4.2 Electricity — printable coverage pack

4.2.1.1 · Standard circuit diagram symbols

Circuit diagrams represent components with standard symbols rather than pictures of the apparatus.
Place an ammeter in series with the component whose current is measured and a voltmeter in parallel across that component.
A valid measuring circuit has a source of potential difference, a complete conducting loop and correctly joined component symbols.
A common error is to put the voltmeter in series or the ammeter in parallel, which changes or prevents the intended measurement.

Tier 1 · Easy

1. Draw a circuit diagram containing one cell, one open switch and one lamp, all connected in a single loop.[3 marks]
Answer
- A single complete loop using the standard cell, open-switch and lamp symbols, with no extra branches.
Method: Draw one cell symbol, then an open-switch symbol and a lamp symbol in series. Join the components with straight conducting lines to make one loop; the switch contacts remain separated.

Tier 2 · Standard

1. Draw a circuit that can measure both the current through a fixed resistor and the potential difference across it. Include a battery and a switch.[4 marks]
Answer
- Battery, switch, ammeter and resistor in series, with a voltmeter connected in parallel across the resistor; all components shown by standard symbols.
Method: Make the main loop from the battery, switch, ammeter and resistor so the ammeter carries the resistor current. Add a separate branch containing the voltmeter between the two ends of the resistor, so it measures the resistor's potential difference.

Tier 3 · Hard

1. A student draws a resistance-measuring circuit with the ammeter connected across the resistor and the voltmeter inserted in the main loop. State both corrections and describe the corrected circuit.[4 marks]
Answer
- Move the ammeter into series with the resistor.
- Connect the voltmeter in parallel across the resistor; the source, switch, ammeter and resistor form the main loop.
Method: Current through a component is measured by making that current pass through the ammeter, so the ammeter belongs in the main series path. Potential difference is measured between the component's two terminals, so the voltmeter must form a parallel branch across the resistor.

4.2.1.2 · Electrical charge and current

Electric current is the rate of flow of electrical charge, linked by $Q=It$ , where charge is in coulombs, current in amperes and time in seconds.
A source of potential difference and a closed circuit are needed for charge to flow; current has the same value at every point in one closed loop.
For example, a current of $0.50\,\text{A}$ for $20\,\text{s}$ transfers $Q=(0.50)(20)=10\,\text{C}$ .
A common error is to treat current as an amount of charge rather than a rate, or to use minutes directly in $Q=It$ .

Tier 1 · Easy

1. A current of $0.35\,\text{A}$ flows for $40\,\text{s}$ . Calculate the charge that flows.[2 marks]
Answer
- $14\,\text{C}$
Method: Use $Q=It$ : $Q=(0.35)(40)=14\,\text{C}$ .

Tier 2 · Standard

1. $240\,\text{C}$ of charge passes a point in a circuit in $3.0$ minutes. Calculate the current.[3 marks]
Answer
- $1.3\,\text{A}$
Method: Convert the time: $3.0\,\text{min}=180\,\text{s}$ . Rearrange $Q=It$ to $I=Q/t$ , so $I=240/180=1.33\,\text{A}$ , which is $1.3\,\text{A}$ to two significant figures.

Tier 3 · Hard

1. A device carries $0.75\,\text{A}$ for $80\,\text{s}$ and then $0.30\,\text{A}$ for a further $150\,\text{s}$ . Determine the total charge transferred.[4 marks]
Answer
- $105\,\text{C}$
Method: Calculate each charge separately. During the first interval, $Q_1=(0.75)(80)=60\,\text{C}$ . During the second, $Q_2=(0.30)(150)=45\,\text{C}$ . Therefore $Q_{\text{total}}=60+45=105\,\text{C}$ .

4.2.1.3 · Current, resistance and potential difference

Potential difference, current and resistance are linked by $V=IR$ ; at a fixed potential difference, a greater resistance gives a smaller current.
Measure current with an ammeter in series and potential difference with a voltmeter in parallel, then calculate $R=V/I$ .
For a $12\,\Omega$ component carrying $0.25\,\text{A}$ , the potential difference is $V=(0.25)(12)=3.0\,\text{V}$ .
A common error is to divide in the wrong order when finding resistance: it is potential difference divided by current, not current divided by potential difference.

