Solved Problems · Set 16

No-Load and On-Load Phasor Analysis

Part 3 · Transformers — the exciting current resolved into its two components, and the phasor addition that turns a secondary load into a primary current.

Prof. Mithun Mondal 7 solved problems GATE · ESE · University

Set 16 — No-Load and On-Load Phasor Analysis

A transformer draws current even with nothing connected to its secondary, and that current is the key to the whole phasor picture. This set splits it into the iron-loss component that carries the core loss and the magnetising component that carries the flux, obtains both from open-circuit readings and from raw iron data, and then adds the load component to find what the supply actually sees when the secondary is working.

The single method drilled throughout is phasor addition with the applied voltage as reference: resolve every current into a component along \(\bar{V}_1\) and one at right angles to it, add the parts separately, and convert back. Problems 4 to 7 are the same three lines with the unknown moved around.

Part 3 · Single-Phase Transformers · 7 solved problems

i Method Recap

  • The no-load current has two components at right angles. One supplies the core losses and is in phase with the applied voltage; the other magnetises the core and lags it by 90°.

    \[ I_0 = \sqrt{I_w^2 + I_\mu^2}, \qquad I_w = I_0\cos\phi_0, \qquad I_\mu = I_0\sin\phi_0 \]

  • The no-load wattmeter reads the iron loss. With the secondary open the primary carries only \(I_0\), and \(I_0^2R_1\) is a fraction of a percent of the reading, so

    \[ P_i = V_1 I_0\cos\phi_0 = V_1 I_w \quad\Longrightarrow\quad I_w = \frac{P_i}{V_1} \]

  • The magnetising current comes from the magnetic circuit. Total ampere-turns for the required \(B_{\max}\) give the peak value of \(I_\mu\); divide by \(\sqrt{2}\) for the rms value used in the phasor diagram.

    \[ \hat{I}_\mu = \frac{\mathcal{F}_{\text{total}}}{N_1}, \qquad I_\mu = \frac{\hat{I}_\mu}{\sqrt{2}} \]

  • On load the primary current is a phasor sum, not an arithmetic one. The load component \(I_2'\) balances the secondary ampere-turns; the no-load component persists unchanged:

    \[ \bar{I}_1 = \bar{I}_0 + \bar{I}_2', \qquad I_2' = \frac{I_2}{a}, \qquad a = \frac{N_1}{N_2} \]

  • Resolve along and across \(V_1\). Taking \(\bar{V}_1\) as reference and both currents lagging, each phasor becomes \(I(\cos\phi - j\sin\phi)\); add the real and imaginary parts separately, then

    \[ I_1 = \sqrt{(\Sigma\text{real})^2 + (\Sigma\text{imag})^2}, \qquad \cos\phi_1 = \frac{\Sigma\text{real}}{I_1} \]

  • The cosine rule is the same statement in polar form. With \(\theta\) the angle between \(\bar{I}_0\) and \(\bar{I}_2'\), that is \(\theta = \phi_0 - \phi_2\):

    \[ I_1^2 = I_0^2 + I_2'^2 + 2I_0I_2'\cos\theta \]

  • The EMF equation still governs the core. Problem 3 needs it to size the core before any current can be found:

    \[ E_1 = 4.44\,f\,N_1\,B_{\max}A \]
VideoWalkthrough
Problem 1CoreNo-Load Current Components

A 2200/200 V single-phase transformer draws a no-load primary current of 0.6 A and absorbs 400 W. Find

  1. the iron-loss (working) component of the no-load current
  2. the magnetising component
  3. the no-load power factor
Solution

Read the wattmeter as iron loss. With the secondary open the primary carries only \(I_0 = 0.6\ \text{A}\), so the primary copper loss is \(0.6^2R_1\) — a few tenths of a watt against a 400 W reading. The whole 400 W is therefore core loss, and it is delivered by the component of \(I_0\) in phase with \(V_1\):

\[ I_w = \frac{P_i}{V_1} = \frac{400}{2200} = 0.182\ \text{A} \]

Note which voltage is used: the no-load current is drawn on the 2200 V side, so 2200 V is the divisor. The 200 V rating plays no part in this calculation.

