Recall and use the equation for 100% efficiency in a transformer I_p V_p = I_s V_s where p and s refer to primary and secondary

4.5.6 The Transformer

Objective

Recall and use the ideal‑transformer equation (100 % efficient):

\$Ip Vp = Is Vs\$

where the subscripts p and s denote the primary (input) and secondary (output) windings respectively.

1. What is a Transformer?

  • A static electrical device that transfers electrical energy between two or more circuits by electromagnetic induction.

  • It works only with an alternating current (a.c.) supply – a changing magnetic flux is required.

Construction

  • Core – laminated iron (laminations suppress eddy‑current losses).

  • Primary winding – the input side, connected to the a.c. source.

  • Secondary winding – the output side, delivers the transformed voltage/current.

Simple transformer cross‑section

Cross‑section of a transformer showing the laminated iron core, primary winding ( \(Np\) turns) and secondary winding ( \(Ns\) turns). Terminals are labelled \(Vp, Ip\) and \(Vs, Is\).

2. Principle of Operation

When an a.c. flows through the primary winding it produces a time‑varying magnetic flux \(\Phi\) in the core. This same flux links the secondary winding, inducing an emf in each winding (Faraday’s law):

\$\$\mathcal{E}p = -Np\frac{d\Phi}{dt},\qquad

\mathcal{E}s = -Ns\frac{d\Phi}{dt}\$\$

Because the same flux links both windings, the induced voltages are directly proportional to the number of turns:

Turns‑ratio (voltage) relationship

\$\frac{Vs}{Vp}= \frac{Ns}{Np}\$

Current‑ratio relationship (ideal transformer)

\$\frac{Is}{Ip}= \frac{Np}{Ns}\$

3. Ideal (100 % Efficient) Transformer

  • No losses as heat, sound or stray magnetic fields.

  • All input power equals output power.

Therefore

\$P{\text{in}} = P{\text{out}}\quad\Longrightarrow\quad Ip Vp = Is Vs\$

Derivation link: Combining the voltage‑ratio \(\displaystyle Vs/Vp=Ns/Np\) with the power definition \(P=IV\) gives the current‑ratio \(\displaystyle Is/Ip=Np/Ns\), which together lead directly to \(IpVp=IsVs\).

Step‑up vs. Step‑down

  • Step‑up – \(Ns > Np\): voltage ↑, current ↓.

  • Step‑down – \(Ns < Np\): voltage ↓, current ↑.

Real‑world considerations (separate box)

Note: Real transformers are not perfectly efficient. Typical efficiencies are 95–98 % due to core (hysteresis, eddy‑current) and copper losses. In IGCSE questions the transformer is assumed ideal unless losses are explicitly mentioned.

Safety & handling (one‑sentence reminder)

When disconnecting a transformer, always switch off the live (primary) side first; the core can become hot during operation.

4. Summary of Ideal‑Transformer Relationships

QuantityIdeal‑Transformer Expression
Turns ratio\(\displaystyle\frac{Np}{Ns}= \frac{Vp}{Vs}= \frac{Is}{Ip}\)
Voltage ratio\(\displaystyle\frac{Vs}{Vp}= \frac{Ns}{Np}\)
Current ratio\(\displaystyle\frac{Is}{Ip}= \frac{Np}{Ns}\)
Power balance (ideal)\(\displaystyle Ip Vp = Is Vs\)

5. Using the Power Equation \(\;Ip Vp = Is Vs\)

  1. Identify which two of the four quantities (\(Ip, Vp, Is, Vs\)) are given.

  2. If the turns ratio is also supplied, first find the missing voltage or current using the ratio formulas.

  3. Insert the known values into the power equation and solve for the unknown.

  4. Check that the result is consistent with the turns‑ratio relationship.

6. Worked Example

Question: A step‑down transformer has \(Np = 500\) turns and \(Ns = 100\) turns. The primary is connected to a 240 V, 50 Hz supply and draws a current of 2 A. Assuming 100 % efficiency, find the secondary voltage and current.

Solution:

  1. Turns ratio: \(\displaystyle\frac{Np}{Ns}= \frac{500}{100}=5\)

  2. Voltage ratio (step‑down): \(\displaystyle\frac{Vs}{Vp}= \frac{Ns}{Np}= \frac{1}{5}\)

  3. Secondary voltage: \(\displaystyle Vs = Vp \times \frac{1}{5}= 240\;\text{V}\times0.2 = 48\;\text{V}\)

  4. Power balance: \(\displaystyle Ip Vp = Is Vs\)

  5. Secondary current: \(\displaystyle Is = \frac{Ip Vp}{Vs}= \frac{2\;\text{A}\times240\;\text{V}}{48\;\text{V}} = 10\;\text{A}\)

Result: the secondary delivers 48 V at 10 A.

7. Common Mistakes to Avoid

  • Direction of the turns ratio: Remember that a step‑up transformer has \(Ns > Np\) (voltage ↑, current ↓) and a step‑down transformer has \(Ns < Np\) (voltage ↓, current ↑).

  • Assuming ideal behaviour: Use the power equation only when the question states the transformer is ideal (or when no loss information is given).

  • Symbol consistency: Keep the notation \(Vp, Vs, Ip, Is\) throughout; avoid mixed symbols such as \(V_{primary}\).

  • Forgetting the same flux link: The voltage ratio derives from the fact that the *same* magnetic flux links both windings.

8. Practice Questions

  1. A transformer has 250 turns on the primary and 500 turns on the secondary. The primary is connected to a 120 V source and draws 0.8 A. Assuming 100 % efficiency, calculate the secondary voltage and current.

  2. A step‑down transformer reduces 240 V to 30 V. If the secondary supplies a lamp that draws 3 A, what is the primary current?

  3. In an ideal transformer, the primary voltage is 230 V and the secondary voltage is 115 V. If the secondary current is 4 A, find the primary current.

  4. Explain why a transformer will not work with a direct‑current (d.c.) supply.

9. Quick Summary

  • For an ideal transformer, input power = output power: \(Ip Vp = Is Vs\).

  • Turns ratio links voltage and current:

    \(\displaystyle\frac{Np}{Ns}= \frac{Vp}{Vs}= \frac{Is}{Ip}\).

  • Use the voltage‑ratio to find missing voltages, the current‑ratio for missing currents, and verify with the power equation.

  • Transformers operate only with a.c.; they are static devices and must be switched off at the live side for safety.

  • Real devices are slightly less efficient (≈95–98 %); treat them as ideal unless losses are mentioned in the question.