Electricity

Press a switch and a lamp lights. Behind that instant is a chain: charge pushed by a potential difference, through wires and components, transferring energy from power station to bulb. This reference covers AQA GCSE Physics 8463, topic 4.2 Electricity.

Start with a simple question: a cell, switch and lamp form a series loop. Why does opening the switch stop sustained current? When it is closed, does the lamp use up charge?

The switch must complete a conducting path, and the cell provides a potential difference 电势差. In a steady series loop the current is the same before and after the lamp. The lamp transfers energy; charge is not consumed. Later calculations link $Q=It$, $E=QV$, $P=VI$ and $E=Pt$.

This reference uses standard circuit symbols 电路符号 and the Physics Equations Sheet 物理公式表 when supplied for the examination. Use the sheet issued for your examination series; practise choosing and rearranging equations rather than assuming every future paper has the same support. AQA uses “potential difference” in questions and accepts correct use of “voltage”. Static electricity and electric fields are physics-only content.

A circuit diagram uses standard symbols. Know these: cell, battery, switch (open, closed), lamp, resistor, variable resistor, ammeter, voltmeter, diode, LED, thermistor, LDR and fuse. Ammeters sit in series 串联; voltmeters sit in parallel 并联 across the component.

For charge to flow, the circuit must be closed and include a source of potential difference. Electric current 电流 is a flow of electrical charge 电荷, and its size is the rate of flow:

• $Q$ charge flow in coulombs, C; $I$ current in amperes, A; $t$ time in seconds, s.
• Current has the same value at every point of a single series loop.
• Conventional current flows from + to −; electrons flow the opposite way.

Worked reasoning: charge is not current

Teacher-written: 4.0 C passes a point in 2.0 s in a steady series circuit. Current is charge flow per second:

One ampere means one coulomb per second. The same 4.0 C passes another point of that steady loop in the same 2.0 s. If the same charge takes 4.0 s instead:

Doubling the time for the same charge halves the current.

Charge-flow practice: attempt before checking

Teacher-written: a charger supplies a constant 0.90 A for 25 minutes. Find charge in coulombs, then predict the effect of doubling the time at the same current.

Check: use seconds, because amperes measure coulombs per second.

Twice the time gives twice the charge, at the same current.

Actual exam calculation: current from charge flow

AQA GCSE Physics June 2024 Paper 1H Q10.3 gives a fuse wire melting when 2.0 C flows in 400 ms. Calculate current before checking.

Check: known charge and time mean use $Q=It$, rearranged for current.

This agrees with the official scheme. Teacher extension: treating 400 ms as 400 s would make the denominator 1000 times too large and current 1000 times too small. Check the time unit before substituting.

The current through a component depends on both the potential difference across it and its resistance 电阻:

• $V$ potential difference in volts, V; $I$ current in amperes, A; $R$ resistance in ohms, Ω.
• The greater the resistance, the smaller the current for a given potential difference.

Worked example. A 0.45 V potential difference drives 0.0075 A through a coin. Find the coin's resistance.

• Known: $V$ and $I$; rearrange before substituting.
  $$R = \frac{V}{I} = \frac{0.45\ \text{V}}{0.0075\ \text{A}} = 60\ \Omega$$

Required practical 3: resistance of a wire and resistor combinations

Attach a resistance wire (nichrome or constantan) along a metre rule. Measure the selected length between the actual contact points of a fixed clip and a movable clip. The ammeter is in series with that length; the voltmeter is connected across the same two contact points.

Use a low potential difference and switch off between readings to limit heating. Change length only: keep the wire material, cross-sectional area and temperature constant. For each length record the measured potential difference and current, then calculate $R = V/I$. Repeat readings and investigate inconsistent results.

For example, these teacher-written ideal data illustrate the calculation; they are not experimental measurements:

At constant temperature, for the same material and cross-sectional area, resistance is directly proportional to wire length. Plot calculated resistance against length; a straight line through the origin supports this relationship. The measured potential difference need not be identical at each length, so calculate each resistance from its own paired readings.

A non-zero intercept needs investigation. Check that length was measured between the contact points; contact and lead resistance can also affect results. Do not force the graph through the origin or subtract every intercept as a zero error without identifying its cause.

In the second part of this practical, connect two equal resistors in series, then in parallel. With the ammeter measuring total current and the voltmeter across the whole combination, measure total potential difference and current and calculate total resistance. Compare with one resistor: series has greater total resistance; parallel has smaller total resistance. For two identical 10 Ω resistors, ideal totals are 20 Ω in series and 5 Ω in parallel. The parallel result can be explained from the doubled total current at the same potential difference, without needing a reciprocal-resistance formula.