The magnetising component is in quadrature, so the two combine as the legs of a right-angled triangle with \(I_0\) as hypotenuse:

\[ I_0^2 = I_w^2 + I_\mu^2 \;\Longrightarrow\; I_\mu = \sqrt{0.6^2 - 0.182^2} = \sqrt{0.3269} = 0.572\ \text{A} \]

The no-load power factor follows from the same triangle:

\[ \cos\phi_0 = \frac{I_w}{I_0} = \frac{0.182}{0.6} = 0.303\ \text{lagging}, \qquad \phi_0 = 72.4^\circ \]

A no-load power factor of 0.3 is high for a modern transformer — 0.1 to 0.2 is typical — which tells you this core is comparatively lossy for the flux it carries.

Almost every no-load question is this one triangle read in a different order. Two of the four quantities \(I_0\), \(I_w\), \(I_\mu\), \(\cos\phi_0\) fix the other two, and the power reading always enters through \(I_w = P_i/V_1\).
Answer\(I_w = 0.182\ \text{A}\), \(I_\mu = 0.572\ \text{A}\), \(\cos\phi_0 = 0.303\) lagging
Problem 2CoreComponents From Power Factor

A 2200/2540 V single-phase transformer takes 0.5 A at a power factor of 0.3 on open circuit. Find the magnetising and working components of the no-load primary current, and the iron loss.

Solution

The power factor resolves the current directly. Here the phase angle is given rather than the power, so no division is needed — project \(\bar{I}_0\) onto \(\bar{V}_1\) and onto the perpendicular:

\[ I_w = I_0\cos\phi_0 = 0.5 \times 0.3 = 0.15\ \text{A} \]

The quadrature component uses \(\sin\phi_0 = \sqrt{1-0.3^2} = 0.954\):

\[ I_\mu = I_0\sin\phi_0 = 0.5 \times 0.954 = 0.477\ \text{A} \]

Equivalently \(I_\mu = \sqrt{0.5^2 - 0.15^2} = \sqrt{0.2275} = 0.477\ \text{A}\) — the same triangle from the other side.

The iron loss follows, taken on the 2200 V winding because that is the winding the 0.5 A flows in:

\[ P_i = V_1 I_0 \cos\phi_0 = 2200 \times 0.5 \times 0.3 = 330\ \text{W} \]

This is a step-up transformer, 2200 V to 2540 V. The higher secondary voltage never enters a no-load current calculation; it only tells you the turns ratio, \(a = 2200/2540 = 0.866\).

\(I_\mu\) dominates \(I_0\) in every practical transformer. Because \(\cos\phi_0\) is small, \(I_\mu \approx I_0\) to within a few percent, while \(I_w\) is the small in-phase residue that carries all the core loss. That asymmetry is why the shunt branch is drawn with a large \(R_0\) and a much smaller \(X_0\).
Answer\(I_w = 0.15\ \text{A}\), \(I_\mu = 0.477\ \text{A}\), \(P_i = 330\ \text{W}\)
Problem 3ChallengeNo-Load From Core Data

A single-phase transformer has 500 turns on the primary and 40 turns on the secondary. The mean length of the magnetic path in the iron core is 150 cm and the joints are equivalent to an air-gap of 0.1 mm. When 3000 V at 50 Hz is applied to the primary, the maximum flux density is 1.2 Wb/m2. Calculate

  1. the cross-sectional area of the core
  2. the no-load secondary voltage
  3. the no-load current drawn by the primary
  4. the power factor on no load

Take the magnetising force in the iron as 5 AT/cm at a flux density of 1.2 Wb/m2, the corresponding iron loss as 2 W/kg at 50 Hz, and the density of iron as 7.8 g/cm3.

Solution

Size the core from the EMF equation. Everything else in the problem depends on the core area, so it must come first. With \(E_1 \approx V_1 = 3000\ \text{V}\) on no load:

\[ \begin{aligned} E_1 &= 4.44\,f\,N_1\,B_{\max}A \\ 3000 &= 4.44 \times 50 \times 500 \times 1.2 \times A \\ A &= \frac{3000}{133200} = 0.0225\ \text{m}^2 = 225\ \text{cm}^2 \end{aligned} \]

The no-load secondary voltage is set by the turns ratio alone, since no current flows and there is no impedance drop:

\[ a = \frac{N_1}{N_2} = \frac{500}{40} = 12.5, \qquad E_2 = \frac{E_1}{a} = \frac{3000}{12.5} = 240\ \text{V} \]

