← T-02: Construction | 🏠 Index | T-03b: Phasor Diagrams Under Load →

---

T-03a: No-Load Operation

Section: A | Priority: 🟠 HIGH | Exam Frequency: 3/7 years Sources: Theraja Ch-32 (Art. 32.6–32.9), VK Mehta Ch-7 (Art. 7.4–7.6), Slides L-09 S04–S08

Why This Topic Matters

No-load operation is tested in 3/7 papers (2019, 2020, 2024). It appears as "explain no-load operation with phasor diagram" (3–6 marks) or as a numerical where you decompose $I_0$ into its components. Understanding no-load current is also the foundation for the OC test (T-06a) and the shunt branch of the equivalent circuit (T-04). A near-identical numerical (220V/110V, 0.5A, 30W) was repeated in 2020 and 2024.

---

📝 Key Definitions

No-load current ($I_0$): "When the primary of a transformer is connected to the supply and the secondary is open, the transformer is on no-load. The small current $I_0$ taken by the primary under no-load conditions is called no-load current." — VK Mehta, Art. 7.6. It is typically 2–10% of rated current.

Magnetizing current ($I_{\mu }$ or $I_m$): "The component $I_{\mu }$ of $I_0$ which establishes the alternating flux ${\Phi }_m$ in the core is called the magnetizing current. It is in phase with the flux ${\Phi }_m$ and lags $V_1$ by 90°." — VK Mehta, Art. 7.6

Core-loss current ($I_w$ or $I_c$): "The component $I_w$ of $I_0$ which supplies the iron losses (hysteresis + eddy current) is called the active or working or iron-loss component. It is in phase with $V_1$." — VK Mehta, Art. 7.6

---

What Happens at No-Load

When you apply $V_1$ to the primary and leave the secondary open ($I_2=0$):

1. A small no-load current $I_0$ flows in the primary.
2. This current does two jobs:
   • Magnetic job: The magnetizing component $I_{\mu }$ sets up the alternating core flux ${\Phi }_m$.
   • Thermal job: The core-loss component $I_w$ supplies power for hysteresis and eddy current losses.
3. The flux ${\Phi }_m$ induces EMFs in both windings: $E_1=4.44f{N}_1{\Phi }_m$ and $E_2=4.44f{N}_2{\Phi }_m$.
4. Since the secondary is open, no current flows there. $V_2=E_2$.

---

No-Load Current Components

The no-load current $I_0$ splits into two perpendicular components:

$$I_0=\sqrt{I_w^2+I_{\mu }^2}$$

where:

• $I_w=I_0\cos {\varphi }_0$ (active component, in phase with $V_1$)
• $I_{\mu }=I_0\sin {\varphi }_0$ (reactive component, 90° lagging from $V_1$)

The no-load power factor is very low (0.1–0.3) because $I_{\mu }\gg I_w$. The magnetizing current dominates.

No-load power input:

$$W_0=V_1{I}_0\cos {\varphi }_0=V_1{I}_w$$

This power equals the iron loss (core loss) because copper loss at no-load is negligible ($I_0$ is tiny, so $I_0^2{R}_1\approx 0$).

$$\cos {\varphi }_0=\frac{W_0}{V_1{I}_0}$$

---

No-Load Phasor Diagram

Step-by-step construction:

1. Draw ${\Phi }_m$ as the reference (horizontal, pointing right).
2. $E_1$ and $E_2$ lag ${\Phi }_m$ by 90° (pointing downward). The induced EMF is maximum when flux passes through zero.
3. $V_1=-E_1$ (pointing upward). The applied voltage must balance the back-EMF.
4. $I_w$ is in phase with $V_1$ (pointing upward). It supplies real power for core losses.
5. $I_{\mu }$ lags $V_1$ by 90°. It points in the same direction as ${\Phi }_m$ (horizontal right). This makes sense: the magnetizing current creates the flux, so they are in phase.
6. $I_0=I_w+I_{\mu }$ (phasor sum). The angle between $V_1$ and $I_0$ is ${\varphi }_0$.
7. Since secondary is open: $V_2=E_2$ (no drops).

---

Why Primary Current Increases with Load

When you connect a load and secondary current $I_2$ flows:

1. $I_2$ through $N_2$ turns creates a demagnetizing MMF = $N_2{I}_2$.
2. This tends to reduce the core flux.
3. But $V_1$ is fixed by the supply. Since $V_1\approx E_1=4.44f{N}_1{\Phi }_m$, the flux must stay constant.
4. The primary draws additional current $I_2^{\prime }$ to cancel the secondary MMF:

$$N_1{I}_1=N_1{I}_0+N_2{I}_2$$

$$I_1=I_0+\frac{N_2}{N_1}{I}_2=I_0+I_2^{\prime }$$

"The increased primary current brings in more energy from the supply to match the energy delivered to the load. The transformer does not generate energy: it regulates primary current to always match the secondary load." — From explanation notes

---

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Explain the no-load operation of a 1-phase transformer with a neat phasor diagram.

Appeared: 2020 Q1(c) — 3 marks, 2024 Q1(b) — 6 marks

Full Answer:

When a transformer primary is connected to AC supply with secondary open-circuited, a small no-load current $I_0$ flows (2–10% of rated current). This current has two components:

1. Magnetizing current ($I_{\mu }$): This component establishes the alternating mutual flux ${\Phi }_m$ in the core. It is purely reactive and lags the applied voltage $V_1$ by 90°. It is in phase with the flux ${\Phi }_m$. It does no real work — it simply oscillates the magnetic field back and forth.

2. Core-loss (active) current ($I_w$): The core heats up due to hysteresis and eddy current losses. These losses require real power input. $I_w$ is in phase with $V_1$ and supplies this power.

