S2 Electricity

Take a torch apart and you will see a battery, a bulb, a metal strip and a sliding clip. The moment the clip touches the battery terminal, the bulb glows. What you have built is a circuit: a route laid out so that charge can leave one terminal of a supply, travel through the components, and arrive back at the other terminal of that same supply.

The word to hold on to is continuously. Charge only keeps moving while the loop stays unbroken. Break it anywhere — lift a switch, loosen a wire, take the cell out — and the movement stops at once. Every bulb in that loop goes dark, no matter how far it sits from the break.

What every circuit must have

Four jobs have to be done in any circuit you will meet at this level, whether it is the wiring inside a hairdryer or the circuit board of a handheld fan.

The conducting path is usually copper wire because copper lets charge pass through it easily. Materials that do this well are called conductors; materials that block it almost completely, such as the plastic sleeving around a wire, are insulators.

Charge is a property that particles of matter carry, in the same way that mass is a property they carry. There are two kinds, and they behave in opposite ways.

Charge is measured in coulombs (C). One coulomb is a large amount of charge: it takes roughly 6 × 10¹⁸ electrons to make it up. In the wires of a circuit it is electrons — the negative charges — that are free to move, which is why a metal can conduct at all.

Whenever charge moves, it carries energy with it. That is the point of a circuit: the supply hands energy to the charge, the charge delivers some of it to the load, and the charge returns to the supply to be topped up again.

One electron drifting along a wire is not something you would notice. What matters is how much charge passes a point, and how quickly.

Current is given the symbol I and is measured in amperes, written A and shortened to amps in everyday speech. One ampere is one coulomb of charge passing a point each second (1 A = 1 C/s). A current of 0.5 A means half a coulomb sweeps past every second — a steady, orderly procession of an enormous number of electrons.

Measuring current: the ammeter

Current is measured with an ammeter. To measure the current through a component, the ammeter is placed in series with that component, so the charge it is measuring must pass through the meter as well. An ammeter has a very low resistance, so adding it to the loop barely changes the current it is there to measure.

Conventional current and electron flow

Long before anyone knew that electrons exist, scientists decided that current flows from the positive terminal of a supply, round the circuit, back to the negative one. They guessed wrong about which particles move, but the convention stuck, and it is still used in every circuit diagram today.

If a current of I amperes flows for t seconds, the charge that passes any point in that time is found by multiplying the two together.

Rearranged, this also gives I = Q / t, the definition of current in symbols, and t = Q / I, which tells you how long a given charge will take to pass. The formula assumes the current stays steady for the whole time t.

Worked example

A hand-held vacuum cleaner draws a steady current of 0.40 A. It is switched on for 2.5 minutes. Calculate the charge that passes through its motor in that time.

Sixty coulombs is about 3.8 × 10²⁰ electrons streaming through that motor in two and a half minutes.

Charge will not make its way round a circuit unless something keeps supplying it with energy. That job is done by the source — a dry cell, a battery or a power supply — which changes chemical, light or mechanical energy into electrical energy. How much energy the source hands over for each coulomb it drives round is what we call the electromotive force, shortened to e.m.f.

Despite the name, e.m.f. is not a force — nothing is being pushed in newtons. It is an energy-per-charge quantity, which is why it is measured in volts (V), the same unit as potential difference.

Since it is energy divided by charge, we can write it as

e.m.f. = energy supplied ÷ charge driven round

V = W / Q

where V = e.m.f. (V), W = energy supplied by the source (J) and Q = charge driven round the circuit (C).

The SI unit of e.m.f. is the volt (V), and one volt is one joule of energy per coulomb of charge: 1 V = 1 J/C.

A short calculation. A cell supplies 24 J of energy while driving 4.0 C round a complete circuit. Its e.m.f. is 24 J ÷ 4.0 C = 6.0 V.

Cells joined together

Cells in parallel do not raise the voltage, but they share the load between them, so the supply lasts longer before it goes flat.

Measuring e.m.f. Place a voltmeter in parallel with the source itself, directly across its terminals — not across any component. A voltmeter has a very high internal resistance, so only a tiny current passes through it and the circuit it is measuring is barely disturbed.

Once charge leaves the source it passes through the components, and a bulb, heater or motor takes energy back out of it. The energy converted per coulomb by one component is its potential difference, shortened to p.d.

