Chapter 6: Review

The chapter problem, solved

The cabin runs from the 230 V mains supply and from a 12 V battery. Work through the facts one at a time.

1. The heater. The heater takes a current of 5.0 A from the 230 V supply.

\begin{aligned} P &= VI \\ &= 230 \times 5.0 \\ &= 1150\,\text{W} \end{aligned}

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{230}{5.0} \\ &= 46\,\Omega \end{aligned}

The heater has a power of 1150 W and a resistance of 46 $\Omega$ .

2. The lamp. The lamp takes a current of 0.20 A from the same supply.

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

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{230}{0.20} \\ &= 1150\,\Omega \end{aligned}

The lamp has a power of 46 W and a resistance of 1150 $\Omega$ .

3. The total current. The heater and the lamp are in parallel, so the potential difference across each one is 230 V. The total current is the sum of the branch currents.

\begin{aligned} I_{total} &= 5.0 + 0.20 \\ &= 5.2\,\text{A} \end{aligned}

The supply provides 5.2 A.

4. The energy the heater transfers. The heater runs for 30 minutes.

\begin{aligned} t &= 30 \times 60 = 1800\,\text{s} \\ E &= Pt \\ &= 1150 \times 1800 \\ &= 2\,070\,000\,\text{J} \\ &= 2.1 \times 10^6\,\text{J} \end{aligned}

To check, find the charge that flows through the heater and use $E = QV$ .

\begin{aligned} Q &= It \\ &= 5.0 \times 1800 \\ &= 9000\,\text{C} \\ E &= QV \\ &= 9000 \times 230 \\ &= 2\,070\,000\,\text{J} \end{aligned}

Both methods give $2.1 \times 10^6\,\text{J}$ to two significant figures.

5. The pump circuit. The pump circuit has a 12 V battery and two resistors of 6.0 $\Omega$ and 2.0 $\Omega$ in series.

\begin{aligned} R_{total} &= 6.0 + 2.0 = 8.0\,\Omega \\ I &= \frac{V}{R_{total}} = \frac{12}{8.0} = 1.5\,\text{A} \end{aligned}

The current is 1.5 A everywhere in the series circuit. Find the potential difference across each resistor.

\begin{aligned} V_{6.0} &= IR = 1.5 \times 6.0 = 9.0\,\text{V} \\ V_{2.0} &= IR = 1.5 \times 2.0 = 3.0\,\text{V} \end{aligned}

Check: $9.0\,\text{V} + 3.0\,\text{V} = 12\,\text{V}$ , which is the battery potential difference.

6. Why the appliances are in parallel. Each appliance is designed to work at 230 V. In parallel, the potential difference across each one is the full 230 V, so each works at its correct power. In series they would share the 230 V, so each would receive less than 230 V and would not work properly. In parallel the two appliances can also be switched on and off separately.

7. Safety. The metal frame of the cabin is connected to the earth wire. If a fault brought the live wire into contact with the frame, the current would flow to earth along the low-resistance path instead of through a person. The frame would not stay at 230 V, so it would be much safer to touch.

8. The charged window. Rubbing the plastic window with a dry cloth transfers electrons from the cloth to the window. The window gains electrons, so it becomes negatively charged. The charged window creates an electric field around itself. The field exerts a force on the nearby dust, so the dust is pulled towards the window and collects on it.

Mixed practice — choose the method first

For each question, say which relationship or idea you will use before you calculate. Keep your units with your numbers.

The cabin heater takes a current of 5.0 A from the 230 V supply. Calculate its power.
The cabin heater takes a current of 5.0 A from the 230 V supply. Calculate its resistance.
The cabin lamp takes a current of 0.20 A from the 230 V supply. Calculate its power.
The cabin lamp takes a current of 0.20 A from the 230 V supply. Calculate its resistance.
The heater takes 5.0 A and the lamp takes 0.20 A. They are connected in parallel. Calculate the total current from the supply.
The 1150 W heater is switched on for 30 minutes. Calculate the energy it transfers.
The heater takes a current of 5.0 A for 30 minutes. Calculate the charge that flows through it.
The pump circuit contains a 6.0 Ω resistor and a 2.0 Ω resistor in series. Calculate the total resistance.
The pump circuit has a total resistance of 8.0 Ω and is connected to a 12 V battery. Calculate the current.
A current of 1.5 A flows through the 2.0 Ω resistor in the pump circuit. Calculate the potential difference across it.
Explain why the heater and the lamp in the cabin are connected in parallel rather than in series.
Explain how the earth wire makes the metal frame of the cabin safer if a fault connects the live wire to the frame.
The plastic window is rubbed with a dry cloth and afterwards attracts dust. Explain how the window becomes charged and how it attracts the dust.
A ball of mass 0.20 kg moves at 15 m/s. Calculate the energy in its kinetic store. This uses Chapter 1.
A force of 20 N moves an object 3.0 m in the direction of the force. Calculate the work done. This uses Chapter 2.
A sound wave travels through air at 340 m/s and has a frequency of 170 Hz. Calculate its wavelength. This uses Chapter 4.
A motor transfers 2000 J of energy, and 1500 J is transferred usefully. Calculate the efficiency of the motor. This uses Chapter 1.
A 920 W appliance is connected to the 230 V mains supply. Say which relationship you will use, then calculate the current through it.

