How unit 3 is examined
Magnetic-circuit basics, B-H curve, inductance, energy and electromagnetic induction, then the single phase transformer, which carries most of the marks (EMF equation, losses, efficiency, OC/SC tests).
Basic definitions
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Definition. A magnetic circuit is the closed path followed by magnetic flux, and it obeys the law $\text{mmf} = \text{flux} \times \text{reluctance}$.
Key points.
- Magnetomotive force $\text{mmf} = NI$ ampere-turns drives the flux, like emf drives current.
- Flux density is $B = \phi/A$ in tesla, and field intensity is $H = NI/l$ in A/m, with $B = \mu_0\mu_r H$.
- Reluctance $S = l/(\mu A)$ opposes flux, so $\phi = NI/S$.
- Permeability $\mu$ measures how easily a material carries flux.
Magnetization characteristics of ferromagnetic materials
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Definition. The magnetization characteristic is the B-H curve showing how flux density $B$ in a ferromagnetic material rises with field intensity $H$.
Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u3-01" viewBox="0 0 553 355.2" width="553" height="355.2" role="img" aria-label="B-H curve: B on y-axis, H on x-axis; slow start, steep rise, knee, saturation"><style>#dsfig-u3-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u3-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u3-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u3-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u3-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u3-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u3-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u3-01 .t{fill:#16181D;font-weight:500}#dsfig-u3-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u3-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u3-01 .dot{fill:#16181D}#dsfig-u3-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u3-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u3-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u3-01 .ah{fill:#454C5A}#dsfig-u3-01 .ah.hi{fill:#2340B8}#dsfig-u3-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u3-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u3-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u3-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u3-01 .e{stroke:#B1B7C3}html.dark #dsfig-u3-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u3-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u3-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u3-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u3-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u3-01 .t{fill:#E6E8ED}html.dark #dsfig-u3-01 .t.inv{fill:#0F1115}html.dark #dsfig-u3-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u3-01 .dot{fill:#E6E8ED}html.dark #dsfig-u3-01 .ann{fill:#8FA3FF}html.dark #dsfig-u3-01 .lbl{fill:#858D9C}html.dark #dsfig-u3-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u3-01 .ah{fill:#B1B7C3}html.dark #dsfig-u3-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u3-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u3-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u3-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah8" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh8" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M57,306.7 L107.2,281.6" marker-end="url(#ah8)"/><path class="e" d="M134.5,255.2 L202.6,119" marker-end="url(#ah8)"/><path class="e" d="M230.4,95.6 L363.6,62.3" marker-end="url(#ah8)"/><path class="e" d="M402.8,54.7 L492.2,42.8" marker-end="url(#ah8)"/><g class="wl"><rect x="145.5" y="177.2" width="47.1" height="18" rx="9"/><text class="t" x="169" y="186.2" dy=".35em" text-anchor="middle">steep</text></g><g class="wl"><rect x="277.6" y="69.7" width="40.8" height="18" rx="9"/><text class="t" x="298" y="78.7" dy=".35em" text-anchor="middle">knee</text></g><g class="wl"><rect x="407.4" y="39.6" width="82.2" height="18" rx="9"/><text class="t" x="448.5" y="48.6" dy=".35em" text-anchor="middle">saturation</text></g><circle class="n" cx="40" cy="315.2" r="18"/><text class="t" x="40" y="315.2" dy=".35em" text-anchor="middle">O</text><circle class="n" cx="126" cy="272.2" r="18"/><text class="t" x="126" y="272.2" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="212" cy="100.2" r="18"/><text class="t" x="212" y="100.2" dy=".35em" text-anchor="middle">K</text><circle class="n" cx="384" cy="57.2" r="18"/><text class="t" x="384" y="57.2" dy=".35em" text-anchor="middle">S</text><circle class="n" cx="513" cy="40" r="18"/><text class="t" x="513" y="40" dy=".35em" text-anchor="middle">H</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">B-H curve: B on y-axis, H on x-axis; slow start, steep rise, knee, saturation</figcaption></figure>
Key points.
- In the initial region B rises slowly because domain walls move only a little and reversibly.
- In the steep linear region small increases in H align many domains, so B rises rapidly.
- Near the knee point most domains are aligned and the slope falls.
- In saturation all domains are aligned, so B hardly increases and the material behaves like air ($\mu_r \to 1$).
- On reducing H to zero the material keeps residual flux (retentivity), and a reverse H called coercive force is needed to remove it, giving the hysteresis loop.
- Permeability $\mu = B/H$ is therefore not constant; it is highest near the knee.
Answer frame. Open with the definition; draw the B-H curve with origin, initial region, steep region, knee and saturation; then develop points 1-4, add hysteresis (point 5); close with "operate transformers below the knee to avoid saturation".