The arithmetic matches the e.m.f. case — energy divided by charge — but now the energy is being spent by a component rather than given by a source:

p.d. = energy converted ÷ charge passing through

V = W / Q

where V = potential difference (V), W = energy converted by the component (J) and Q = charge passing through it (C). The unit is again the volt (V), with 1 V = 1 J/C.

A short calculation. A lamp converts 45 J of electrical energy while 15 C passes through it. Its potential difference is 45 J ÷ 15 C = 3.0 V.

Measuring a potential difference

• A voltmeter measures p.d. and goes in parallel with the component you are testing — across the same two points that bracket the bulb or resistor.
• Join the positive terminal of the voltmeter to the positive side of the circuit, otherwise the pointer will swing the wrong way.
• A voltmeter is designed to have high internal resistance, so it draws almost no current and hardly alters the circuit it is measuring.
• An ammeter is different: it must go in series, because it has to measure the current actually flowing through the component.

Both quantities are measured in volts and both compare energy with charge, which is exactly why they are so easily confused. The difference lies in where the energy is going.

Components do not let charge through equally easily. A short thick copper wire lets a large current flow; a long thin heating coil holds the current back badly. This opposition is resistance.

Resistance links the p.d. across a component to the current through it:

V = I × R

and rearranged for resistance itself:

R = V / I

where V = p.d. across the component (V), I = current through it (A) and R = resistance (Ω).

The SI unit of resistance is the ohm (Ω), which is one volt per ampere: 1 Ω = 1 V/A.

For a given p.d. the relationship runs both ways: a larger resistance means a smaller current, and a smaller resistance lets a larger current through. Length and thickness matter too — a long thin wire has high resistance, a short thick wire has low resistance.

Worked example — finding resistance from two meter readings

Step 1 — List what the question gives you.
V = 6.0 V    I = 0.75 A    R = ?

Step 2 — Choose the right arrangement of the equation.
The unknown is resistance, so use the form with R on its own: R = V / I.

Step 3 — Substitute the values.
R = 6.0 V ÷ 0.75 A

Step 4 — Carry out the arithmetic.
R = 8.0

Step 5 — Give the unit and answer the question.
The resistance of the resistor is 8.0 Ω.

Check the unit as you go: volts ÷ amperes gives ohms, since 1 Ω = 1 V/A, so no conversion is needed.

Recognise the reverse pattern too. When the p.d. is wanted, give V = I × R a turn: a 24 Ω resistor carrying 0.15 A has V = I × R = 0.15 A × 24 Ω = 3.6 V across it.

Once you can measure current and potential difference, the next question is how the parts of a circuit are wired to one another. There are only two arrangements to know.

• Series: the components sit one after another along a single loop, so the charges have no choice but to pass through every component in turn.
• Parallel: the components sit on separate branches that meet at the junctions on either side of the supply, so the charges choose a route.

Before touching a calculator, get into the habit of asking three questions about any circuit diagram:

1. How many routes can the charge take? One route means series; more than one means parallel; a mixture of both is treated section by section.
2. Where does the current divide? Current is the same value at every point of a single loop, but it splits at a junction and rejoins afterwards.
3. Which quantity does the source fix? The cell or battery sets the total potential difference. How that total is used up depends entirely on the arrangement.

Take a battery, a switch and three lamps joined end to end. The same stream of charge passes through all three, so the ammeter reading is unchanged wherever you clip it in.

• Current: I = I₁ = I₂ = I₃
• Potential difference: V = V₁ + V₂ + V₃ (the source's total is shared out)
• Effective resistance: R = R₁ + R₂ + R₃

Resistances add in series because each component obstructs the moving charge in turn. Three obstacles in a row obstruct more than one obstacle alone, so the combined resistance is always larger than the biggest single value in the loop.

Now add a fourth lamp to the same battery. The effective resistance rises, and since the supply's potential difference is unchanged, the current falls. Every lamp therefore receives less potential difference than before and all four glow more dimly than three did. There is a second consequence: the loop is a single path, so breaking it anywhere — a filament snapping, a switch opening — stops the current everywhere at once, and every lamp goes dark together. This is a real limitation of series wiring.

Now join three lamps so that each sits on its own branch, with the branches meeting at the junctions either side of the battery. The supply's full potential difference appears across each branch, because every branch is connected directly to both terminals of the battery.