Tier 1 · Easy

1. A $15\,\Omega$ resistor has a potential difference of $6.0\,\text{V}$ across it. Calculate the current.[2 marks]
Answer
- $0.40\,\text{A}$
Method: Rearrange $V=IR$ to $I=V/R$ . Therefore $I=6.0/15=0.40\,\text{A}$ .

Tier 2 · Standard

1. A component carries $0.24\,\text{A}$ when the potential difference across it is $3.6\,\text{V}$ . Its resistance remains constant. Calculate its resistance and the current when the potential difference is $9.0\,\text{V}$ .[4 marks]
Answer
- Resistance $=15\,\Omega$
- Current $=0.60\,\text{A}$
Method: First use $R=V/I=3.6/0.24=15\,\Omega$ . With the resistance unchanged, $I=V/R=9.0/15=0.60\,\text{A}$ .

Tier 3 · Hard

1. A $0.80\,\text{m}$ uniform wire at constant temperature has $3.0\,\text{V}$ across it and carries $0.25\,\text{A}$ . The wire is replaced by $0.50\,\text{m}$ of the same wire. Assume resistance is proportional to length. Calculate the new resistance and current at $3.0\,\text{V}$ .[5 marks]
Answer
- New resistance $=7.5\,\Omega$
- New current $=0.40\,\text{A}$
Method: The original resistance is $R=V/I=3.0/0.25=12\,\Omega$ . Scale by the length ratio: $R_{\text{new}}=12(0.50/0.80)=7.5\,\Omega$ . Then $I_{\text{new}}=V/R=3.0/7.5=0.40\,\text{A}$ .

4.2.1.4 · Resistors

At constant temperature an ohmic conductor has current directly proportional to potential difference; a diode conducts in one direction and has very high resistance in reverse.
Use an ammeter in series and a voltmeter in parallel, vary the supply, and calculate $R=V/I$ from corresponding readings to investigate a component.
A filament lamp's resistance rises as its filament gets hotter; thermistor resistance falls as temperature rises, and LDR resistance falls as light intensity rises.
A common error is to call every curved current-potential difference graph an experimental mistake; curvature can show that the component is non-linear.

Tier 1 · Easy

1. An automatic garden light must switch on when it becomes dark. State how the resistance of its LDR changes as darkness increases.[1 mark]
Answer
- The LDR's resistance increases as the light intensity decreases.
Method: An LDR has lower resistance in brighter light. Therefore reducing the light intensity makes its resistance increase.

Tier 2 · Standard

1. A filament lamp carries $0.40\,\text{A}$ at $2.0\,\text{V}$ and $0.80\,\text{A}$ at $6.0\,\text{V}$ . Calculate its resistance at each potential difference and explain the change.[4 marks]
Answer
- Resistance rises from $5.0\,\Omega$ to $7.5\,\Omega$ because the filament becomes hotter.
Method: At $2.0\,\text{V}$ , $R=V/I=2.0/0.40=5.0\,\Omega$ . At $6.0\,\text{V}$ , $R=6.0/0.80=7.5\,\Omega$ . The larger current heats the filament more, so its resistance increases.

Tier 3 · Hard

1. Describe an investigation of how the resistance of a thermistor changes with temperature. Include the circuit, measurements, one control and the processing of results.[6 marks]
Answer
- Use a thermistor with an ammeter in series and voltmeter in parallel, measure temperature and paired current-potential difference readings over a range, control the supply potential difference, calculate $R=V/I$ , and plot resistance against temperature.
Method: Place the thermistor in a water bath with a thermometer. Connect it in a circuit with an ammeter in series and a voltmeter across it. Keep the supply potential difference constant. Change the bath temperature gradually, allow the reading to settle, and record temperature, current and potential difference at each point. Calculate each resistance using $R=V/I$ and plot resistance against temperature; the downward trend shows that thermistor resistance decreases as temperature increases.

