7.3 Induced potential, transformers and the national grid — Practise with help

1. The size of the current, the magnetic flux density, and the length of the conductor in the field.
2. $P = VI = 12 \times 2.0 = 24\,\text{W}$.

If question 1 went wrong, look again at the motor effect in Section 7.2. If question 2 went wrong, look at electrical power in Section 6.4.

The effect needs a change: either the conductor moves, or the field changes.

Say what is induced across the ends of the conductor, and what happens if the conductor is in a complete circuit.

If an electrical conductor moves relative to a magnetic field, or the magnetic 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 original change. This is a model answer for you to compare with your own wording.

Think about how fast the change happens, and how strong the field is.

Name a change to the movement and a change to the field for the size. For the direction, name a change that turns the effect around.

To increase the size: move the conductor faster through the field, or use a stronger magnetic field, or use a coil with more turns. To reverse the direction: reverse the direction of movement, or reverse the direction of the magnetic field.

The difference is what happens to the connections as the coil turns.

Say that the coil rotates in a field in both cases. Then say what the alternator does to the current each half turn, and what the commutator does in a dynamo.

In both machines a coil rotates in a magnetic field, so a changing field passes through the coil 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.

The ratio of the potential differences equals the ratio of the numbers of turns.

Write $V_s = V_p \times \dfrac{n_s}{n_p}$, then put in the four numbers.

This is a step-down transformer.

You want the primary turns, so rearrange the turns equation first.

Write $n_p = n_s \times \dfrac{V_p}{V_s}$, then work out $2000 \times \dfrac{12}{240}$.

If the transformer is 100% efficient, the power input equals the power output.

Write $V_s I_s = V_p I_p$, then work out $\dfrac{11\,000 \times 0.50}{250}$.

A transformer needs a changing magnetic field in its core.

Say what a steady d.c. does to the field in the core, then say whether a potential difference is induced in the secondary coil.

A transformer works because the alternating current in the primary coil makes a changing magnetic field in the core. The changing field induces a potential difference in the secondary coil. A steady direct current makes a steady field, so the field in the core does not change and no potential difference is induced.

Use $P = VI$. For the same power, what happens to the current when the potential difference is large?

Say that the current is smaller, then link a smaller current to less heating in the cables.

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 transmission cables, so less energy is wasted as the cables heat up.