-Load Currents and Flux

A 25 kVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to a 3000 V, 50 Hz supply. Neglecting the leakage drops and the no-load primary current, find

  1. the full-load primary and secondary currents
  2. the secondary e.m.f.
  3. the maximum flux in the core.
Solution

The turns ratio, written the same way every time:

\[ a = \frac{N_1}{N_2} = \frac{500}{50} = 10 \]

A step-down transformer, so expect the secondary voltage to be ten times smaller and the secondary current ten times larger.

The secondary e.m.f., most simply through the e.m.f. per turn:

\[ \frac{E_1}{N_1} = \frac{3000}{500} = 6\ \text{V/turn}, \qquad E_2 = 6 \times 50 = 300\ \text{V} \]
\[ \text{or } E_2 = \frac{E_1}{a} = \frac{3000}{10} = 300\ \text{V}\ \checkmark \]

The rated currents, from the fact that the 25 kVA is the same on both sides:

\[ I_1 = \frac{25{,}000}{3000} = 8.33\ \text{A}, \qquad I_2 = \frac{25{,}000}{300} = 83.3\ \text{A} \]

And indeed \(I_2 = aI_1 = 10 \times 8.33\) — current transforms inversely to voltage, which is what keeps the product constant.

The core flux, from the e.m.f. equation with the area not needed:

\[ 3000 = 4.44 \times 50 \times 500 \times \Phi_m \;\Longrightarrow\; \Phi_m = \frac{3000}{111{,}000} = 27.0\ \text{mWb} \]
Neglecting the no-load current is what makes \(I_1 = I_2/a\) exact. In a real transformer the primary also carries the magnetising and core-loss components, but at full load these are two or three percent of the rated current and the phasor addition barely moves the magnitude — which is why this approximation survives all the way into the equivalent circuit of Set 17.
Answera\(I_1 = 8.33\ \text{A},\ I_2 = 83.3\ \text{A}\) b\(E_2 = 300\ \text{V}\) c\(\Phi_m = 27.0\ \text{mWb}\)
Problem 5Exam levelThree-Phase Delta/Star Core

The core of a three-phase, 50 Hz, 11000/550 V delta/star, 300 kVA core-type transformer operates with a maximum flux of 0.05 Wb. Find

  1. the number of high-voltage and low-voltage turns per phase
  2. the e.m.f. per turn
  3. the full-load high-voltage and low-voltage phase currents.
Solution

The e.m.f. per turn depends only on the flux and the frequency, so it is the same on every limb and in both windings:

\[ \frac{E}{\text{turn}} = 4.44\,f\,\Phi_m = 4.44 \times 50 \times 0.05 = 11.1\ \text{V} \]

Convert the nameplate line voltages to phase voltages, which is the only place the connection matters. The 11000 V winding is in delta and the 550 V winding in star:

\[ V_{ph,\text{HV}} = 11{,}000\ \text{V}, \qquad V_{ph,\text{LV}} = \frac{550}{\sqrt3} = 317.5\ \text{V} \]

In delta the phase winding sees the full line voltage; in star it sees the line voltage divided by \(\sqrt3\).

Turns per phase are then phase voltage divided by e.m.f. per turn:

\[ N_1 = \frac{11{,}000}{11.1} = 991, \qquad N_2 = \frac{317.5}{11.1} = 28.6 \approx 29 \]

The low-voltage winding would be wound with 29 turns; the resulting flux is then 1.4% below the stated 0.05 Wb, which is the usual consequence of rounding a small turn count.

Phase currents come from the rating per phase, which is one third of the total:

\[ S_{ph} = \frac{300}{3} = 100\ \text{kVA} \]
\[ I_{ph,\text{HV}} = \frac{100 \times 10^3}{11{,}000} = 9.09\ \text{A}, \qquad I_{ph,\text{LV}} = \frac{100 \times 10^3}{317.5} = 315\ \text{A} \]

Check against the line-current formula, which must agree:

\[ I_{L} = \frac{S}{\sqrt3\,V_L} \;\Longrightarrow\; I_{L,\text{LV}} = \frac{300 \times 10^3}{\sqrt3 \times 550} = 315\ \text{A}\ \checkmark \]
\[ I_{L,\text{HV}} = \frac{300 \times 10^3}{\sqrt3 \times 11{,}000} = 15.7\ \text{A} = \sqrt3 \times 9.09\ \checkmark \]

On the star side phase and line current are equal; on the delta side the line current is \(\sqrt3\) times the phase current. The check catches the commonest error in three-phase transformer work.