4.2.2 · Series and parallel circuits

In series, current is the same through each component, supply potential difference is shared and resistances add: $R_{\text{total}}=R_1+R_2+\ldots$ .
In parallel, potential difference is the same across each branch and total current is the sum of the branch currents.
Adding a series resistor makes charge flow through more resistance, while adding a parallel branch provides another path and reduces total resistance.
A common error is to add parallel resistances as if they were in series; this specification requires only the qualitative result that the parallel total is below the smallest branch resistance.

Tier 1 · Easy

1. Two resistors of $4\,\Omega$ and $7\,\Omega$ are connected in series. Calculate their total resistance.[1 mark]
Answer
- $11\,\Omega$
Method: Series resistances add, so $R_{\text{total}}=4+7=11\,\Omega$ .

Tier 2 · Standard

1. A $2.0\,\Omega$ resistor and a $4.0\,\Omega$ resistor are connected in series to a $12\,\text{V}$ supply. Calculate the circuit current and the potential difference across each resistor.[5 marks]
Answer
- Current $=2.0\,\text{A}$
- Potential differences are $4.0\,\text{V}$ and $8.0\,\text{V}$ respectively.
Method: The total resistance is $2.0+4.0=6.0\,\Omega$ , so $I=V/R=12/6.0=2.0\,\text{A}$ . The drops are $V_1=IR_1=(2.0)(2.0)=4.0\,\text{V}$ and $V_2=(2.0)(4.0)=8.0\,\text{V}$ ; they sum to the $12\,\text{V}$ supply.

Tier 3 · Hard

1. Two branches are connected in parallel across a $12\,\text{V}$ supply. One branch contains a $6.0\,\Omega$ resistor. The other branch carries $3.0\,\text{A}$ . Calculate the resistance in the second branch and the current from the supply.[5 marks]
Answer
- Second-branch resistance $=4.0\,\Omega$
- Supply current $=5.0\,\text{A}$
Method: Each parallel branch has the full $12\,\text{V}$ . For the second branch, $R=V/I=12/3.0=4.0\,\Omega$ . The first-branch current is $I=12/6.0=2.0\,\text{A}$ . Branch currents add, so the supply current is $2.0+3.0=5.0\,\text{A}$ .

4.2.3.1 · Direct and alternating potential difference

A direct potential difference keeps one polarity, so current in the circuit has one direction; cells and batteries supply dc.
An alternating potential difference repeatedly reverses polarity, so the current also repeatedly reverses direction.
The UK mains supply is approximately $230\,\text{V}$ ac at $50\,\text{Hz}$ , meaning $50$ complete cycles each second.
A common error is to say ac merely changes size; its defining feature is that the potential difference reverses direction.

Tier 1 · Easy

1. State whether a battery supplies direct or alternating potential difference, and give the defining feature of that supply.[2 marks]
Answer
- Direct potential difference; its polarity does not reverse.
Method: A battery keeps the same positive and negative terminals, so it supplies a direct potential difference and drives current in one direction.

Tier 2 · Standard

1. The UK mains supply has a frequency of $50\,\text{Hz}$ . Calculate the number of complete cycles in $0.30\,\text{s}$ and describe what happens to the polarity.[3 marks]
Answer
- $15$ complete cycles; the polarity repeatedly reverses.
Method: Frequency is cycles per second, so the number of cycles is $50\times0.30=15$ . Because the supply is alternating, its polarity reverses during every cycle.

Tier 3 · Hard

1. Source A maintains a potential difference of $+6.0\,\text{V}$ . Source B varies between positive and negative values and completes $25$ cycles in $0.50\,\text{s}$ . Identify each supply type, calculate the frequency of B and compare the current directions they produce.[5 marks]
Answer
- A is dc; B is ac at $50\,\text{Hz}$ ; A drives current in one direction whereas B repeatedly reverses it.
Method: A keeps a fixed polarity, so it is direct. B changes polarity, so it is alternating. Its frequency is $f=25/0.50=50\,\text{Hz}$ . A therefore produces current in one direction, while B produces a current that reverses direction.