Magnetic circuit: total ampere-turns. The iron path and the equivalent joint gap are in series, so their MMFs add. The iron contributes at the tabulated 5 AT/cm; the gap must be computed from \(H = B/\mu_0\), since air is not saturable and its relative permeability is unity:

\[ \begin{aligned} \mathcal{F}_{\text{iron}} &= 150 \times 5 = 750\ \text{AT} \\ \mathcal{F}_{\text{gap}} &= \frac{B}{\mu_0}\,l_g = \frac{1.2}{4\pi\times10^{-7}} \times 0.0001 = 95.5\ \text{AT} \\ \mathcal{F}_{\text{total}} &= 750 + 95.5 = 845.5\ \text{AT} \end{aligned} \]

A joint only 0.1 mm wide demands 95.5 AT against the 750 AT taken by 1.5 m of iron — eleven percent of the total from one ten-thousandth of the path length. This is the air-gap dominance seen in Set 2.

Convert ampere-turns to the magnetising current. The ampere-turns above correspond to the peak flux density, so they give the peak of \(i_\mu\); the rms value wanted for the phasor diagram is smaller by \(\sqrt{2}\):

\[ \hat{I}_\mu = \frac{845.5}{500} = 1.691\ \text{A}, \qquad I_\mu = \frac{1.691}{\sqrt{2}} = 1.196\ \text{A} \]

The iron-loss component comes from the mass of the core. Volume is the mean path length times the cross-section found in step 1:

\[ \begin{aligned} \text{Volume} &= 150 \times 225 = 33\,750\ \text{cm}^3 \\ \text{Mass} &= \frac{33\,750 \times 7.8}{1000} = 263.25\ \text{kg} \\ P_i &= 263.25 \times 2 = 526.5\ \text{W} \\ I_w &= \frac{P_i}{V_1} = \frac{526.5}{3000} = 0.176\ \text{A} \end{aligned} \]

Combine the two components to get the no-load current and its power factor:

\[ I_0 = \sqrt{I_\mu^2 + I_w^2} = \sqrt{1.196^2 + 0.176^2} = 1.209\ \text{A} \]
\[ \cos\phi_0 = \frac{I_w}{I_0} = \frac{0.176}{1.209} = 0.145\ \text{lagging} \]

A sanity check on magnitude: \(I_0 = 1.21\ \text{A}\) against a primary current that would be tens of amperes at full load, and a power factor of 0.145 — both exactly what a healthy core gives.

This problem runs the whole no-load model backwards from iron data. \(I_\mu\) comes from the magnetic circuit — ampere-turns per unit length — while \(I_w\) comes from the thermal data — watts per kilogram. They meet only at the phasor sum. Confusing the two routes, or forgetting the \(\sqrt{2}\) that separates peak ampere-turns from rms current, is where this problem is usually lost.
Answer(a)\(A = 225\ \text{cm}^2\) (b)\(E_2 = 240\ \text{V}\) (c)\(I_0 = 1.209\ \text{A}\) (d)\(\cos\phi_0 = 0.145\) lagging
Problem 4Exam levelPrimary Current On Load

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no-load current is 3 A at a power factor of 0.2 lagging. Calculate the primary current and its power factor when the secondary current is 280 A at a power factor of 0.80 lagging. Neglect the winding voltage drops.

Solution

Refer the secondary current to the primary. Ampere-turn balance requires \(N_1I_2' = N_2I_2\), so the load component of primary current is the secondary current divided by the turns ratio:

\[ a = \frac{N_1}{N_2} = \frac{1000}{200} = 5, \qquad I_2' = \frac{I_2}{a} = \frac{280}{5} = 56\ \text{A} \]

Fix the two phase angles with respect to the common reference \(\bar{V}_1\). Neglecting the winding drops makes \(\bar{V}_2\) in phase with \(\bar{V}_1\), so the load angle carries straight through:

\[ \phi_2 = \cos^{-1}0.80 = 36.87^\circ, \qquad \phi_0 = \cos^{-1}0.20 = 78.46^\circ,\ \ \sin\phi_0 = 0.980 \]

Add the two phasors component by component:

\[ \begin{aligned} \bar{I}_1 &= \bar{I}_0 + \bar{I}_2' \\ &= 3\,(0.20 - j\,0.980) + 56\,(0.80 - j\,0.60) \\ &= (0.60 - j\,2.94) + (44.80 - j\,33.60) \\ &= 45.40 - j\,36.54 \end{aligned} \]

Convert back to polar form:

\[ I_1 = \sqrt{45.40^2 + 36.54^2} = 58.28\ \text{A}, \qquad \phi_1 = \tan^{-1}\frac{36.54}{45.40} = 38.83^\circ \]
\[ \cos\phi_1 = \frac{45.40}{58.28} = 0.779\ \text{lagging} \]

Read what the no-load current did. Adding 3 A of \(I_0\) to 56 A of \(I_2'\) raised the magnitude by only 2.28 A, because the two are almost 42° apart — but it dragged the power factor down from 0.80 to 0.779. The magnetising current costs little in current and rather more in power factor.

