On-Load Operation and Phasor Diagrams

In Chapter 42 the secondary was open and the primary drew only the exciting current. Connecting a load allows the secondary induced EMF to drive a current. That current produces its own magnetomotive force (MMF), and the primary must now supply the load as well as excite the core.

1. The secondary current produces an MMF that opposes the original alternating flux, according to Lenz’s law.
2. A small initial reduction in flux reduces the primary back EMF. The resulting voltage imbalance increases the primary current.
3. The additional primary current supplies nearly equal and opposite ampere-turns to those of the secondary.
4. In steady operation, the resultant core MMF remains the small MMF needed for excitation. The source supplies the output power and the losses.

This is a sinusoidal steady-state approximation. More precisely, flux is set by the internal primary EMF, not by the terminal voltage alone.

All bold quantities below are RMS phasors, with counterclockwise rotation representing positive phase. Let \(a=N_1/N_2\). We retain Chapter 42’s induced back-EMF convention: \(\mathbf E_1\) opposes the applied primary voltage. Choose secondary polarity so that \(\mathbf E_2=\mathbf E_1/a\); both induced EMFs lag the core flux by 90°. The secondary current leaves the terminal defined positive for \(\mathbf V_2\).

The minus sign means that the primary load component is drawn opposite to the secondary current in this winding reference. The magnitudes obey \(I_2'=I_2/a\).

Phasors describe the fundamental sinusoidal component. The actual exciting current can contain harmonics, as discussed in Chapter 42.

The first equation represents the net exciting ampere-turns in the adopted winding reference. The second describes the cancellation of load ampere-turns. Therefore, \(\mathbf I_1\) is the vector sum of the load component and the exciting current.

For negligible winding drops, \(\mathbf V_1\approx-a\mathbf V_2\). Thus a lagging secondary load gives a primary load component lagging the primary supply by approximately the same angle. The 180° reversal of both voltage and current preserves the power factor.

In the excitation-branch model, \(\mathbf I_w\) is in phase with \(\mathbf U_1=-\mathbf E_1\), and \(\mathbf I_\mu\) lags it by 90°. Replacing \(\mathbf U_1\) by \(\mathbf V_1\) is the usual small-drop approximation. At light load, the exciting current noticeably changes input power factor; at substantial load, the load component usually dominates.

Leakage reactance causes a reactive voltage drop; an ideal reactance does not consume average real power. Leakage flux is distinct from the mutual core flux that induces EMF in both windings.

1. Choose \(\mathbf V_2\) as the horizontal reference.
2. Place \(\mathbf I_2\) at the load power-factor angle.
3. From the tip of \(\mathbf V_2\), add \(\mathbf I_2R_2\) parallel to current. From its tip, add \(j\mathbf I_2X_2\) 90° ahead of current. The origin-to-endpoint vector is \(\mathbf E_2\).
4. Draw \(\mathbf E_1=a\mathbf E_2\) in the same direction. Draw flux 90° ahead of the induced EMFs.
5. Draw \(\mathbf I_2'\) opposite \(\mathbf I_2\). Add \(\mathbf I_0\) to obtain \(\mathbf I_1\).
6. Reverse \(\mathbf E_1\) to obtain \(-\mathbf E_1\), then add the primary resistance and reactance drops to obtain \(\mathbf V_1\).

Voltage and current arrows require separate scales. The following diagrams enlarge winding drops to make their construction visible.

Take \(\mathbf V_2=V_2\angle0^\circ\). An inductive load draws \(\mathbf I_2=I_2\angle(-\phi_2)\). The resistance drop points below the horizontal; the reactance drop points upward and to the right.

For positive resistance and inductive leakage reactance, the real component of \(\mathbf E_2\) exceeds \(V_2\), so \(E_2>V_2\). The angle of \(\mathbf E_2\) depends on both drops; it lies above \(\mathbf V_2\) only when \(X_2\cos\phi_2>R_2\sin\phi_2\).

Here current and terminal voltage are in phase. Add the resistance drop horizontally, then the reactance drop vertically upward.

A resistive load does not make the transformer’s leakage reactance vanish. Also, unity output power factor does not guarantee exactly unity input power factor: excitation and leakage require reactive power.