• Potential difference: V = V₁ = V₂ = V₃
• Current: I = I₁ + I₂ + I₃ (charge is conserved at the junction — what leaves the source must return to it)
• Effective resistance: 1/R = 1/R₁ + 1/R₂ + 1/R₃

Because each branch is fed the same potential difference, a branch with a smaller resistance carries a larger current: I = V/R at fixed V means current and resistance pull in opposite directions. A short, thick, low-resistance branch is the easy route, and it takes the greater share of the total current.

Adding a further branch opens one more route, so the combined resistance falls to less than the smallest branch resistance and the total current drawn from the battery rises. If two lamps are wired in parallel and one filament breaks, the other branch is still a complete loop, so it continues to glow at unchanged brightness — the lamps are independent of each other. A parallel circuit therefore needs a switch in each branch if you want to control the branches separately.

Four resistors of 4 Ω, 5 Ω, 3 Ω and another 3 Ω are joined in series across a 12 V battery of negligible internal resistance.

Step 1 — Classify the circuit. There is a single route from one terminal of the battery to the other, with no junction anywhere, so this is a series circuit. Current is the same in every resistor and the 12 V is shared.

Step 2 — Combine the resistances.
R = 4 + 5 + 3 + 3 = 15 Ω

Step 3 — Find the current using Ohm's law on the whole circuit.
I = V / R = 12 / 15 = 0.8 A
This is the current in each resistor, and the reading a series ammeter would give.

Step 4 — Work out the potential difference across each resistor using V = IR:

• 4 Ω: V = 0.8 × 4 = 3.2 V
• 5 Ω: V = 0.8 × 5 = 4.0 V
• 3 Ω: V = 0.8 × 3 = 2.4 V
• 3 Ω (second one): V = 0.8 × 3 = 2.4 V

Step 5 — Check against the source. 3.2 + 4.0 + 2.4 + 2.4 = 12.0 V, which equals the battery's e.m.f. This is conservation of energy: the potential difference the source provides must be used up across the components, no more and no less. Note also that the largest resistance took the largest share of the potential difference — 4.0 V against 2.4 V for an equal current — and every value is smaller than the total, as it must be in series.

Mains wiring in a building is parallel throughout, and each socket or lighting point sits on its own branch with its own switch. Three practical reasons follow directly from the comparison above.

• Every appliance receives the full mains potential difference. A lamp, a kettle and a television are all designed for the same supply of about 240 V, so each must be connected across the full supply rather than in a chain where the supply would be divided among them.
• Appliances work independently. Switching off the kitchen light must not switch off the refrigerator, and a blown filament must not leave the rest of the house dark. In a parallel installation one branch failing leaves the others complete.
• Each appliance draws only the current it needs. Currents from the branches add up, so a high-power appliance can take a large current from the supply without that current having to pass through any other appliance.

The price of this design is that the combined resistance of the installation is low and the total current drawn is high, which is why a heavy load heats the supply cables. That is the reason for fuses, circuit breakers and correctly chosen wire thickness, and it is also why overloading a single socket by plugging in too many appliances is dangerous: each extra branch lowers the combined resistance and raises the total current further.

Electrical power is the rate at which electrical energy is transferred or converted into other forms. A high-power appliance converts a large amount of energy each second; a low-power appliance converts the same energy much more slowly. The SI unit of power is the watt (W), and one watt equals one joule per second (1 W = 1 J/s).

Because power is a rate, it is defined as energy converted per unit time:

P = E / t

where P = power (W), E = energy converted (J) and t = time taken (s). In a circuit, power can also be found from the potential difference across the component and the current through it:

P = V I

where V = potential difference (V) and I = current (A). Substituting V = I R gives P = I²R and P = V²/R; use those two only when the resistance is known or can be found first.

Worked example. Our air fryer is labelled 230 V, 1250 W. Find (i) the current it draws and (ii) the energy it converts in 45 minutes.

(i) I = P / V = 1250 / 230 = 5.43 A (3 s.f.).
(ii) Time first: 45 × 60 = 2700 s. Then E = P t = 1250 × 2700 = 3 375 000 J, that is 3.375 MJ.

Whenever current flows through a wire, some electrical energy becomes thermal energy inside the wire. What we notice depends on whether that heat is wanted, wasted or damaging.

1. It can be the whole point of the appliance. Heating components are built to have high resistance, so a large amount of thermal energy is produced in a small space. A toaster's metal ribbon and a hair dryer's coil glow orange because their resistance is deliberately high and the current through them is large.