Never add \(I_0\) and \(I_2'\) arithmetically. Here \(56 + 3 = 59\ \text{A}\) would be 1.2% high, which looks harmless — but at light load, where \(I_2'\) and \(I_0\) are comparable, the same shortcut is catastrophically wrong. Always resolve into components first.
Answer\(I_1 = 58.28\ \text{A}\) at \(\cos\phi_1 = 0.779\) lagging \((38.83^\circ)\)
Problem 5Exam levelCosine-Rule Combination

A single-phase transformer with a ratio of 440/110 V takes a no-load current of 5 A at 0.2 power factor lagging. If the secondary supplies 120 A at a power factor of 0.8 lagging, estimate the current taken by the primary.

Solution

Refer the load current, using the voltage ratio in place of the turns ratio:

\[ a = \frac{V_1}{V_2} = \frac{440}{110} = 4, \qquad I_2' = \frac{I_2}{a} = \frac{120}{4} = 30\ \text{A} \]

Find the angle between the two phasors. Both lag \(\bar{V}_1\), so the angle between them is the difference of the two lags:

\[ \begin{aligned} \phi_0 &= \cos^{-1}0.2 = 78^\circ 28' \\ \phi_2 &= \cos^{-1}0.8 = 36^\circ 52' \\ \theta &= \phi_0 - \phi_2 = 41^\circ 36' = 41.59^\circ \end{aligned} \]

Apply the cosine rule to the triangle of phasors. Since \(\bar{I}_1\) is the sum, the enclosed angle appears with a plus sign:

\[ \begin{aligned} I_1 &= \sqrt{I_0^2 + I_2'^2 + 2I_0I_2'\cos\theta} \\ &= \sqrt{5^2 + 30^2 + 2(5)(30)\cos 41.59^\circ} \\ &= \sqrt{25 + 900 + 224.3} = \sqrt{1149.3} = 33.90\ \text{A} \end{aligned} \]

Confirm with the component method, which is worth doing whenever a cosine rule is used, because a sign or an angle slip there is invisible:

\[ \begin{aligned} \bar{I}_0 &= 5(0.2 - j\,0.980) = 1.00 - j\,4.90 \\ \bar{I}_2' &= 30(0.8 - j\,0.60) = 24.00 - j\,18.00 \\ \bar{I}_1 &= 25.00 - j\,22.90 = 33.90\angle-42.49^\circ\ \checkmark \end{aligned} \]

Both routes agree, and the second one also delivers the input power factor free of charge: \(\cos\phi_1 = 25.00/33.90 = 0.737\) lagging.

The cosine rule and the component method are the same calculation in different coordinates. Polar is quicker when only a magnitude is wanted; rectangular is safer when a phase angle, a power factor or a further phasor addition is coming. Choosing rectangular by default costs one extra line and removes an entire class of error.
Answer\(I_1 = 33.90\ \text{A}\) at \(\cos\phi_1 = 0.737\) lagging
Problem 6Exam levelRecovering The No-Load Current

A transformer has a primary winding of 800 turns and a secondary winding of 200 turns. When the load current on the secondary is 80 A at 0.8 power factor lagging, the primary current is 25 A at 0.707 power factor lagging. Determine the no-load current of the transformer and its phase with respect to the applied voltage.

Solution

Run the phasor equation backwards. The relation \(\bar{I}_1 = \bar{I}_0 + \bar{I}_2'\) is being used to find \(\bar{I}_0\) from the other two, so the subtraction must be done as a phasor difference:

\[ \bar{I}_0 = \bar{I}_1 - \bar{I}_2', \qquad a = \frac{N_1}{N_2} = \frac{800}{200} = 4, \qquad I_2' = \frac{80}{4} = 20\ \text{A} \]

Write both known phasors in rectangular form, with \(\bar{V}_1\) as reference:

\[ \begin{aligned} \bar{I}_1 &= 25\angle-45.01^\circ = 17.68 - j\,17.68 \\ \bar{I}_2' &= 20\angle-36.87^\circ = 16.00 - j\,12.00 \end{aligned} \]

A power factor of 0.707 is 45° to three decimal places, which is why the real and imaginary parts of \(\bar{I}_1\) come out equal.