Asked: [7 marks] (Nov 2022, Dec 2023) Draw the typical normal magnetization curve of ferromagnetic material. Discuss magnetization characteristics of ferromagnetic material.
Self inductance and mutual inductance
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Definition. Self inductance $L$ is the property of a coil by which it opposes a change in its own current, $L = N\phi/I$; mutual inductance $M$ is the property by which a changing current in one coil induces an emf in a neighbouring coil, $M = N_2\phi_{12}/I_1$.
Key points.
- With reluctance $S$, $L_1 = N_1^2/S$ and $L_2 = N_2^2/S$.
- Induced emf is $e = -L\,di/dt$ for self and $e_2 = -M\,di_1/dt$ for mutual induction.
- Let $k$ be the coefficient of coupling, the fraction of one coil's flux linking the other, $0 \le k \le 1$.
- Mutual flux gives $M = kN_1N_2/S$.
Derivation. Current $I_1$ in coil 1 produces flux $\phi_1 = N_1I_1/S$, of which $k\phi_1$ links coil 2, so $M = N_2k\phi_1/I_1 = kN_1N_2/S$. Multiply the expressions for $L_1$ and $L_2$:
$$L_1L_2 = \frac{N_1^2N_2^2}{S^2} \Rightarrow \frac{N_1N_2}{S} = \sqrt{L_1L_2}$$
$$\boxed{M = k\sqrt{L_1L_2}}$$
For $k=1$, $M=\sqrt{L_1L_2}$.
Answer frame. Open with both definitions; write points 1-2; derive $M$ as above; close with $M = k\sqrt{L_1L_2}$ and its meaning for $k$.
Asked: [7 marks] (Jun 2023, Dec 2024) What do you understand by self inductance and mutual inductance? Derive the relation between self inductance and mutual inductance.
Energy in linear magnetic systems
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Definition. Energy stored in a magnetic field is $W = \int ei\,dt = \int Li\,di = \tfrac12 LI^2$.
Key points.
- Put $L = N^2\mu A/l$ and $H = NI/l$, so $W = \tfrac12\mu H^2\,Al$.
- Since $B = \mu H$ and volume $= Al$, $W = \tfrac12 BH \times \text{volume}$.
- Energy density is ==$w = W/\text{volume} = \tfrac12 BH = B^2/2\mu$== J/m$^3$.
- The energy is released when the field collapses.
Asked: [7 marks] (Jun 2022) Derive an expression for an energy density in magnetic circuits.
Coils connected in series
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Definition. Two coupled coils in series have total inductance $L = L_1 + L_2 \pm 2M$.
Key points.
- In series aiding, the fluxes add, so $L = L_1 + L_2 + 2M$.
- In series opposing, the fluxes subtract, so $L = L_1 + L_2 - 2M$.
- Hence $M = (L_{aid} - L_{opp})/4$.
AC excitation in magnetic circuits
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Definition. When a coil on an iron core is fed from an AC source, the flux is alternating, $\phi = \phi_m\sin\omega t$, and it induces a back emf $E = 4.44f\phi_mN$.
Key points.
- The applied voltage sets the flux, since $V \approx E$.
- The excitation current has a magnetizing component that makes the flux.
- It has a core-loss component that supplies hysteresis and eddy loss, so it is non-sinusoidal and lags the voltage by less than $90^\circ$.
Magnetic field produced by current carrying conductor
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Definition. A current-carrying conductor produces a magnetic field around it, whose direction follows the right-hand grip rule.
Key points.
- Biot-Savart law: $dB = \mu_0 I\,dl\sin\theta/(4\pi r^2)$.
- For a long straight wire, $B = \mu_0 I/(2\pi r)$.
- For a solenoid, $B = \mu_0 NI/l$.
- Field lines are concentric circles about the wire.
Force on a current carrying conductor
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Definition. A conductor of length $l$ carrying current $I$ in a field $B$ experiences the force $F = BIl\sin\theta$.
Key points.
- The force is maximum when the conductor is perpendicular to the field, $F = BIl$.
- Its direction is given by Fleming's left-hand rule.
- It is the basis of motor action.
- It is zero when the conductor is parallel to the field.
Induced voltage, laws of electromagnetic induction, direction of induced EMF
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Definition. Electromagnetic induction is the production of an emf in a circuit whenever the flux linking it changes.
Key points.
- Faraday's first law: whenever the flux linking a circuit changes, an emf is induced in it.
- Faraday's second law: the magnitude of the induced emf equals the rate of change of flux linkage, $e = -N\,d\Phi/dt$.
- Lenz's law: the direction of the induced emf is such that the current it drives opposes the change producing it.
- The negative sign expresses Lenz's law and is a consequence of conservation of energy, since otherwise energy would be created.
- Statically induced emf arises from a changing field in a fixed coil (transformer); dynamically induced emf arises from a conductor moving in a field, $e = Blv\sin\theta$ (generator).
