GCSE Physics (9-1)

7.3 Induced potential, transformers and the national grid — Practise

Work through these on your own. Keep your working neat, and keep the units with your numbers. Every question here is for Higher tier.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

  1. H Describe the generator effect, including what the induced current does to the magnetic field around it.
  2. H State three factors that increase the size of an induced potential difference.
  3. H Explain how an alternator generates a.c. and a dynamo generates d.c.
  4. H A transformer has 400 turns on the primary coil and 100 turns on the secondary coil. The primary potential difference is 12 V. Calculate the secondary potential difference.
  5. H A transformer has 800 turns on the primary coil and 40 turns on the secondary coil. The primary potential difference is 240 V. Calculate the secondary potential difference.
  6. H A step-up transformer changes 12 V to 240 V. The primary coil has 100 turns. Calculate the number of turns on the secondary coil.
  7. H A transformer has a primary potential difference of 240 V and a primary current of 0.25 A. The secondary potential difference is 12 V. Assume the transformer is 100% efficient. Calculate the secondary current.
  8. H Explain why a transformer does not work when it is connected to a steady direct current supply.
  9. H Explain why the national grid transmits electrical power at a very high potential difference.
  10. H Say which equation you would use to find the secondary current of a transformer, and name the quantities you need.
  11. H State two ways to reverse the direction of an induced potential difference.
  12. H A transmission line carries 5.0 MW of power at a potential difference of 25 000 V. Calculate the current in the line.
Answers and concise solutions

1. The generator effect

  • If a conductor moves relative to a magnetic field, or the field around it changes, a potential difference is induced across its ends.
  • If the conductor is part of a complete circuit, the induced potential difference drives a current.
  • The induced current makes a magnetic field that opposes the change that made it.

2. Factors that increase the induced p.d.

  • Move the conductor faster through the magnetic field.
  • Use a stronger magnetic field.
  • Use a coil with more turns.

3. Alternator and dynamo

  • In both machines a coil rotates in a magnetic field, so the field through the coil keeps changing and a potential difference is induced.
  • In an alternator the connections stay the same, so the current in the outside circuit reverses every half turn. The output is alternating current.
  • In a dynamo a commutator reverses the connections every half turn, so the current in the outside circuit always flows the same way. The output is direct current.

4. Secondary potential difference

Answer: Vs=Vp×nsnp=12×100400=3.0 VV_s = V_p \times \dfrac{n_s}{n_p} = 12 \times \dfrac{100}{400} = 3.0\,\text{V}. This is a step-down transformer.

5. Secondary potential difference

Answer: Vs=Vp×nsnp=240×40800=12 VV_s = V_p \times \dfrac{n_s}{n_p} = 240 \times \dfrac{40}{800} = 12\,\text{V}. This is a step-down transformer.

6. Number of turns on the secondary coil

Answer: ns=np×VsVp=100×24012=2000n_s = n_p \times \dfrac{V_s}{V_p} = 100 \times \dfrac{240}{12} = 2000 turns.

7. Secondary current

Answer: VsIs=VpIpV_s I_s = V_p I_p, so Is=VpIpVs=240×0.2512=5.0 AI_s = \dfrac{V_p I_p}{V_s} = \dfrac{240 \times 0.25}{12} = 5.0\,\text{A}.

8. A transformer on d.c.

  • A transformer needs a changing magnetic field in its core.
  • A steady direct current makes a steady field, so the field in the core does not change and no potential difference is induced in the secondary coil.

9. Why the grid uses high voltage

  • For the same power, a higher potential difference means a smaller current, because P = VI.
  • A smaller current transfers less energy to the thermal store of the cables, so less energy is wasted as the cables heat up.

10. Choosing the method

  • Use V_s I_s = V_p I_p, the power equation for a 100% efficient transformer.
  • I need the primary potential difference, the primary current and the secondary potential difference.

11. Reversing the induced p.d.

  • Reverse the direction in which the conductor moves through the field.
  • Reverse the direction of the magnetic field.

12. Current in the transmission line

Answer: I=PV=5.0×10625 000=200 A=2.0×102 AI = \dfrac{P}{V} = \dfrac{5.0 \times 10^6}{25\,000} = 200\,\text{A} = 2.0 \times 10^2\,\text{A}.

Settings

Colour scheme

Applies to every book you read here.

Reading theme

Applies to every book you read here.