Everything magnetic is per phase; only the nameplate is per line. Turns, e.m.f. per turn and flux belong to one limb, so the first move in any three-phase transformer problem is to convert the line voltages and the total kVA into per-phase quantities and forget the connection until the very end.
Answera\(N_1 = 991,\ N_2 = 28.6 \approx 29\) per phase b\(11.1\ \text{V/turn}\) c\(I_{ph,\text{HV}} = 9.09\ \text{A},\ I_{ph,\text{LV}} = 315\ \text{A}\)
Problem 6CorePeak Flux Density

A single-phase transformer has 500 turns in the primary and 1200 turns in the secondary. The cross-sectional area of the core is 80 cm2. If the primary winding is connected to a 50 Hz supply at 500 V, calculate

  1. the peak flux density
  2. the voltage induced in the secondary.
Solution

Find the flux first, then divide by the area — it keeps the arithmetic in two clean steps:

\[ 500 = 4.44 \times 50 \times 500 \times \Phi_m \;\Longrightarrow\; \Phi_m = \frac{500}{111{,}000} = 4.50\ \text{mWb} \]

The peak flux density:

\[ B_m = \frac{\Phi_m}{A} = \frac{4.50 \times 10^{-3}}{80 \times 10^{-4}} = 0.563\ \text{Wb/m}^2 \]

Comfortably below the 1.2 – 1.5 T at which silicon steel begins to saturate, so this core is conservatively rated.

The secondary voltage, with \(a = N_1/N_2 = 500/1200\):

\[ E_2 = \frac{E_1}{a} = 500 \times \frac{1200}{500} = 1200\ \text{V} \]

The e.m.f. per turn here is exactly 1 V, so the secondary voltage is simply its turn count.

The area is needed only for the flux density, never for the voltage ratio. Problems 3 and 6 are the same problem with different numbers, and in both the second part could be answered without knowing anything about the core at all — the turns ratio carries it.
Answera\(\Phi_m = 4.50\ \text{mWb},\ B_m = 0.563\ \text{T}\) b\(E_2 = 1200\ \text{V}\)
Problem 7CoreCurrents, EMF and Flux

A 25 kVA single-phase transformer has 250 turns on the primary and 40 turns on the secondary winding. The primary is connected to 1500 V, 50 Hz mains. Calculate

  1. the primary and secondary currents on full load
  2. the secondary e.m.f.
  3. the maximum flux in the core.
Solution

The secondary voltage from the turns ratio, \(a = 250/40 = 6.25\):

\[ E_2 = \frac{E_1}{a} = 1500 \times \frac{40}{250} = 240\ \text{V} \]

The full-load currents, each from the rating and the voltage on its own side:

\[ I_1 = \frac{25{,}000}{1500} = 16.67\ \text{A}, \qquad I_2 = \frac{25{,}000}{240} = 104.2\ \text{A} \]

Check: \(I_2/I_1 = 104.2/16.67 = 6.25 = a\), as it must be.

The maximum core flux, from the primary e.m.f. equation:

\[ 1500 = 4.44 \times 50 \times 250 \times \Phi_m \;\Longrightarrow\; \Phi_m = \frac{1500}{55{,}500} = 27.0\ \text{mWb} \]

The same 27 mWb as Problem 4 — not a coincidence, since both have an e.m.f. per turn of 6 V at 50 Hz.

Three questions, three independent routes into the same data. The voltage comes from the turns, the currents from the kVA, and the flux from the e.m.f. equation; none of them needs the answer to another. Recognising that structure is what makes these problems a two-minute exercise rather than a chain of substitutions.
Answera\(I_1 = 16.67\ \text{A},\ I_2 = 104.2\ \text{A}\) b\(E_2 = 240\ \text{V}\) c\(\Phi_m = 27.0\ \text{mWb}\)
Problem 8Exam levelIdeal Transformer on Load

An ideal transformer has a 150-turn primary and a 750-turn secondary. The primary is connected to a 240 V, 50 Hz source, and the secondary supplies a load drawing 4 A at a lagging power factor of 0.8. Determine

  1. the transformation ratio \(a\)
  2. the current in the primary winding
  3. the power supplied to the load
  4. the flux in the core.
Solution

The transformation ratio, defined as primary to secondary turns:

\[ a = \frac{N_1}{N_2} = \frac{150}{750} = 0.2 \]

Less than one, so this is a step-up transformer: the secondary has the higher voltage and the smaller current.

The secondary voltage:

\[ V_2 = \frac{V_1}{a} = \frac{240}{0.2} = 1200\ \text{V} \]

The primary current. An ideal transformer has no magnetising current, so the ampere-turns balance exactly, \(N_1I_1 = N_2I_2\):

\[ I_1 = \frac{I_2}{a} = \frac{4}{0.2} = 20\ \text{A} \]

Voltage steps up by five, current steps down by five, and the volt-amperes are 4800 on both sides.

The power delivered to the load is the only place the power factor appears:

\[ P_L = V_2I_2\cos\theta = 1200 \times 4 \times 0.8 = 3840\ \text{W} \]

The transformer passes it on unchanged: an ideal machine has no loss, so the primary also draws 3840 W, at 240 V and 20 A with the same 0.8 power factor.