- Direction of the dynamic emf is found by Fleming's right-hand rule.
Answer frame. Open with the definition of induction; state the two laws and Lenz's law with the formula; explain the negative sign; give one static and one dynamic example; close with applications (transformer, generator).
Asked: [7 marks] (Dec 2023, Dec 2024, Jun 2025) Discuss / explain Faraday's laws of electromagnetic induction.
Single phase transformer- General construction, working principle, e.m.f. equation, equivalent circuits, phasor diagram, voltage regulation, losses and efficiency, open circuit and short circuit test
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Definition. A transformer is a static device that transfers electrical energy from one AC circuit to another at the same frequency by mutual induction, changing voltage and current levels.
Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u3-02" viewBox="0 0 596 80" width="596" height="80" role="img" aria-label="Transformer: V1 source, P primary winding N1, C laminated core, S secondary winding N2, L load"><style>#dsfig-u3-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u3-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u3-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u3-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u3-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u3-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u3-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u3-02 .t{fill:#16181D;font-weight:500}#dsfig-u3-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u3-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u3-02 .dot{fill:#16181D}#dsfig-u3-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u3-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u3-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u3-02 .ah{fill:#454C5A}#dsfig-u3-02 .ah.hi{fill:#2340B8}#dsfig-u3-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u3-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u3-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u3-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u3-02 .e{stroke:#B1B7C3}html.dark #dsfig-u3-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u3-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u3-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u3-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u3-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u3-02 .t{fill:#E6E8ED}html.dark #dsfig-u3-02 .t.inv{fill:#0F1115}html.dark #dsfig-u3-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u3-02 .dot{fill:#E6E8ED}html.dark #dsfig-u3-02 .ann{fill:#8FA3FF}html.dark #dsfig-u3-02 .lbl{fill:#858D9C}html.dark #dsfig-u3-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u3-02 .ah{fill:#B1B7C3}html.dark #dsfig-u3-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u3-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u3-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u3-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah9" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh9" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,40 L148,40" marker-end="url(#ah9)"/><path class="e" d="M188,40 L279,40"/><path class="e" d="M317,40 L408,40"/><path class="e" d="M446,40 L535,40" marker-end="url(#ah9)"/><g class="wl"><rect x="91.3" y="31" width="26.4" height="18" rx="9"/><text class="t" x="104.5" y="40" dy=".35em" text-anchor="middle">V1</text></g><g class="wl"><rect x="216.7" y="31" width="33.6" height="18" rx="9"/><text class="t" x="233.5" y="40" dy=".35em" text-anchor="middle">phi</text></g><g class="wl"><rect x="345.7" y="31" width="33.6" height="18" rx="9"/><text class="t" x="362.5" y="40" dy=".35em" text-anchor="middle">phi</text></g><g class="wl"><rect x="478.3" y="31" width="26.4" height="18" rx="9"/><text class="t" x="491.5" y="40" dy=".35em" text-anchor="middle">V2</text></g><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">V</text><circle class="n" cx="169" cy="40" r="18"/><text class="t" x="169" y="40" dy=".35em" text-anchor="middle">P</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">C</text><circle class="n" cx="427" cy="40" r="18"/><text class="t" x="427" y="40" dy=".35em" text-anchor="middle">S</text><circle class="n" cx="556" cy="40" r="18"/><text class="t" x="556" y="40" dy=".35em" text-anchor="middle">L</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Transformer: V1 source, P primary winding N1, C laminated core, S secondary winding N2, L load</figcaption></figure>
Key points.
- Construction: a laminated silicon-steel core (core type or shell type) carries the flux, with primary and secondary windings on it and oil or air cooling.
- Working: an AC voltage $V_1$ on the primary drives an alternating flux in the core, which links the secondary and induces $E_2$ by mutual induction (Faraday's law).
- Transformation ratio: $E_2/E_1 = N_2/N_1 = K = I_1/I_2$; $K>1$ is step-up, $K<1$ step-down.
- EMF equation: ==$E = 4.44\,f\,\Phi_m N$==, so $E_1 = 4.44f\Phi_mN_1$ and $E_2 = 4.44f\Phi_mN_2$.
- Losses: iron loss (hysteresis, reduced by silicon steel; eddy current, reduced by thin laminations), copper loss $I^2R$ (thicker conductor), stray and dielectric loss (shielding, good oil).
- Efficiency $\eta = \dfrac{\text{output}}{\text{output} + P_i + P_{cu}}$; regulation $= \dfrac{E_2 - V_2}{E_2}\times100$.
- Tests: the OC test gives iron loss and $R_0, X_0$; the SC test gives copper loss and $R_{01}, X_{01}$; Sumpner's back-to-back test gives temperature rise at full load with little power.
- Types: by construction (core, shell), phases, use (power, distribution) and voltage (step-up, step-down).