4.2.3.2 · Mains electricity

In a three-core cable the live wire is brown, the neutral wire is blue and the earth wire has green-and-yellow stripes.
The live wire carries the alternating potential difference, the neutral completes the circuit, and the earth is a safety wire that carries current only during a fault.
Live-to-earth is about $230\,\text{V}$ , while neutral and earth are at or close to $0\,\text{V}$ .
A common error is to assume an open switch makes the live wire safe; if the switch is in the neutral wire, internal parts can remain connected to the live supply.

Tier 1 · Easy

1. State the insulation colour of the live, neutral and earth wires in a UK three-core mains cable.[3 marks]
Answer
- Live: brown
- Neutral: blue
- Earth: green and yellow stripes
Method: Recall the standard identification: brown is live, blue is neutral, and green-and-yellow striped insulation identifies earth.

Tier 2 · Standard

1. A fault makes the metal case of a mains appliance touch the live wire. Explain how the earth wire reduces the danger to a user.[4 marks]
Answer
- The earth wire provides a low-resistance path for a large fault current, keeps the case near $0\,\text{V}$ and causes the protective device to disconnect the supply.
Method: The earth wire connects the case to earth potential. If live contacts the case, current flows through the earth wire rather than through a person. The large fault current makes a fuse or circuit breaker disconnect the live supply, so the case does not remain live.

Tier 3 · Hard

1. A mains lamp is wired so that its switch opens the neutral wire rather than the live wire. The lamp goes out when the switch is opened. Explain why the lamp holder may still be dangerous to touch.[4 marks]
Answer
- The open switch stops the current but leaves the lamp holder connected to the live wire at about $230\,\text{V}$ relative to earth, so touching it while earthed could complete a dangerous path.
Method: Opening the neutral wire breaks the normal circuit, so the lamp turns off. However, the live connection has not been isolated. Parts of the holder can therefore remain at the live potential relative to earth, and a person touching them could provide a current path to earth.

4.2.4.1 · Power

Electrical power is the rate of energy transfer and can be calculated using $P=VI$ or $P=I^2R$ .
Choose $P=VI$ when potential difference and current are known, and $P=I^2R$ when current and resistance are known.
A device with $12\,\text{V}$ across it and current $2.0\,\text{A}$ transfers energy at $P=(12)(2.0)=24\,\text{W}$ .
A common error in $P=I^2R$ is to forget to square the current; one watt means one joule transferred per second.

Tier 1 · Easy

1. A motor has a potential difference of $12\,\text{V}$ across it and a current of $3.0\,\text{A}$ . Calculate its power.[2 marks]
Answer
- $36\,\text{W}$
Method: Use $P=VI$ : $P=(12)(3.0)=36\,\text{W}$ .

Tier 2 · Standard

1. A heating resistor has resistance $8.0\,\Omega$ and carries $2.5\,\text{A}$ . Calculate the power transferred.[2 marks]
Answer
- $50\,\text{W}$
Method: Use $P=I^2R$ : $P=(2.5)^2(8.0)=6.25\times8.0=50\,\text{W}$ .

Tier 3 · Hard

1. A mains heating element is rated at $1.8\,\text{kW}$ when connected to $230\,\text{V}$ . Calculate its current and resistance at this operating point.[5 marks]
Answer
- Current $=7.8\,\text{A}$
- Resistance $=29\,\Omega$
Method: Convert power: $1.8\,\text{kW}=1800\,\text{W}$ . From $P=VI$ , $I=P/V=1800/230=7.83\,\text{A}$ . Then $P=I^2R$ gives $R=P/I^2=1800/(7.83)^2=29.4\,\Omega$ . To two significant figures, the results are $7.8\,\text{A}$ and $29\,\Omega$ .

