Interactive practical

Output of a transformer

Wind different numbers of turns on the secondary of an iron-cored transformer, measure the output with an a.c. voltmeter, and compare the voltage ratio with the turns ratio.

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GCSE Physics

Investigating transformer output

WJEC 3420QS — matched to your course.

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At a glance

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Practical focuses1
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Practical focus

Practical method

Aim

Investigate how the number of turns on the secondary coil of a transformer affects the output potential difference.

Variables

Independent variable: number of turns on the secondary coil
Dependent variable: secondary (output) potential difference
Control variables: primary turns (100); primary ac potential difference (2.0 V); the same iron core.

Apparatus

Method

  1. Wind 100 turns on one side of the iron core (primary) and connect it to a 2.0 V ac supply.
  2. Wind 50 turns on the other side (secondary) and connect an ac voltmeter across it.
  3. Switch on and record the primary and secondary potential differences.
  4. Repeat for 100, 150, 200 and 250 secondary turns.
  5. Compare \(\dfrac{Vs}{Vp}\) with \(\dfrac{Ns}{Np}\).

Sample results

Primary coil 100 turns at 2.0 V ac.

Secondary turnsSecondary pd / VPredicted pd (Vp × Ns/Np) / V
500.951.0
1001.92.0
1502.83.0
2003.74.0
2504.65.0

Analysis

The output pd is directly proportional to the number of secondary turns: \(\dfrac{Vs}{Vp}\) ≈ \(\dfrac{Ns}{Np}\). Each measured value is slightly below the prediction because a real transformer is not 100% efficient: some energy is lost heating the coils and core, and not all the magnetic field links the two coils.

Errors and improvements

ProblemEffectImprovement
Energy lossesVs below predictionUse a laminated core and thicker wire; ensure the core is tightly closed
Turns miscountedWrong ratioCount carefully and recount
DC supply usedNo outputUse an ac supply

Risk assessment

HazardRiskControl measure
Coils warm upBurnsSwitch off between readings
Higher output potential differencesElectric shockKeep the supply pd low (2.0 V) and do not exceed about 12 V output

Common mistakes

Worked analysis

Np = 100, Vp = 2.0 V, Ns = 200. Ideal Vs = 2.0 × 200 ÷ 100 = 4.0 V. Measured 3.7 V. If the primary current is 0.50 A, ideal secondary current = VpIp ÷ Vs = (2.0 × 0.50) ÷ 4.0 = 0.25 A.

Common misconception

A transformer does not work with direct current; it needs an alternating current to produce a changing magnetic field.

Assessment coverage

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Practical 2
Foundation 1Standard 1
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What is the transformer equation?

\(\dfrac{Vp}{Vs}\) = \(\dfrac{Np}{Ns}\).

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