Required practical 4 measures current through a resistor, filament lamp and diode at a range of measured potential differences across each component. Connect an ammeter in series and a voltmeter in parallel with the component. Vary the pd using a variable dc supply, or a variable resistor in series. Start at zero and stay within component ratings. Record paired readings across a suitable range; repeat and investigate inconsistent readings. Switch off before reversing the supply connections to obtain negative values, using meters that can read the reversed polarity. Plot current vertically against potential difference horizontally.

For the lamp investigation in AQA June 2023 8463/1H Q06.1, Figure 6 covers −6 V to +6 V with readings at 1 V intervals. Collect positive values, then reverse the supply to obtain the negative values; these settings belong to that lamp investigation, rather than every possible component.

For a diode, use a suitable protective resistor in series to limit current and a milliammeter to measure the small current. The protective resistor, not the milliammeter, protects the diode. Measure pd across the diode alone, excluding the protective resistor. Keep the ohmic resistor near constant temperature; the lamp's changing filament temperature is part of the effect being investigated.

• Ohmic conductor 欧姆导体 (fixed resistor at constant temperature): current is directly proportional to potential difference; resistance is constant. Straight line through the origin.
• Filament lamp 白炽灯: resistance increases as its filament temperature rises. The current increases less than proportionally with pd, so the I–V curve flattens away from the origin in both directions.
• Diode 二极管: conducts in the forward direction; reverse current is negligible in this model, so reverse resistance is very high. Do not assume every diode has exactly the same forward voltage.

At a chosen operating point, calculate resistance using $R=V/I$. On a current-against-voltage graph, resistance is not the gradient. For a straight line through the origin, the gradient is $I/V=1/R$; for a curved characteristic use the coordinates of the chosen point, rather than a tangent gradient.

Worked example, adapted from AQA June 2023 8463/1H Q06.2. At +3.0 V, the official lamp graph gives approximately 0.16 A:

At 6.0 V the same paper gives 0.21 A (Q06.3). As a teacher extension, compare the resistance:

The larger resistance is consistent with a hotter filament: increased lattice vibrations make electron motion more difficult. Current still increases, but by a smaller proportion than pd.

• Thermistor 热敏电阻: in the type required here, resistance falls as temperature rises — used as a temperature sensor in a thermostat.
• LDR 光敏电阻: resistance falls as light intensity rises — used as a light sensor in an automatic lighting circuit.

A sensor does not by itself specify when an appliance switches on. For example, a controller set to switch a lamp on when LDR resistance is high will turn it on in darkness. A cooling controller can be arranged to switch on as thermistor resistance falls with rising temperature. State the given controller rule and trace the change through it. Only for the same pd across the sensor does falling resistance imply rising current by $I=V/R$; a fixed supply does not guarantee fixed sensor pd in a series circuit.

In series, the components share one unbranched loop. In parallel, components are connected on separate branches between the same two junctions. Trace these paths in the diagram before applying the current and potential-difference rules.

For components in series:

• the current is the same through each component;
• the supply potential difference is shared between components;
• total resistance is the sum: $R_{total} = R_1 + R_2$.

For components in parallel:

• the potential difference across each component is the same;
• the total current is the sum of the branch currents;
• the total resistance of two resistors is less than the smallest single one.

You must explain both directions: adding resistors in series puts extra opposition in the same unbranched conducting path, so total resistance rises; in parallel each resistor opens an extra path for charge, so more current flows for the same potential difference and the total resistance falls.

You are not required to calculate the combined resistance of two parallel resistors — only to compare and explain.

Worked example. A 6.0 V battery drives a lamp in series with a variable resistor set to 6.0 Ω. The lamp has a resistance of 12 Ω at this operating point.

• Known: supply pd and both resistances at this operating point. Keep the unrounded current when finding the voltage shares.

The shares add to 6.0 V. Equal shares occur only for equal resistances at the operating point; series components do not always share voltage equally.

Teacher-written comparison with fixed resistors. Two 8.0 Ω resistors are connected to an ideal 12 V supply. In series, $R_{total}=R_1+R_2=16\ \Omega$ and $I=V/R_{total}=0.75\ \text{A}$; each resistor has $V=IR=6.0\ \text{V}$. In parallel, each branch has 12 V, so each branch current is $I=V/R=1.5\ \text{A}$ and $I_{total}=I_1+I_2=3.0\ \text{A}$. Adding another parallel resistor gives another current path and increases total current at the same supply pd. No reciprocal-resistance formula is needed here.

