Electromagnetic Induction and Alternating Current • Topic 3 of 3

Transformers & LC Oscillations

AC generator. An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A coil of $N$ turns and area $A$ is rotated with angular speed $\omega$ in a uniform field $B$. The flux through it is $\Phi = NBA\cos\omega t$, so by Faraday's law the induced EMF is $\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$, where the peak EMF is $\varepsilon_0 = NBA\omega$. The output is sinusoidal AC, delivered through slip rings.

Transformer. A transformer changes the voltage of an AC supply using mutual induction. Two coils — the primary ($N_p$ turns) and secondary ($N_s$ turns) — are wound on a common laminated soft-iron core. An alternating current in the primary creates a changing flux that links the secondary and induces an EMF in it.

Turns ratio. For an ideal transformer, the same flux links each turn, so the voltages are in the ratio of the turns:

  • $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ — the transformation ratio.
  • A step-up transformer has $N_s > N_p$, so $V_s > V_p$ (raises voltage, lowers current).
  • A step-down transformer has $N_s < N_p$, so $V_s < V_p$ (lowers voltage, raises current).
  • For an ideal (lossless) transformer, power is conserved: $V_pI_p = V_sI_s$, so $\frac{I_s}{I_p} = \frac{N_p}{N_s}$.

Why transformers matter. Power is transmitted over long distances at very high voltage and low current to minimise $I^2R$ heat losses in the cables; step-down transformers then reduce the voltage to a safe $220\,\text{V}$ before it reaches homes. A transformer works only on AC, since a steady DC produces no changing flux.

Transformer losses. Real transformers are not perfect:

  • Copper loss: $I^2R$ heating in the windings — reduced by using thick low-resistance wire.
  • Eddy-current loss: currents induced in the core — reduced by laminating the core.
  • Hysteresis loss: energy spent repeatedly magnetising and demagnetising the core — reduced by using soft iron with a narrow hysteresis loop.
  • Flux leakage: not all flux links both coils — reduced by good core design.

LC oscillations. When a charged capacitor is connected to an inductor, energy oscillates back and forth between the electric field of the capacitor and the magnetic field of the inductor — just like a mechanical oscillator exchanges potential and kinetic energy. The charge oscillates sinusoidally at the natural frequency $f = \frac{1}{2\pi\sqrt{LC}}$. In an ideal (resistance-free) LC circuit, the total energy $U = \frac{q^2}{2C} + \frac{1}{2}LI^2$ stays constant; in practice, resistance damps the oscillations.

Transformer: primary and secondary coils wound on a common soft-iron corelaminatedcorePrimaryNp turnsSecondaryNs turns~AC inVs outVs/Vp = Ns/Np
1
Worked Example
A transformer has 100 turns in the primary and 500 turns in the secondary. If the primary is fed $220\,\text{V}$, find the secondary voltage. Is it step-up or step-down?
Solution
  1. Step 1: Use the turns ratio $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$.
  2. Step 2: $V_s = V_p \times \dfrac{N_s}{N_p} = 220 \times \dfrac{500}{100}$.
  3. Step 3: $V_s = 220 \times 5 = 1100\,\text{V}$.
  4. Step 4: Since $V_s > V_p$ (and $N_s > N_p$), it is a step-up transformer.

Answer: Secondary voltage $= 1100\,\text{V}$; it is a step-up transformer.

2
Worked Example
An ideal step-down transformer reduces $2200\,\text{V}$ to $220\,\text{V}$. If the output current is $10\,\text{A}$, find the input current.
Solution
  1. Step 1: For an ideal transformer, power is conserved: $V_pI_p = V_sI_s$.
  2. Step 2: $I_p = \dfrac{V_sI_s}{V_p} = \dfrac{220 \times 10}{2200}$.
  3. Step 3: $I_p = \dfrac{2200}{2200} = 1\,\text{A}$.

Answer: The input (primary) current is $1\,\text{A}$ — a step-down transformer raises the current as it lowers the voltage.