4.2.4.2 · Energy transfers in everyday appliances

Electrical appliances transfer energy to other stores, such as the kinetic energy of a motor or the thermal energy of a heating device.
Calculate transferred energy with $E=Pt$ or $E=QV$ , using power in watts and time in seconds for an answer in joules.
A $500\,\text{W}$ appliance used for $30\,\text{s}$ transfers $E=(500)(30)=15\,000\,\text{J}$ .
A common error is to compare appliance power ratings as if they were amounts of energy; power states how quickly energy is transferred.

Tier 1 · Easy

1. A $60\,\text{W}$ fan runs for $5.0$ minutes. Calculate the energy transferred.[3 marks]
Answer
- $18\,000\,\text{J}$ or $18\,\text{kJ}$
Method: Convert the time: $5.0\,\text{min}=300\,\text{s}$ . Then $E=Pt=(60)(300)=18\,000\,\text{J}=18\,\text{kJ}$ .

Tier 2 · Standard

1. $9000\,\text{C}$ of charge flows through a $12\,\text{V}$ motor. Calculate the energy transferred and name the main useful energy transfer.[3 marks]
Answer
- $108\,000\,\text{J}$
- Electrical energy is transferred to kinetic energy.
Method: Use $E=QV$ : $E=(9000)(12)=108\,000\,\text{J}$ . A motor's intended output is movement, so the useful transfer is to kinetic energy.

Tier 3 · Hard

1. A $1.5\,\text{kW}$ appliance operates for $18$ minutes from a $230\,\text{V}$ supply. Calculate the energy transferred and the charge that flows through the appliance.[5 marks]
Answer
- Energy $=1.62\times10^6\,\text{J}$
- Charge $=7.0\times10^3\,\text{C}$
Method: Convert the quantities: $P=1500\,\text{W}$ and $t=18\times60=1080\,\text{s}$ . Then $E=Pt=(1500)(1080)=1.62\times10^6\,\text{J}$ . From $E=QV$ , $Q=E/V=(1.62\times10^6)/230=7.04\times10^3\,\text{C}$ , which is $7.0\times10^3\,\text{C}$ to two significant figures.

4.2.4.3 · The National Grid

The National Grid is the system of cables and transformers that links power stations to consumers.
Step-up transformers increase potential difference for transmission, and step-down transformers reduce it to a much lower value for consumers.
For the same transferred power, $P=VI$ shows that increasing potential difference reduces current, so $P_{\text{loss}}=I^2R$ gives smaller heating losses in the cables.
A common error is to claim that a step-up transformer creates energy; it changes potential difference and current while allowing efficient energy transfer.

Tier 1 · Easy

1. Name the transformer used before long-distance transmission and the transformer used before electricity enters homes.[2 marks]
Answer
- Step-up transformer before transmission
- Step-down transformer before homes
Method: The transmission potential difference is raised by a step-up transformer. Near consumers it is reduced to a much lower domestic value by a step-down transformer.

Tier 2 · Standard

1. A transmission line transfers $6.0\,\text{MW}$ at $300\,\text{kV}$ . Calculate the current in the line and explain why this is preferable to transmitting the same power at a much lower potential difference.[4 marks]
Answer
- Current $=20\,\text{A}$
- The smaller current causes less heating loss in the cables.
Method: Convert units and use $P=VI$ : $I=P/V=(6.0\times10^6)/(300\times10^3)=20\,\text{A}$ . For the same power, a lower potential difference would require a larger current. Since cable heating is proportional to $I^2R$ , the high-potential-difference transmission wastes less power.

Tier 3 · Hard

1. A cable of resistance $0.80\,\Omega$ transfers $2.4\,\text{MW}$ . Compare the power lost in the cable when transmission is at $12\,\text{kV}$ and at $240\,\text{kV}$ .[6 marks]
Answer
- At $12\,\text{kV}$ the loss is $32\,\text{kW}$ ; at $240\,\text{kV}$ it is $80\,\text{W}$ , so the higher potential difference reduces the loss by a factor of $400$ .
Method: At $12\,\text{kV}$ , $I=P/V=(2.4\times10^6)/(12\times10^3)=200\,\text{A}$ , so $P_{\text{loss}}=I^2R=(200)^2(0.80)=32\,000\,\text{W}=32\,\text{kW}$ . At $240\,\text{kV}$ , $I=10\,\text{A}$ and $P_{\text{loss}}=(10)^2(0.80)=80\,\text{W}$ . The loss ratio is $32\,000/80=400$ .