2. It is wasted energy everywhere else. The cable feeding a television warms up slightly, and that energy reaches the surroundings doing no useful job. The same happened in an old-style filament lamp, hot to touch because a large share of the energy supplied became heat instead of light. A modern LED gives the same light using far less energy, so it is cheaper to run.

3. It can damage the installation. If resistance or current becomes too large, a wire can become hot enough to soften or melt its plastic insulation. Bare metal is then exposed and can give a shock or start a short circuit, while the heat itself can set light to nearby material and start a fire. Long, thin wires carrying too many devices are the usual culprits.

Energy converted is found from E = P t, but the joule is an awkward unit for a household bill. One kilowatt-hour is 1000 W running for 3600 s:

1 kWh = 1000 × 3600 = 3 600 000 J = 3.6 MJ

The kilowatt-hour (kWh) is simply a larger, more convenient energy unit, and it is the unit electricity is charged in. Convert the power to kilowatts and the time to hours:

Energy in kWh = Power in kW × Time in hours

Keep the two routes separate: watts with seconds give joules, kilowatts with hours give kilowatt-hours.

An electricity bill is an energy bill. The supplier states a tariff in dollars per kilowatt-hour, the meter records the total energy used, and the charge is the two multiplied together:

Cost = Energy used (kWh) × Tariff ($/kWh)

Here is a calculation using our own home and a tariff of $0.28 per kWh. Our air fryer is rated 1250 W and we use it for 45 minutes each day. Find the energy used in a month of 30 days and its cost.

1. Power in kilowatts: 1250 W ÷ 1000 = 1.25 kW.
2. Time in hours: 45 min ÷ 60 = 0.75 h.
3. Energy for one use = 1.25 × 0.75 = 0.9375 kWh.
4. Energy for 30 days = 0.9375 × 30 = 28.125 kWh.
5. Cost = 28.125 × $0.28 = $7.875, about $7.88 for the month.

Four appliances in our home, used as shown over a 30-day month, give this comparison:

Note the dehumidifier: lowest power of the three big items, highest cost, because it runs six hours a day. Frugal choices attack both numbers at once — lower power, fewer hours.

Most marks lost here come from a unit slip, a missing formula or an unfinished comparison. In calculations, write the formula before substituting, show the substituted line, and give the unit on your answer. In explanations, name the quantity that changes and say what it does to another: "the current is larger, so the heating rate P = I²R is larger, so more thermal energy is produced in the wire each second" earns more than "it gets hotter".

Question (10 marks). A study desk has a dehumidifier rated 320 W running 6 hours every day, and an LED desk lamp rated 9 W running 4 hours every day. Electricity costs $0.28 per kWh.

(a) State the formula relating power, energy and time, and give the unit of each quantity in it. (2 marks)
(b) Calculate the energy used by the dehumidifier in one day, in kilowatt-hours. (2 marks)
(c) Calculate the cost of running the dehumidifier for 30 days at $0.28 per kWh. (3 marks)
(d) A student claims that adding one hour a day to the lamp saves more money than removing one hour a day from the dehumidifier. Use calculation to show whether the claim is correct. (2 marks)
(e) Explain why the wires warm up while these appliances are running. (1 mark)

Model answers.

1. (a) P = E / t, with P = power in watts (W), E = energy converted in joules (J) and t = time in seconds (s). (1 mark formula, 1 mark units.)
2. (b) 320 ÷ 1000 = 0.32 kW; t = 6 h. E = 0.32 × 6 = 1.92 kWh. (1 mark conversion, 1 mark answer.)
3. (c) Energy = 1.92 × 30 = 57.6 kWh. Cost = 57.6 × $0.28 = $16.128, about $16.13. (1 mark energy, 1 mark × tariff, 1 mark answer with unit and rounding.)
4. (d) The claim is not correct. The extra lamp hour uses 0.009 kW × 1 h = 0.009 kWh a day, about 0.27 kWh a month, costing 0.27 × $0.28 = $0.08. Removing the dehumidifier hour saves 0.32 kWh a day, about 9.6 kWh a month, costing 9.6 × $0.28 = $2.69. The dehumidifier hour is worth far more because its power rating is about 36 times larger. (1 mark both calculations, 1 mark correct conclusion.)
5. (e) The appliances drive a current through wires that have resistance, so electrical energy is converted to thermal energy in them (P = I²R) and lost to the surroundings instead of doing useful work. (1 mark heating, 1 mark energy wasted; maximum 1 mark.)