6.4 Energy transfers and the national grid — Practise

Work through these on your own. Keep the units with your numbers. Some questions ask you to choose the relationship, so read each one before you calculate.

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

A lamp on the 230 V mains supply takes a current of 0.20 A. Calculate its power.
A current of 3.0 A flows through an 8.0 Ω resistor. Calculate the power transferred to the resistor.
A current of 0.50 A flows through a 40 Ω resistor. Calculate the power transferred to the resistor.
A 46 W lamp is switched on for 200 s. Calculate the energy it transfers.
A charge of 40 C flows through an appliance at a potential difference of 230 V. Calculate the energy transferred.
A 460 W appliance is connected to the 230 V mains supply. Calculate the current through it.
A 138 W appliance is switched on for 5.0 minutes. Calculate the energy it transfers.
A 1150 W heater is switched on for 30 minutes. Calculate the energy it transfers.
Explain why the National Grid transfers electrical power at a very high potential difference.
Describe the job of the step-up transformers and the step-down transformers in the National Grid.
Describe how a kettle and an electric motor transfer energy from the supply to a useful store.
An electric motor transfers 2000 J of energy, and 1500 J is transferred usefully. Calculate the efficiency of the motor. This uses the idea of efficiency from Chapter 1.
A heater is connected to the 230 V mains supply and takes a current of 4.0 A. Say which relationships you will use, then calculate the power and the energy transferred in 10 minutes.

Answers and concise solutions

1. Power of the lamp

Answer: $P = VI = 230 \times 0.20 = 46\,\text{W}$ .

2. Power transferred to the resistor

Answer: $P = I^2R = 3.0^2 \times 8.0 = 9.0 \times 8.0 = 72\,\text{W}$ .

3. Power transferred to the resistor

Answer: $P = I^2R = 0.50^2 \times 40 = 0.25 \times 40 = 10\,\text{W} = 1.0 \times 10^1\,\text{W}$ .

4. Energy transferred by the lamp

Answer: $E = Pt = 46 \times 200 = 9200\,\text{J} = 9.2 \times 10^3\,\text{J}$ .

5. Energy transferred to the appliance

Answer: $E = QV = 40 \times 230 = 9200\,\text{J} = 9.2 \times 10^3\,\text{J}$ .

6. Current through the appliance

Answer: $I = \dfrac{P}{V} = \dfrac{460}{230} = 2.0\,\text{A}$ .

7. Energy transferred by the appliance

Answer: Change the time to seconds: $5.0 \times 60 = 300\,\text{s}$ . Then $E = Pt = 138 \times 300 = 41\,400\,\text{J} = 4.1 \times 10^4\,\text{J}$ .

8. Energy transferred by the heater

Answer: $30\,\text{minutes} = 1800\,\text{s}$ , so $E = Pt = 1150 \times 1800 = 2\,070\,000\,\text{J} = 2.1 \times 10^6\,\text{J}$ .

9. Why the National Grid uses a high potential difference

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 squared R, so a lower current wastes much less energy. This makes the transfer efficient.

10. The two kinds of transformer

Step-up transformers increase the potential difference from the power station to the transmission cables.
Step-down transformers decrease the potential difference to a much lower value for homes and businesses.

11. The energy transfers in the kettle and the motor

In a kettle, energy is transferred from the supply to the thermal store of the water, in the heating element.
In an electric motor, energy is transferred from the supply to the kinetic store of the moving parts.

12. Efficiency of the motor

Answer: efficiency $= \dfrac{\text{useful output}}{\text{total input}} = \dfrac{1500}{2000} = 0.75$ , which is 75%.

13. Power and energy transferred

Answer: First use $P = VI = 230 \times 4.0 = 920\,\text{W}$ . Then use $E = Pt$ , with $10\,\text{minutes} = 600\,\text{s}$ : $E = 920 \times 600 = 552\,000\,\text{J} = 5.5 \times 10^5\,\text{J}$ to two significant figures.