Subtract:

\[ \bar{I}_0 = (17.68 - 16.00) - j\,(17.68 - 12.00) = 1.68 - j\,5.68 \]
\[ I_0 = \sqrt{1.68^2 + 5.68^2} = 5.92\ \text{A}, \qquad \phi_0 = \tan^{-1}\frac{5.68}{1.68} = 73.57^\circ\ \text{lagging} \]

Check the answer against physical expectation. A no-load current should be small and strongly lagging:

QuantityValueIs it plausible?
\(I_0\)5.92 A24% of \(I_1\) — large, so this is a small or lightly loaded unit
\(\cos\phi_0\)0.283 laggingCorrectly low; the branch is mostly reactive
\(I_w\)1.68 AIn phase with \(V_1\), supplies the iron loss
\(I_\mu\)5.68 AQuadrature, magnetises the core

The real and imaginary parts of \(\bar{I}_0\) are \(I_w\) and \(I_\mu\) — no extra work is needed to separate them.

The graphical construction asked for in the original wording is exactly this arithmetic drawn to scale: set out \(\bar{V}_1\) horizontally, draw \(\bar{I}_1\) at 45° below it and \(\bar{I}_2'\) at 36.87° below it, and close the triangle from the tip of \(\bar{I}_2'\) to the tip of \(\bar{I}_1\). That closing side is \(\bar{I}_0\).

Subtraction is where sign errors live. Both currents lag, so both imaginary parts are negative, and the difference of two negatives is easy to get backwards. If \(\bar{I}_0\) comes out leading, or larger than \(\bar{I}_1\), the subtraction was performed in the wrong order.
Answer\(I_0 = 5.92\ \text{A}\) lagging \(V_1\) by \(73.57^\circ\) \((\cos\phi_0 = 0.283)\)
Problem 7CoreInput Current And Power Factor

A single-phase transformer takes 10 A on no load at a power factor of 0.2 lagging. The turns ratio is 4:1 (step down). If the load on the secondary is 200 A at a power factor of 0.85 lagging, find the primary current and power factor. Neglect the voltage drop in the windings.

Solution

Refer the load current:

\[ a = 4, \qquad I_2' = \frac{200}{4} = 50\ \text{A} \]

Resolve both currents onto \(\bar{V}_1\) and its quadrature axis:

\[ \begin{aligned} \bar{I}_0 &= 10\angle-78.46^\circ = 2.00 - j\,9.80 \\ \bar{I}_2' &= 50\angle-31.79^\circ = 42.50 - j\,26.34 \end{aligned} \]

The real part of \(\bar{I}_2'\) is simply \(50 \times 0.85\); the imaginary part uses \(\sin(\cos^{-1}0.85) = 0.527\).

Add and convert:

\[ \bar{I}_1 = 44.50 - j\,36.14 = 57.32\angle-39.08^\circ\ \text{A} \]
\[ \cos\phi_1 = \frac{44.50}{57.32} = 0.776\ \text{lagging} \]

Notice the direction of the shift. The load is at 0.85 lagging but the supply sees 0.776 lagging: the magnetising current, being almost purely reactive, always worsens the input power factor. It can never improve it, whatever the load, because \(I_0\) adds reactive amperes and almost no real ones.

The transformer is not power-factor neutral. Between secondary and primary the power factor degrades by an amount that depends on \(I_0/I_2'\) — negligible at full load on a large unit, appreciable here where \(I_0\) is a fifth of \(I_2'\). This is one of the reasons transformers are not left energised on no load.
Answer\(I_1 = 57.32\ \text{A}\) at \(\cos\phi_1 = 0.776\) lagging \((39.08^\circ)\)
Formulas