The core flux, set by the applied primary voltage alone:

\[ \Phi_m = \frac{V_1}{4.44\,f\,N_1} = \frac{240}{4.44 \times 50 \times 150} = 7.21\ \text{mWb} \]

Neither the 4 A of load current nor its power factor changes this number — the load's ampere-turns are cancelled by the extra primary ampere-turns that answer them.

Power factor affects the watts, never the volts, the amperes or the flux. The turns ratio fixes the voltage and current magnitudes; \(\cos\theta\) only decides how much of the 4800 VA is real power. Keeping the two ideas separate prevents the standard error of dividing a current by the power factor somewhere it does not belong.
Answera\(a = 0.2\) b\(I_1 = 20\ \text{A}\) c\(P_L = 3840\ \text{W}\) d\(\Phi_m = 7.21\ \text{mWb}\)
Formulas

Key Formulas

QuantityRelationNotes
E.m.f. equation\(E = 4.44fN\Phi_m\)R.m.s. value; the 4.44 is \(4 \times 1.11\) — every problem
In terms of flux density\(E = 4.44fNB_mA\)\(A\) is the net iron area — Problems 1, 2, 3
E.m.f. per turn\(E/N = 4.44f\Phi_m\)Identical on both windings — Problems 1, 2, 5
Net core area\(A = k_sA_{\text{gross}}\)Stacking factor 0.9 in Problem 2
Flux and flux density\(\Phi_m = B_mA\)Problems 3, 6
Turns ratio\(a = N_1/N_2 = E_1/E_2\)\(a > 1\) step-down, \(a < 1\) step-up
Current ratio\(I_1/I_2 = 1/a\)From \(N_1I_1 = N_2I_2\) — Problems 4, 7, 8
Rated currents\(I_1 = S/V_1,\ \ I_2 = S/V_2\)The kVA is the same on both sides
Load power\(P_L = V_2I_2\cos\theta\)Power factor enters here only — Problem 8
Star connection\(V_{ph} = V_L/\sqrt3,\ \ I_{ph} = I_L\)LV side of Problem 5
Delta connection\(V_{ph} = V_L,\ \ I_{ph} = I_L/\sqrt3\)HV side of Problem 5
Three-phase rating\(S_{ph} = S/3,\ \ S = \sqrt3V_LI_L\)Work per phase throughout — Problem 5
Volts per hertz\(\Phi_m \propto V/f\)Why reduced frequency saturates the core
Pitfalls

Common Mistakes

  1. Using the gross core area instead of the net. The stacking factor must be applied before the e.m.f. equation; ignoring the 0.9 in Problem 2 gives 953 turns and a real flux density of 1.44 T.

  2. Writing 4.44 as 4 or as 2π. The factor is \(4f\) for the average value times the 1.11 form factor of a sine wave — every problem on this page.

  3. Mixing \(\Phi_m\) and \(B_m\). The e.m.f. equation takes flux in webers; using a flux density in the same slot is out by the core area — Problems 3 and 6.

  4. Rounding turns down when a flux-density limit is given. Fewer turns means more flux, so the limit is breached; round up — Problems 1 and 2.

  5. Inverting the turns ratio. With \(a = N_1/N_2\) the voltage divides by \(a\) and the current multiplies by it; Problem 8 has \(a = 0.2\), and reading it as 5 turns a step-up into a step-down.

  6. Using line voltage for turns on a three-phase transformer. Turns per phase need the phase voltage, so the star winding of Problem 5 uses 317.5 V and not 550 V.

  7. Dividing the total kVA by the line voltage to get a phase current. Use one third of the rating with the phase voltage, or the full rating with \(\sqrt3V_L\) — never a mixture — Problem 5.

  8. Letting the power factor into the current calculation. The 4 A of Problem 8 is already the load current; \(\cos\theta\) belongs only in the wattage.

  9. Believing the load changes the core flux. Flux is fixed by \(V/f\) and the turns, which is why iron loss is treated as constant — Problems 3 and 8.

→ Looking Ahead

Everything on this page assumed an ideal transformer: no winding resistance, no leakage flux, no magnetising current and no core loss. On those assumptions the e.m.f. equation and the turns ratio are the whole theory, and they are enough to size a core, count turns and find every current at full load.

Each assumption now has to be paid for. A real primary draws a small current even with the secondary open, and that current is neither in phase with the voltage nor purely reactive — it has a component supplying the iron loss and a much larger one magnetising the core. Drawing it correctly, and then watching what happens to it when load is applied, is the next step.

Next: Set 16 — No-Load and On-Load Phasor Analysis, where \(I_0\) is resolved into \(I_w\) and \(I_\mu\), and the primary current is built up as a phasor sum.