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Transformers questions
Two coils share one core, and Faraday's law does the rest. The turns ratio sets the voltage ratio, imperfections drain a little power away as heat, and the national grid rests on one consequence, that high voltage means small current and tiny transmission losses.
18 original questions · 48 marks · the transformers notes · Magnetic fields
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Explain how a transformer transfers energy from its primary coil to its secondary coil.
Mark scheme
The alternating current in the primary produces a continually changing magnetic flux in the iron core, which links the secondary coil (1); by Faraday's law the changing flux linkage induces an alternating emf in the secondary, so energy is transferred without an electrical connection (1).A transformer has 1000 turns on its primary coil and 100 turns on its secondary coil. The primary voltage is 230 V. Calculate the secondary voltage.
Mark scheme
Vs = Vp × Ns/Np = 230 × 100/1000 (1)
Vs = 23 V (1)A transformer has fewer turns on its secondary coil than on its primary coil. State whether it is a step-up or a step-down transformer.
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Step-down: the secondary voltage is lower than the primary voltage (1).Explain why a transformer produces no output when its primary coil is connected to a steady dc supply.
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A steady current produces a constant flux in the core (1); with no change in the flux linkage of the secondary, no emf is induced (1).State two causes of energy loss in a real transformer.
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Any two of the following, one mark each (2): resistance of the windings heating the coils; eddy currents induced in the core; flux leakage, with some flux missing the secondary; energy used in repeatedly magnetising the core each cycle.A phone charger contains a transformer that steps 230 V down to 5.0 V. Calculate the turns ratio Np:Ns.
Mark scheme
Np/Ns = Vp/Vs = 230/5.0 (1)
Np:Ns = 46:1 (1)A transformer steps 240 V down to 12 V. The primary coil has 2000 turns. Calculate the number of turns on the secondary coil.
Mark scheme
Ns = Np × Vs/Vp = 2000 × 12/240 (1)
Ns = 100 turns (1)An ideal transformer has a primary voltage of 230 V and draws a primary current of 2.0 A. The secondary voltage is 23 V. Calculate the power transferred and the secondary current.
Mark scheme
P = VpIp = 230 × 2.0 = 460 W (1)
Ideal, so VsIs = VpIp (1)
Is = 460/23 = 20 A (1)A transformer has an input power of 500 W and a useful output power of 450 W. Calculate its efficiency.
An ideal step-up transformer raises 230 V to 1150 V. The secondary current is 0.40 A. Calculate the primary current.
Explain what eddy currents are, why they waste energy in a transformer, and how the design of the core reduces them.
The alternating flux in a transformer core links every turn of both coils. Explain why the induced emf per turn is the same for both coils, and how this leads to the transformer equation Vs/Vp = Ns/Np.
A student tests a transformer with four meters. The primary takes 0.26 A at 230 V; the secondary delivers 4.6 A at 11.5 V. Calculate the efficiency of the transformer and the power dissipated inside it.
A power of 100 kW is transmitted along a cable of resistance 5.0 Ω. Calculate the power lost in the cable when the transmission voltage is 10 kV, and when it is 100 kV. Comment on the results.
A transformer is 96% efficient. The primary coil takes 5.0 A at 230 V and the secondary output is at 46 V. Calculate the input power, the output power and the secondary current.
Explain why electrical energy is transmitted across the National Grid at very high voltage.
A garden lighting system needs an rms supply of between 11.5 V and 12.5 V, taken from the 230 V mains. Three transformers are available.
T1: 1000 primary turns, 52 secondary turns
T2: 1000 primary turns, 60 secondary turns
T3: 750 primary turns, 30 secondary turns
Deduce which transformer, if any, is suitable.An ideal transformer steps the 230 V mains down with a turns ratio of 10:1. The secondary is connected to a 4.6 Ω resistor. Calculate the secondary voltage, the secondary current and the primary current.
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