6.4 Energy transfers and the national grid — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Chapter 1 and Section 6.1.

A motor transfers 600 J of energy in 5.0 s. Calculate its power.
Use $Q = It$ to find the charge when $I = 0.30\,\text{A}$ and $t = 60\,\text{s}$ .
A 12 $\Omega$ resistor carries a current of 0.50 A. Calculate the potential difference across it.

Answers

$P = \dfrac{E}{t} = \dfrac{600}{5.0} = 120\,\text{W}$ .
$Q = 0.30 \times 60 = 18\,\text{C}$ .
$V = IR = 0.50 \times 12 = 6.0\,\text{V}$ .

If question 1 went wrong, remember that power is the rate of energy transfer. If question 3 went wrong, check that you used $V = IR$ .

Practise with help

1. A lamp is connected to the 230 V mains supply and takes a current of 0.20 A. Calculate its power.

Hint

You know the potential difference and the current, and you want the power. Pick the form of the power equation that uses both.

Next step

Write $P = VI$ and multiply 230 V by 0.20 A.

Full working

\begin{aligned} P &= VI \\ &= 230 \times 0.20 \\ &= 46\,\text{W} \end{aligned}

2. A current of 3.0 A flows through an 8.0 $\Omega$ resistor. Calculate the power transferred to the resistor.

Hint

You know the current and the resistance, so use the form of the power equation with those two quantities.

Next step

Write $P = I^2R$ . Square the current first, then multiply by the resistance.

Full working

\begin{aligned} P &= I^2R \\ &= 3.0^2 \times 8.0 \\ &= 9.0 \times 8.0 \\ &= 72\,\text{W} \end{aligned}

3. The lamp of question 1 has a power of 46 W. It is switched on for 200 s. Calculate the energy it transfers.

Hint

Use the energy equation that links power and time. The time is already in seconds.

Next step

Write $E = Pt$ and substitute $P = 46\,\text{W}$ and $t = 200\,\text{s}$ .

Full working

\begin{aligned} E &= Pt \\ &= 46 \times 200 \\ &= 9200\,\text{J} \\ &= 9.2 \times 10^3\,\text{J} \end{aligned}

4. A charge of 40 C flows through an appliance connected to the 230 V supply. Calculate the energy transferred.

Hint

This question gives you charge and potential difference, not current and time. Use the energy equation with those two quantities.

Next step

Write $E = QV$ and multiply 40 C by 230 V.

Full working

\begin{aligned} E &= QV \\ &= 40 \times 230 \\ &= 9200\,\text{J} \\ &= 9.2 \times 10^3\,\text{J} \end{aligned}

This is the same energy as in question 3, because the charge and the power describe the same lamp.

5. A 460 W appliance is connected to the 230 V mains supply. Calculate the current through it.

Hint

Start from $P = VI$ and make the current the subject.

Next step

Write $I = \dfrac{P}{V}$ and divide 460 W by 230 V.

Full working

\begin{aligned} I &= \frac{P}{V} \\ &= \frac{460}{230} \\ &= 2.0\,\text{A} \end{aligned}

6. Explain why the National Grid transfers electrical power at a very high potential difference, and state what the step-down transformers do.

Hint

Think about the current that is needed for a given power, and then about the energy wasted as heating in the cables.

Next step

Use $P = VI$ to link the potential difference to the current. Then use $P = I^2R$ for the power wasted in the cables.

Full working

For a given power, a higher potential difference means a lower current, because $P = VI$ . The power wasted as heating in the cables is $P = I^2R$ , so a lower current wastes much less energy. Transferring power at a high potential difference therefore makes the National Grid efficient.

Step-down transformers reduce the potential difference from the transmission cables to a much lower value, so that the supply is safe to use in homes and businesses.

This answer is a model answer for you to compare with your own wording.

7. An appliance is connected to the 230 V mains supply. It takes a current of 0.60 A for 5.0 minutes. Say which relationships you will use, then calculate the power and the energy transferred.

Hint

Find the power first, from the potential difference and the current. Then use that power with the time.

Next step

Use $P = VI$ , then change 5.0 minutes to seconds and use $E = Pt$ .

Full working

\begin{aligned} P &= VI = 230 \times 0.60 = 138\,\text{W} \\ t &= 5.0 \times 60 = 300\,\text{s} \\ E &= Pt = 138 \times 300 = 41\,400\,\text{J} \\ &= 4.1 \times 10^4\,\text{J} \end{aligned}

To two significant figures the energy is $4.1 \times 10^4\,\text{J}$ .