The total no-load current is the phasor sum:

$$I_0=\sqrt{I_w^2+I_{\mu }^2}$$

The no-load power factor is very low ($\cos {\varphi }_0=0.1$ to $0.3$) because $I_{\mu }\gg I_w$.

No-load power input: $W_0=V_1{I}_0\cos {\varphi }_0=V_1{I}_w=P_{\text{iron}}$ (copper loss is negligible)

Phasor Diagram Construction:

1. Draw ${\Phi }_m$ horizontal as reference.
2. $E_1$, $E_2$ lag ${\Phi }_m$ by 90° (downward). By Faraday's law: $e=-Nd\Phi /dt$, so EMF is maximum when flux passes through zero.
3. $V_1=-E_1$ (upward). Applied voltage balances the back-EMF.
4. $I_w$ is in phase with $V_1$ (upward).
5. $I_{\mu }$ is in phase with ${\Phi }_m$ (horizontal right). Lags $V_1$ by 90°.
6. $I_0=I_w+I_{\mu }$ at angle ${\varphi }_0$ from $V_1$.
7. Since secondary is open: $V_2=E_2$ (no load drops).

---

🎯 Q2: No-load test data: 220V/110V, 0.5 A, 30 W. Find magnetizing and core-loss components.

Appeared: 2020 Q2(d), 2024 Q3(c) — identical data in both years

Full Answer:

Given: $V_1=220$ V, $V_2=110$ V, $I_0=0.5$ A, $W_0=30$ W.

No-load power factor:

$$\cos {\varphi }_0=\frac{W_0}{V_1{I}_0}=\frac{30}{220\times 0.5}=\frac{30}{110}=0.2727$$

$${\varphi }_0={\cos }^{-1}(0.2727)=74.17°$$

Core-loss (active) current:

$$I_w=I_0\cos {\varphi }_0=0.5\times 0.2727=\boxed{0.136\text{ A}}$$

Magnetizing (reactive) current:

$$I_{\mu }=I_0\sin {\varphi }_0=0.5\times \sin (74.17°)=0.5\times 0.9621=\boxed{0.481\text{ A}}$$

Iron loss: $P_{Fe}=W_0=30$ W (copper loss at no-load is negligible since $I_0$ is very small)

Check: $I_0=\sqrt{0.136^2+0.481^2}=\sqrt{0.0185+0.2314}=\sqrt{0.2499}=0.5$ A ✓

Note that $I_{\mu }\gg I_w$. The no-load current is mostly magnetizing.

---

🎯 Q3: Why does primary current increase as secondary load increases?

Appeared: 2019 Q1(c) — 4 marks

Full Answer:

An ideal transformer maintains the core flux ${\Phi }_m$ at a constant value set by the supply voltage: ${\Phi }_m\approx V_1/(4.44f{N}_1)$. This flux requires a certain MMF to sustain it.

At no-load, the primary alone provides all the required MMF: $N_1{I}_0=\text{const}$.

When a load draws secondary current $I_2$, the secondary winding (with $N_2$ turns) creates a demagnetizing MMF $=N_2{I}_2$. This tends to reduce the core flux.

But $V_1$ is fixed by the supply. Since $V_1\approx E_1=4.44f{N}_1{\Phi }_m$, the flux cannot change (it is locked to the supply voltage). Therefore, the primary current must increase to cancel the secondary demagnetization:

$$N_1{I}_1-N_2{I}_2=N_1{I}_0\approx \text{const}$$

$$N_1{I}_1=N_1{I}_0+N_2{I}_2$$

$$I_1=I_0+\frac{N_2}{N_1}{I}_2=I_0+I_2^{\prime }$$

As $I_2$ increases (more load), $I_1$ increases proportionally. The increased primary current brings in more energy from the supply to match the energy delivered to the load. The transformer does not generate energy; it acts as an automatic current regulator that adjusts $I_1$ to always balance the secondary MMF.

---

Exam Variants

Year	Question	Data Given	Key Answer
2020 Q2(d)	Find $I_{\mu }$, $I_w$	220V/110V, $I_0=0.5$ A, $W_0=30$ W	$I_w=0.136$ A, $I_{\mu }=0.481$ A
2024 Q3(c)	Same problem	Identical data	Same answers
2019 Q1(c)	Why does $I_1$ increase with load?	Conceptual	MMF balance argument
2020 Q1(c)	Explain no-load with sketch	Conceptual	Phasor diagram + components
2024 Q1(b)	No-load with phasor diagram	Conceptual, 6 marks	Full phasor construction

---

⚡ Exam Tips & Common Mistakes

1. $I_{\mu }$ is the bigger component. Students often swap $I_w$ and $I_{\mu }$. Remember: most of the no-load current goes into creating flux ($I_{\mu }$), not into losses ($I_w$). The no-load pf is low (0.1–0.3).
2. No-load power = iron loss only. Copper loss at no-load is negligible because $I_0$ is tiny. So $W_0=P_{\text{iron}}$.
3. The phasor diagram uses ${\Phi }_m$ as reference, not $V_1$. Start with flux horizontal. Everything else follows from there.
4. The 220V/110V, 0.5A, 30W numerical is a guaranteed repeat. Memorize the answers.

🔗 Related Topics

• T-01: Fundamentals — EMF equation that determines the flux
• T-03b: Phasor Under Load — Adding load to the no-load diagram
• T-04: Equivalent Circuit — $R_0$ and $X_0$ come from no-load data
• T-06a: OC Test — The OC test IS the no-load test

---

← T-02: Construction | 🏠 Index | T-03b: Phasor Diagrams Under Load →