3
Worked Example
Why can a transformer not operate on a DC supply?
Solution
  1. Step 1: A transformer works by mutual induction, which needs a changing magnetic flux in the core.
  2. Step 2: A steady DC produces a constant current and hence a constant, unchanging flux.
  3. Step 3: With no change in flux, no EMF is induced in the secondary.

Answer: DC gives a constant flux, so no EMF is induced; a transformer needs the changing flux that only AC provides.

4
Worked Example
List the four main energy losses in a real transformer and how each is reduced.
Solution
  1. Step 1: Copper loss — $I^2R$ heating in windings; reduced by thick, low-resistance wire.
  2. Step 2: Eddy-current loss — induced currents in the core; reduced by laminating the core.
  3. Step 3: Hysteresis loss — repeated magnetisation of the core; reduced by using soft iron (narrow loop).
  4. Step 4: Flux leakage — not all flux links both coils; reduced by careful winding and core design.

Answer: Copper loss (thick wire), eddy-current loss (lamination), hysteresis loss (soft iron) and flux leakage (good design).

5
Worked Example
An LC circuit has $L = 0.5\,\text{H}$ and $C = 2\,\mu\text{F}$. Find the frequency of its oscillations.
Solution
  1. Step 1: The natural frequency is $f = \dfrac{1}{2\pi\sqrt{LC}}$.
  2. Step 2: $LC = 0.5 \times 2 \times 10^{-6} = 1 \times 10^{-6}$, so $\sqrt{LC} = 1 \times 10^{-3}\,\text{s}$.
  3. Step 3: $f = \dfrac{1}{2\pi \times 10^{-3}} = \dfrac{1}{6.28 \times 10^{-3}}$.
  4. Step 4: $f \approx 159\,\text{Hz}$.

Answer: The LC oscillation frequency is about $159\,\text{Hz}$.

6
Worked Example
Describe the energy exchange during one cycle of LC oscillations.
Solution
  1. Step 1: Initially the capacitor is fully charged, so all energy is electrical, stored in its electric field ($q^2/2C$).
  2. Step 2: As it discharges through the inductor, the current builds up; energy transfers to the magnetic field of the inductor ($\frac{1}{2}LI^2$).
  3. Step 3: When the capacitor is fully discharged, all energy is magnetic and current is maximum.
  4. Step 4: The inductor then recharges the capacitor with opposite polarity, and the cycle repeats; total energy stays constant in an ideal circuit.

Answer: Energy oscillates between the capacitor's electric field and the inductor's magnetic field, staying constant in an ideal (lossless) LC circuit.

Key Points

  • An AC generator gives $\varepsilon = NBA\omega\sin\omega t$ with peak EMF $\varepsilon_0 = NBA\omega$, delivered through slip rings.
  • A transformer changes AC voltage by mutual induction: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$; step-up raises voltage, step-down lowers it.
  • An ideal transformer conserves power ($V_pI_p = V_sI_s$); it works only on AC, not DC.
  • Transformer losses: copper ($I^2R$), eddy current (cut by lamination), hysteresis (soft iron) and flux leakage.
  • In LC oscillations, energy swaps between the capacitor's electric field and the inductor's magnetic field at $f = \frac{1}{2\pi\sqrt{LC}}$.
Tap an option to check your answer0 / 4
Q1.For a transformer the voltage ratio equals:
Explanation: $V_s/V_p = N_s/N_p$ — the ratio of secondary to primary turns.
Q2.A transformer cannot work on DC because DC produces:
Explanation: Mutual induction needs a changing flux, which steady DC does not give.
Q3.Eddy-current loss in a transformer core is reduced by:
Explanation: Lamination breaks up eddy-current loops, cutting heat loss.
Q4.The frequency of oscillation of an LC circuit is:
Explanation: $f = \frac{1}{2\pi\sqrt{LC}}$, the natural frequency of LC oscillations.