A capacitive load draws \(\mathbf I_2=I_2\angle(+\phi_2)\). The resistance drop points upward and right; rotating current by a further 90° places the leakage-reactance drop upward and left.

The reactance drop now has a negative horizontal component. It can offset the resistive drop and permit a loaded terminal voltage above the no-load value at the same primary supply. Leading power factor permits negative regulation; it does not guarantee it. The exact result follows from the complete voltage polygon, including primary drops.

Neglecting exciting current for the series-drop calculation, refer the primary impedance to the secondary:

The secondary-referred supply phasor is \(\mathbf V_{s,2}=-\mathbf V_1/a\) in our reference convention. It satisfies

For small series drops, the first-order change in voltage magnitude is

Use + for lagging loads and − for leading loads; \(\phi_2\) is a positive angle magnitude. At unity power factor, the first-order result is \(I_2R_{\mathrm{eq},2}\); reactance still contributes a second-order change in magnitude.

The two voltages are compared at the same primary supply voltage and frequency. This is the loaded-voltage or up-regulation definition. Lecture 45 uses down-regulation, with no-load voltage in the denominator: \(\mathrm{VR}_{\mathrm{down}}\%=100(V_{2,\mathrm{NL}}-V_{2,\mathrm{load}})/V_{2,\mathrm{NL}}\). If r_up and r_down are fractions, \(r_{\mathrm{down}}=r_{\mathrm{up}}/(1+r_{\mathrm{up}})\). Thus 3.70% up-regulation is approximately 3.57% down-regulation. State the convention with each numerical result.

A 2300/230 V transformer has \(a=10\). Neglect winding drops for this example. It supplies 100 A at 0.8 lagging power factor, and its exciting current is 2 A at 0.2 lagging power factor. Choose \(\mathbf V_1\) as reference.

The primary current is neither 12 A nor simply 10 A. Exciting current must be added as a phasor. In this primary-voltage reference, \(\mathbf V_2\) is at approximately 180°, and the secondary current is at 143.13°, consistent with the earlier minus sign.

Let \(V_2=230\) V, \(I_2=50\) A, \(R_2=0.08\) Ω and \(X_2=0.12\) Ω. Calculate the internal secondary EMF needed to maintain this terminal voltage for three loads.

With the leading load, the required \(E_2\) is below 230 V in this example. These rows hold terminal voltage and current fixed while comparing required internal EMFs; they are not three operating points at one fixed primary voltage.

A transformer delivers 50 A at 230 V and has total secondary-referred series impedance \(0.10+j0.15\) Ω. Ignore exciting current.

The exact series-model leading result uses \(|229.5+j9|=229.676\) V, giving approximately −0.141%. Close to zero regulation, a small absolute error from the first-order approximation can be a large relative error.

• The source supplies additional primary current to balance secondary load ampere-turns.

• Resistance drops are parallel to current; inductive reactance drops lead current by 90°.

• Construct voltage polygons head to tail and add currents as vectors.

• Leading loads can produce a terminal-voltage rise, depending on impedance and power factor.

The EMF and current signs belong to the winding convention in Section 43-2. The minus sign in the current equation denotes direction, not a negative current magnitude.

1. Adding current magnitudes. Use \(\mathbf I_1=\mathbf I_0+\mathbf I_2'\), not \(I_1=I_0+I_2'\) unless the phasors happen to align.

2. Placing the reactance drop 90° from voltage. It is 90° ahead of the winding current.

3. Assuming flux grows with load current. Load ampere-turns largely cancel; internal EMF and frequency govern mutual flux.

4. Mixing back-EMF and circuit-voltage conventions. Define winding polarities before deciding signs.

5. Claiming every leading load has negative regulation. Resistance, leakage reactance and power factor decide the outcome.

6. Using winding-only drops for transformer regulation. Include both windings on one reference side.

7. Treating no-load input as zero. Excitation and core losses remain when the load is removed.

1. P43.1 A transformer has 1000 primary turns and 200 secondary turns. The load draws 40 A. Find the magnitude of the primary load component and the load ampere-turns on each winding.