4.2.5.1 · Static charge (physics only)

Rubbing certain insulating materials transfers electrons: the material gaining electrons becomes negative and the material losing them is left equally positive.
Determine the sign from electron movement, remembering that the charged particles transferred between the materials are electrons.
Like charges repel and unlike charges attract, and both effects are non-contact forces; a large build-up of charge can discharge as a spark.
A common error is to say positive charge moves onto an object during rubbing; in this model, electrons move and the positive charge is due to an electron deficit.

Tier 1 · Easy

1. During rubbing, material X gains electrons from material Y. State the charge left on each material.[2 marks]
Answer
- X becomes negatively charged
- Y becomes positively charged
Method: X has gained negative electrons, so it becomes negative. Y has lost the same electrons and is left with an equal positive charge.

Tier 2 · Standard

1. A negatively charged insulating strip repels a second charged strip without touching it. What does this show about the charge on the second strip and the type of force involved?[3 marks]
Answer
- The second strip is negatively charged
- The repulsion is a non-contact force
Method: Repulsion occurs only between charges of the same type. Since the known strip is negative, the other charged strip must also be negative. The strips exert the force while separated, so it is non-contact.

Tier 3 · Hard

1. A student rubs an insulating rod with a cloth and the rod becomes negatively charged. The rod is then brought close to a metal object and a spark occurs. Explain the charging and the spark in terms of electrons.[5 marks]
Answer
- Electrons transfer from the cloth to the rod, leaving the rod negative and the cloth equally positive; when the charged rod approaches the metal, electrons move rapidly through the air gap, producing a spark and reducing the charge difference.
Method: Rubbing transfers electrons onto the insulating rod, where charge can remain because the material does not conduct it away. The cloth loses those electrons. Near the metal object, the large electrical effect across the small gap makes charge move suddenly through the air. This rapid transfer is the spark and partially discharges the rod.

4.2.5.2 · Electric fields (physics only)

A charged object creates an electric field around itself; another charged object placed in the field experiences a force.
Draw the field around an isolated charged sphere with radial lines: arrows point away from a positive sphere and towards a negative sphere.
Closer field lines represent a stronger field, so the field and the force on a given charge are greater nearer the charged object.
A common error is to draw field lines crossing; at any point the force on a positive test charge has only one direction.

Tier 1 · Easy

1. Draw the electric field pattern around an isolated positively charged sphere.[2 marks]
Answer
- Symmetrical radial field lines perpendicular to the sphere's surface, with arrowheads pointing away from the sphere.
Method: Space several straight radial lines evenly around the sphere. Put arrows on every line pointing outwards because the sphere is positive; the lines must not cross.

Tier 2 · Standard

1. A small positive test charge is placed first near a positively charged sphere and then farther away. State the force direction in both positions and compare the force sizes.[3 marks]
Answer
- The force is away from the sphere in both positions and is stronger at the nearer position.
Method: The sphere's field points outwards, so a positive test charge is repelled along that direction. The electric field is strongest close to the charged sphere and weaker farther away, so the nearer force is larger.

Tier 3 · Hard

1. A charged metal dome is brought progressively closer to an earthed metal sphere until a spark crosses the gap. Use the electric-field model to explain why there is a force before contact and why a spark becomes more likely as the gap decreases.[5 marks]
Answer
- The dome's field acts on charges in the other sphere without contact; decreasing the gap increases the field and force, and a sufficiently strong field causes charge to cross the air gap as a spark.
Method: The charged dome creates an electric field throughout the surrounding space. Charges in the earthed sphere are therefore acted on even before the objects touch, which explains the non-contact force. Moving the objects closer makes the field in the gap stronger. Eventually charges can move through the air between the objects; the rapid transfer is observed as a spark and reduces the charge separation.