7.8 Transformers
9 exam questions worth 27 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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Which type of supply does a transformer need to work?
Choose one answer.
Correct Not quite: the answer is Alternating current
Mark scheme
B: Alternating current
Model answer
Alternating current (B).
A transformer has more turns on the secondary coil than on the primary coil. What type is it?
Choose one answer.
Correct Not quite: the answer is Step-up
Mark scheme
B: Step-up
Model answer
Step-up (B).
Explain why a transformer does not work with direct current.
Mark scheme
a direct current produces a steady magnetic field (1); a changing field is needed to induce a p.d. in the secondary coil (1); so there is no output (apart from when switching on and off) (1)
Model answer
A d.c. produces a steady magnetic field. A p.d. is only induced in the secondary coil when the field changes, so no p.d. is induced.
Describe how a transformer changes the potential difference of an alternating supply.
Mark scheme
a.c. in the primary coil creates a changing magnetic field (1); the soft iron core carries the field through the secondary coil (1); the changing field induces a p.d. in the secondary coil (1); the ratio of the p.d.s equals the ratio of the number of turns (1)
Model answer
An alternating current in the primary coil makes a changing magnetic field. The iron core carries it through the secondary coil, which induces a p.d. there. The p.d. is changed in the ratio of the numbers of turns.
A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. The primary p.d. is 230 V. Calculate the secondary p.d.
Mark scheme
Vs = Vp × Ns ÷ Np (1); 230 × 100 ÷ 500 (1); 46 V (1)
Model answer
V_s = 230 × 100 / 500 = 46 V.
A transformer is to change 12 V to 240 V. The primary coil has 50 turns. Calculate the number of turns needed on the secondary coil.
Mark scheme
Ns = Np × Vs ÷ Vp (1); 50 × 240 ÷ 12 (1); 1000 turns (1)
Model answer
N_s = 50 × 240 / 12 = 1000 turns.
A 100% efficient transformer has a primary p.d. of 12 V and a primary current of 5.0 A. The secondary p.d. is 240 V. Calculate the secondary current.
Mark scheme
Vp Ip = Vs Is (1); 12 × 5.0 = 240 × Is (1); 0.25 A (1)
Model answer
12 × 5.0 = 240 × I_s, so I_s = 60 / 240 = 0.25 A.
Explain why a power station uses a step-up transformer before transmitting electricity along the National Grid.
Mark scheme
a step-up transformer increases the potential difference (1); for the same power, the current is smaller (1); power wasted in the cables is I² R, so it falls (1); so less energy is wasted / the National Grid is efficient (1)
Model answer
The step-up transformer increases the p.d., so the current is smaller for the same power. The power wasted in the cables, I² R, is therefore less, so less energy is lost as heat.
Explain why real transformers are not 100% efficient, and describe two ways in which designers reduce the energy wasted.
Mark scheme
energy is wasted as heat in the coils because of resistance (1); and by eddy currents in the iron core (1); use thick copper wire for the coils (1); laminate the core (1); wind the coils on a closed iron core to reduce the loss of the magnetic field (1)
Model answer
Some energy is wasted as heat in the coils (because of their resistance) and by eddy currents in the iron core. Designers use thick copper wire to reduce the resistance, and make the core out of thin laminated sheets to reduce the eddy currents.
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