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BT-104 · Basic Electrical & Electronics Engineering/Quick Revision Short Notes

Basic Electrical & Electronics Engineering (BT-104) - Unit 3 Short Notes

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.

  1. Magnetomotive force $\text{mmf} = NI$ ampere-turns drives the flux, like emf drives current.
  2. Flux density is $B = \phi/A$ in tesla, and field intensity is $H = NI/l$ in A/m, with $B = \mu_0\mu_r H$.
  3. Reluctance $S = l/(\mu A)$ opposes flux, so $\phi = NI/S$.
  4. 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.

  1. In the initial region B rises slowly because domain walls move only a little and reversibly.
  2. In the steep linear region small increases in H align many domains, so B rises rapidly.
  3. Near the knee point most domains are aligned and the slope falls.
  4. In saturation all domains are aligned, so B hardly increases and the material behaves like air ($\mu_r \to 1$).
  5. 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.
  6. 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.

  1. With reluctance $S$, $L_1 = N_1^2/S$ and $L_2 = N_2^2/S$.
  2. Induced emf is $e = -L\,di/dt$ for self and $e_2 = -M\,di_1/dt$ for mutual induction.
  3. Let $k$ be the coefficient of coupling, the fraction of one coil's flux linking the other, $0 \le k \le 1$.
  4. 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.

  1. Put $L = N^2\mu A/l$ and $H = NI/l$, so $W = \tfrac12\mu H^2\,Al$.
  2. Since $B = \mu H$ and volume $= Al$, $W = \tfrac12 BH \times \text{volume}$.
  3. Energy density is ==$w = W/\text{volume} = \tfrac12 BH = B^2/2\mu$== J/m$^3$.
  4. 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

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>

Definition. Two coupled coils in series have total inductance $L = L_1 + L_2 \pm 2M$.

Key points.

  1. In series aiding, the fluxes add, so $L = L_1 + L_2 + 2M$.
  2. In series opposing, the fluxes subtract, so $L = L_1 + L_2 - 2M$.
  3. 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.

  1. The applied voltage sets the flux, since $V \approx E$.
  2. The excitation current has a magnetizing component that makes the flux.
  3. 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.

  1. Biot-Savart law: $dB = \mu_0 I\,dl\sin\theta/(4\pi r^2)$.
  2. For a long straight wire, $B = \mu_0 I/(2\pi r)$.
  3. For a solenoid, $B = \mu_0 NI/l$.
  4. 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.

  1. The force is maximum when the conductor is perpendicular to the field, $F = BIl$.
  2. Its direction is given by Fleming's left-hand rule.
  3. It is the basis of motor action.
  4. 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.

  1. Faraday's first law: whenever the flux linking a circuit changes, an emf is induced in it.
  2. Faraday's second law: the magnitude of the induced emf equals the rate of change of flux linkage, $e = -N\,d\Phi/dt$.
  3. Lenz's law: the direction of the induced emf is such that the current it drives opposes the change producing it.
  4. The negative sign expresses Lenz's law and is a consequence of conservation of energy, since otherwise energy would be created.
  5. 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).
  6. 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.

  1. 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.
  2. 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).
  3. Transformation ratio: $E_2/E_1 = N_2/N_1 = K = I_1/I_2$; $K>1$ is step-up, $K<1$ step-down.
  4. EMF equation: ==$E = 4.44\,f\,\Phi_m N$==, so $E_1 = 4.44f\Phi_mN_1$ and $E_2 = 4.44f\Phi_mN_2$.
  5. 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).
  6. Efficiency $\eta = \dfrac{\text{output}}{\text{output} + P_i + P_{cu}}$; regulation $= \dfrac{E_2 - V_2}{E_2}\times100$.
  7. 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.
  8. 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.
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