For independent parallel branches on an ideal fixed-pd supply, opening one branch stops current in that branch; the other branch still has the same pd. Opening the only series path stops current through both components. If the supply pd changes under load, do not assume the other branch's current is unchanged.

Integrated worked example, adapted from AQA June 2024 8463/1H Q05.5. At 20 °C the question's thermistor graph gives about 80 Ω. It is in series with a 400 Ω resistor across 12 V. Find the pd across the thermistor.

The fixed resistor has the remaining 10 V. This example uses the graph reading supplied above; the complete exam question also requires reading that resistance from the graph. A fixed supply pd does not make the thermistor pd equal to the supply pd.

The UK mains supply is alternating 交流 (ac): the potential difference repeatedly changes direction. Frequency 50 Hz, potential difference about 230 V. Batteries give direct 直流 (dc) potential difference — one direction only.

A three-core cable connects appliances to the mains:

Explain the dangers the exam asks for:

• A live wire is dangerous even when the switch is open: the live side is still at about 230 V relative to earth, and a person touching it completes a circuit to earth.
• Any connection between live and earth is dangerous: a very low resistance path lets a large current flow — through a person or causing a fire. The fuse protects against exactly this.

Electrical appliances transfer energy from batteries or the mains. Work is done when charge flows. The equation chain:

• $P$ power in W; $E$ energy in J; $Q$ charge flow in C; $t$ time in seconds.
• $P = I^2R$ comes from combining $P = VI$ with $V = IR$ — use it when the current and resistance are what you know.

Worked example. A lamp carries 0.21 A at 6.0 V for 30 minutes. Find the energy transferred.

• Convert time: $30\ \text{min} = 1800\ \text{s}$.
  $$P = VI = 6.0\ \text{V} \times 0.21\ \text{A} = 1.26\ \text{W}$$
  $$E = Pt = 1.26\ \text{W} \times 1800\ \text{s} = 2268\ \text{J} \approx 2300\ \text{J}$$

Worked example. A pump motor of resistance 6.0 Ω draws power 4.86 W. Find the current.

• Known: power and resistance; the equation linking exactly these is $P = I^2R$.
  $$I = \sqrt{\frac{P}{R}} = \sqrt{\frac{4.86\ \text{W}}{6.0\ \Omega}} = 0.90\ \text{A}$$

The power rating on an appliance label tells you the energy transferred each second at its working potential difference.

The National Grid 国家电网 is the system of cables and transformers linking power stations to consumers.

• A step-up transformer 升压变压器 raises the potential difference for the transmission cables, so the current is low.
• A step-down transformer 降压变压器 lowers it again for homes.

Why this is efficient: for the same delivered power $P = VI$, a higher potential difference means a smaller current. The power wasted by heating the cables is $P = I^2R$ — halving the current wastes a quarter of the power. So transmitting at very high potential difference keeps the cables' losses small.

(The construction and operation of transformers is Higher Tier, physics only — covered with magnetism in topic 4.7.)

When two insulating materials are rubbed together, electrons — negative charges — are rubbed off one and onto the other:

• the material gaining electrons becomes negatively charged;
• the material losing electrons is left with an equal positive charge.

Charged objects exert forces without contact: like charges repel; unlike charges attract — a non-contact force. A spark jumps when the force is strong enough to make air conduct.

A charged object creates an electric field 电场 around itself: a region where another charge feels a force.

Electrons move; the field tells the force. The field is strongest close to the object and weaker further away.

You must draw the field pattern for an isolated charged sphere: straight radial lines pointing away from a positive charge (or towards a negative one), spaced wider as they get further from the sphere.

• Draw the standard symbols; place ammeters in series, voltmeters in parallel.
• Use $Q = It$, $V = IR$, $P = VI$, $P = I^2R$, $E = Pt$, $E = QV$ — chosen from the words of the question.
• Describe RP3's graph ($R \propto L$) and its zero-error and heating points; describe RP4's circuits and the three I–V shapes.
• State the series and parallel rules for current, potential difference and resistance, and explain both resistance trends.
• Recall the mains values (230 V, 50 Hz, ac) and the three wires with colours and jobs; explain the live-wire dangers.
• Explain the National Grid's efficiency with $P = I^2R$.
• (physics only) Explain charging by friction with electrons, and draw the radial field of a charged sphere.