Key Formulas

QuantityRelationNotes
No-load current\(I_0 = \sqrt{I_w^2 + I_\mu^2}\)Components in quadrature — Problems 1, 3
Iron-loss component\(I_w = I_0\cos\phi_0 = P_i/V_1\)In phase with \(V_1\) — Problems 1, 2
Magnetising component\(I_\mu = I_0\sin\phi_0\)Lags \(V_1\) by 90° — Problem 2
No-load power factor\(\cos\phi_0 = I_w/I_0 = P_i/(V_1I_0)\)Typically 0.1–0.3 — Problems 1, 3
Iron loss\(P_i = V_1I_0\cos\phi_0\)The whole no-load wattmeter reading
EMF equation\(E_1 = 4.44\,f\,N_1\,B_{\max}A\)Sizes the core — Problem 3
Turns ratio\(a = N_1/N_2 = V_1/V_2\)Used in every referral
Gap MMF\(\mathcal{F}_g = (B/\mu_0)\,l_g\)Joints behave as a short air-gap — Problem 3
Peak to rms\(I_\mu = \hat{I}_\mu/\sqrt{2}\), \(\hat{I}_\mu = \mathcal{F}/N_1\)Ampere-turns give the peak — Problem 3
Core mass\(m = A\,l\,\rho\), \(P_i = m \times \text{W/kg}\)Route to \(I_w\) from iron data — Problem 3
Load component\(I_2' = I_2/a\)From \(N_1I_2' = N_2I_2\) — Problems 4–7
Primary current\(\bar{I}_1 = \bar{I}_0 + \bar{I}_2'\)Phasor sum, never arithmetic — Problems 4, 7
Rectangular form\(\bar{I} = I(\cos\phi - j\sin\phi)\)Lagging current, \(\bar{V}_1\) as reference
Cosine rule\(I_1^2 = I_0^2 + I_2'^2 + 2I_0I_2'\cos(\phi_0-\phi_2)\)Polar alternative — Problem 5
Input power factor\(\cos\phi_1 = \text{Re}(\bar{I}_1)/I_1\)Always poorer than \(\cos\phi_2\) — Problem 7
Pitfalls

Common Mistakes

  1. Adding \(I_0\) and \(I_2'\) arithmetically. They are typically 40° apart, so the sum of magnitudes overstates \(I_1\) and gives no phase angle at all — Problems 4 and 7.

  2. Dividing the no-load power by the wrong voltage. \(I_w = P_i/V_1\) uses the voltage of the winding that carries \(I_0\), which is 2200 V in Problem 1 and 3000 V in Problem 3 — never the other winding.

  3. Forgetting the \(\sqrt{2}\) between ampere-turns and rms current. Ampere-turns computed at \(B_{\max}\) give the peak magnetising current; the phasor diagram needs the rms value — Problem 3.

  4. Ignoring the joints in the magnetic circuit. An equivalent gap of 0.1 mm contributed 95.5 of the 845.5 total ampere-turns in Problem 3; dropping it understates \(I_\mu\) by 11%.

  5. Using the volume of iron without converting units. Path length in cm times area in cm2 gives cm3, and the density is in g/cm3, so a factor of 1000 is needed to reach kilograms — Problem 3.

  6. Multiplying by the turns ratio when referring a current. Currents transform inversely: \(I_2' = I_2/a\) for a step-down transformer, so 280 A becomes 56 A, not 1400 A — Problem 4.

  7. Taking the angle between \(\bar{I}_0\) and \(\bar{I}_2'\) as the sum of the lags. Both lag the same reference, so the enclosed angle is the difference, \(\phi_0 - \phi_2\) — Problem 5.

  8. Subtracting in the wrong order when recovering \(\bar{I}_0\). It is \(\bar{I}_1 - \bar{I}_2'\); reversing it produces a leading current with no physical meaning — Problem 6.

  9. Expecting the input power factor to equal the load power factor. The magnetising current adds reactive amperes only, so \(\cos\phi_1 < \cos\phi_2\) always — Problem 7.

  10. Assuming a step-up ratio changes the no-load analysis. In Problem 2 the 2540 V secondary is irrelevant; only the excited winding's voltage enters.

→ Looking Ahead

Every problem in this set treated the transformer as an ideal device with one extra current bolted on. The windings had no resistance and no leakage, so \(E_1 = V_1\) and \(E_2 = V_2\), and the only imperfection admitted was the current the core itself demands. That is enough to explain no-load behaviour and to predict the primary current at any load, but not enough to predict what the secondary terminal voltage actually does when load is applied.

The next step gives the windings their real resistances and leakage reactances, and gathers the shunt branch of this set into a proper parallel \(R_0\) and \(X_0\). Every quantity on the secondary side is then referred through \(a^2\) so that the whole machine becomes one series-parallel network, which can be solved for terminal voltage, regulation and efficiency by ordinary circuit analysis.

Next: Set 17 — Equivalent Circuit Parameters, where \(R_{01}\), \(X_{01}\), \(R_{02}\) and \(X_{02}\) are assembled from winding data and the exact circuit is compared with the approximate one.