Example 1 (Jun 2022 and Nov 2022). Turns: $N_1 = 100\times1100/400 = \mathbf{275}$; with $\Phi_m=0.08$: $N_1 = 6600/(4.44\times50\times0.08) = \mathbf{372}$, $N_2 = 600/17.76 = \mathbf{34}$. For 120/12 V with 900 turns, $N_2 = \mathbf{90}$.
Example 2 (OC/SC test). OC: $\cos\phi_0 = 20/(200\times0.7) = 0.143$, $I_c = 0.1$ A, $I_m = 0.7\sin\phi_0 = 0.693$ A, so $R_0 = 200/0.1 = \mathbf{2000\ \Omega}$, $X_0 = 200/0.693 = \mathbf{288.7\ \Omega}$. SC (secondary): $Z_{02} = 10/10 = 1\ \Omega$, $R_{02} = 40/100 = 0.4\ \Omega$, $X_{02} = \sqrt{1-0.16} = 0.9165\ \Omega$; refer to primary with $R_{01} = R_{02}/K^2$, $X_{01} = X_{02}/K^2$.
Example 3. $\eta = \dfrac{25000}{25000+350+400}\times100 = \mathbf{97.09\%}$ at unity pf.
Answer frame. Principle question: open with mutual induction, draw the core transformer, develop points 2-4. Loss question: list iron, copper, stray, dielectric with the cure for each. Test question: draw OC and SC circuits, then Sumpner's test.
Asked: [7 marks] (Dec 2023, Jun 2023, Dec 2024) OC and SC test results: $V_0=200$ V, $I_0=0.7$ A, $W_0=20$ W (primary); $V_s=10$ V, $I_s=10$ A, $W_s=40$ W (secondary). Find equivalent circuit parameters referred to primary. Asked: [7 marks] (Jun 2022) A 1100/400 V, 50 Hz transformer has 100 secondary turns; find primary turns. Also 120 V/12 V with 900 primary turns: secondary turns? Asked: [7 marks] (Jun 2022) Necessary tests for efficiency, voltage regulation and temperature rise of winding and insulation of a transformer. Asked: [7 marks] (Nov 2022) 6600/600 V, 50 Hz transformer, $\Phi_m = 0.08$ Wb; find turns in each winding. Asked: [7 marks] (Nov 2022) Enumerate the various losses in a transformer. How can they be minimized? Asked: [7 marks] (Jun 2023) State the different types of transformers. Describe construction and general principle of transformer. Asked: [7 marks] (Jun 2025) Explain the working principle of a single-phase transformer. Asked: [7 marks] (Jun 2025) 25 kVA, 2000/200 V transformer, iron loss 350 W, copper loss 400 W; efficiency at full load.
Last-minute revision
- mmf $= NI$; reluctance $S = l/\mu A$; $\phi = NI/S$.
- B-H curve: initial, steep, knee, saturation; hysteresis gives retentivity and coercive force.
- $L = N^2/S$; $M = k\sqrt{L_1L_2}$.
- Energy $\tfrac12LI^2$; energy density $\tfrac12BH = B^2/2\mu$.
- Series coils: $L_1 + L_2 \pm 2M$.
- Faraday: $e = -N\,d\Phi/dt$; Lenz: induced current opposes the cause.
- Force on conductor $F = BIl\sin\theta$.
- Transformer EMF: $E = 4.44f\Phi_mN$; $E_2/E_1 = N_2/N_1$.
- OC test gives iron loss; SC test gives copper loss.
- $\eta = \text{output}/(\text{output}+P_i+P_{cu})$; 25 kVA example gives 97.09%.
- Sumpner's test gives temperature rise.
Memory hooks
- Domains: "slow, steep, knee, stuck" for the B-H curve regions.
- OC test = Open, Core loss; SC test = Short, Copper loss.
- 4.44 f Phi N: "four-four-four, flux for more".
- Lenz's minus sign = nature refuses free energy.
Coverage checklist
- Basic definitions: no past questions.
- magnetization characteristics of Ferro magnetic materials: B-H curve question (Nov 2022, Dec 2023).
- self inductance and mutual inductance: derivation of M (Jun 2023, Dec 2024).
- energy in linear magnetic systems: energy density derivation (Jun 2022).
- coils connected in series: no past questions.
- AC excitation in magnetic circuits: no past questions.
- magnetic field produced by current carrying conductor: no past questions.
- Force on a current carrying conductor: no past questions.
- Induced voltage, laws of electromagnetic Induction, direction of induced E.M.F.: laws of induction (Dec 2023, Dec 2024, Jun 2025).
- Single phase transformer- General construction, working principle, e.m.f. equation, equivalent circuits, phasor diagram, voltage regulation, losses and efficiency, open circuit and short circuit test: all eight transformer questions.