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   \[a=5,\quad I_2^{\prime}=40/5=8~\mathrm A,\quad N_1I_2^{\prime}=N_2I_2=8000~\mathrm{A\!\cdot turns}\]

   The two load MMFs oppose one another.

2. P43.2 With secondary voltage as reference, a 0.6 lagging load draws 20 A. Find the angles of the current, resistance drop and reactance drop.

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   \[\phi_2=\cos^{-1}(0.6)=53.13^\circ\]

   \[\angle\mathbf I_2=\angle(\mathbf I_2R_2)=-53.13^\circ,\quad\angle(j\mathbf I_2X_2)=36.87^\circ\]

3. P43.3 A 200 V secondary supplies 20 A at unity power factor. If R₂ = 0.2 Ω and X₂ = 0.3 Ω, find its induced EMF.

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   \[\mathbf E_2=200+20(0.2+j0.3)=204+j6=204.088\angle1.685^\circ~\mathrm V\]

4. P43.4 Repeat P43.3 for 0.8 leading power factor.

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   \[\mathbf I_2=16+j12,\quad\mathbf E_2=200+(16+j12)(0.2+j0.3)=199.6+j7.2~\mathrm V\]

   \[E_2=199.730~\mathrm V,\quad\angle\mathbf E_2=2.066^\circ\]

5. P43.5 A transformer has R_eq,2 = 0.12 Ω and X_eq,2 = 0.16 Ω. Find the leading power factor for zero first-order regulation.

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   \[R_{\mathrm{eq},2}\cos\phi_2=X_{\mathrm{eq},2}\sin\phi_2\Rightarrow\tan\phi_2=0.75\]

   \[\phi_2=36.87^\circ,\quad\cos\phi_2=0.8\ \text{leading}\]

   This cancels the first-order longitudinal drop. The exact series-model magnitude generally retains a small second-order change.

6. P43.6 In a primary-voltage reference, the load component is 6 − j8 A and exciting current is 0.3 − j1.0 A. Find primary current and input power factor.

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   \[\mathbf I_1=6.3-j9.0~\mathrm A,\quad I_1=10.986~\mathrm A,\quad\cos\phi_1=0.5735\ \text{lagging}\]

1. MCQ 1. When load is added at constant supply voltage and frequency, the primary current generally:
   (a) decreases · (b) increases · (c) stays zero · (d) equals the secondary current

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   (b) increases. It supplies the load ampere-turn balance and input power.

2. MCQ 2. The leakage-reactance drop is:
   (a) in phase with current · (b) 90° behind current · (c) 90° ahead of current · (d) always horizontal

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   (c) 90° ahead of current. Multiplication by j rotates a phasor counterclockwise by 90°.

3. MCQ 3. A unity-power-factor secondary load gives exactly unity input power factor:
   (a) always · (b) only at full load · (c) only at no load · (d) not in general

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   (d) not in general. Excitation and leakage reactance contribute reactive input power.

4. MCQ 4. Negative regulation is possible with:
   (a) a suitable leading load · (b) every leading load · (c) every lagging load · (d) zero winding resistance alone

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   (a) a suitable leading load. The reactive contribution must offset the resistive and remaining magnitude contributions.

5. MCQ 5. Referring primary resistance to the secondary gives:
   (a) aR₁ · (b) a²R₁ · (c) R₁/a² · (d) R₁/a

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   (c) R₁/a². Impedance changes by the square of the turns ratio.

6. MCQ 6. The two load MMFs in steady operation are approximately:
   (a) additive · (b) equal and opposing · (c) both zero · (d) independent of load

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   (b) equal and opposing. Their cancellation leaves the exciting MMF needed by the core.

1. Explain why connecting a secondary load increases primary current while mutual flux changes only slightly.

2. Construct all three secondary voltage polygons and then extend one to include primary voltage and current.

3. Explain why a leading load can raise terminal voltage and why the first-order regulation formula can be inaccurate near zero regulation.

4. Distinguish secondary output power factor from primary input power factor at light load.

Lecture 44 combines the exciting branch and winding impedances into the transformer equivalent circuit and accounts for losses. The voltage polygons developed here provide its physical interpretation.