Answers and concise solutions

1. Power of the heater

Answer: $P = VI = 230 \times 5.0 = 1150\,\text{W} = 1.2 \times 10^3\,\text{W}$ to two significant figures.

2. Resistance of the heater

Answer: $R = \dfrac{V}{I} = \dfrac{230}{5.0} = 46\,\Omega$ .

3. Power of the lamp

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

4. Resistance of the lamp

Answer: $R = \dfrac{V}{I} = \dfrac{230}{0.20} = 1150\,\Omega = 1.2 \times 10^3\,\Omega$ to two significant figures.

5. Total current from the supply

Answer: In parallel the branch currents add, so $I_{total} = 5.0 + 0.20 = 5.2\,\text{A}$ . The potential difference across each appliance is 230 V.

6. Energy transferred by the heater

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

7. Charge that flows through the heater

Answer: $Q = It = 5.0 \times 1800 = 9000\,\text{C} = 9.0 \times 10^3\,\text{C}$ . Multiplying by 230 V gives the same energy as $E = Pt$ .

8. Total resistance of the pump circuit

Answer: $R_{total} = 6.0 + 2.0 = 8.0\,\Omega$ .

9. Current in the pump circuit

Answer: $I = \dfrac{V}{R_{total}} = \dfrac{12}{8.0} = 1.5\,\text{A}$ . The same current flows through both resistors.

10. Potential difference across the 2.0 Ω resistor

Answer: $V = IR = 1.5 \times 2.0 = 3.0\,\text{V}$ . The 6.0 Ω resistor has the remaining 9.0 V across it, and $9.0 + 3.0 = 12\,\text{V}$ .

11. Why the appliances are connected in parallel

Each appliance is designed to work at 230 V. In parallel the potential difference across each one is the full 230 V, so each works at its correct power. In series they would share the 230 V.
In parallel the two appliances can be switched on and off separately.

12. How the earth wire keeps the frame safe

If the live wire touches the metal frame, the frame would become live at about 230 V.
The earth wire gives the fault current a low-resistance path to earth, so a large current flows to earth and the frame does not stay at 230 V.

13. How the window charges and attracts dust

Rubbing transfers electrons from the cloth to the window, so the window becomes negatively charged.
The charged window creates an electric field around itself. The field exerts a force on the nearby dust, pulling it towards the window.

14. Energy in the kinetic store

Answer: $E_k = \tfrac{1}{2}mv^2 = 0.5 \times 0.20 \times 15^2 = 22.5\,\text{J}$ , so $23\,\text{J}$ to two significant figures.

15. Work done by the force

Answer: $W = Fs = 20 \times 3.0 = 60\,\text{J} = 6.0 \times 10^1\,\text{J}$ .

16. Wavelength of the sound wave

Answer: $\lambda = \dfrac{v}{f} = \dfrac{340}{170} = 2.0\,\text{m}$ . The wave equation works for sound as well as for light.

17. Efficiency of the motor

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

18. Current through the appliance

Answer: Use $P = VI$ , rearranged to $I = \dfrac{P}{V}$ . Then $I = \dfrac{920}{230} = 4.0\,\text{A}$ .

What you can now do

describe current as a rate of flow of charge, and use $Q = It$ ;
use $V = IR$ , and find a resistance from a $V$ – $I$ graph;
describe how the resistance of a lamp, a diode, a thermistor and an LDR changes;
describe series and parallel circuits, and calculate currents and potential differences in each;
explain why a series connection increases the total resistance and a parallel connection decreases it;
recall the UK supply values and describe the three wires in a mains cable;
explain the dangers of the live wire and the purpose of the earth wire;
use $P = VI$ , $P = I^2R$ , $E = Pt$ and $E = QV$ ;
explain why the National Grid transfers power at high potential difference;
explain static charging, the forces between charges, and electric fields.

Your relevance reflection

Return to the prompt from the chapter opener: where in your own day do you use electricity? Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a plug, a power rating on a label, or a spark when you touch a metal door handle?