Electric charge and current
Key idea: Charge interactions, current and direction, Q = It, source e.m.f. in volts, potential difference, resistance and wire dimensions.
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The core idea
Syllabus and review details
Use the distinction between energy supplied by a source and energy transferred in a component. This Science lesson does not require combinations of sources, resistivity calculations or I–V characteristics.
- K326 / K327 Science Physics componentK326 / K327 · 2027Checked against the syllabus · complete topic coverageK326/K327 2027 syllabus, Electric Charge and Current of Electricity topic 12
Charge and interactions
Charge can be positive, negative or balanced overall
Electric charge, Q, is measured in coulombs (C). An object may have positive charge or negative charge. The sign identifies the kind of charge; it is not a judgement about whether the object is safe.
A neutral object still contains positive and negative charges: their totals balance. In ordinary charging of solid objects, electrons can move from one object to another; the protons remain bound in atomic nuclei. An object that gains electrons becomes negatively charged. Losing electrons leaves a net positive charge.
Worked charge account: two initially neutral objects are rubbed together and electrons move from A to B. A loses negative charge and becomes positive; B gains the same negative charge. The transfer separates charge rather than creating it, so the combined charge remains zero.
Try first: a neutral bead gains four electrons. What is its charge sign, and what happened to its protons?
It is negative because it now has four excess electrons. Its protons did not move between objects.
Use the signs to predict the interaction
Unlike charges
Positive and negative charges attract.
Like charges
Two positive or two negative charges repel.
Use the charge account before applying the interaction rule. Two objects that both gain electrons are both negative, so they repel; the objects A and B in the transfer example have opposite net charges, so they attract.
Independent check: a student says a positive sphere has gained protons and a neutral sphere contains no charged particles. Correct both claims. Predict the interaction between two spheres that have each lost electrons.
Ordinary charging leaves a sphere positive when it loses electrons. A neutral sphere has balanced positive and negative charge. Both spheres that lost electrons are positive, so they repel.
Current, charge flow and direction
Current is the rate of flow of charge
I = Q / t
Current, I, is measured in amperes (A). One ampere means one coulomb of charge passes a point each second. In a metal, electrons move from the negative terminal towards the positive terminal, while conventional currentis defined in the opposite direction, from positive towards negative.
Imagine counting the charge passing a point in a wire during one second. At a steady current, charge does not accumulate in a lamp: the same charge per second enters and leaves it. A smaller current after the lamp would mean charge was continually piling up there.
Apply Q = It with time in seconds
A current of 0.40 A flows for 3.0 minutes. Convert time first:t = 3.0 × 60 = 180 s. ThenQ = It = 0.40 × 180 = 72 C.
Try first: 12 C passes a wire point in 4.0 s. Find the current. If the electrons travel to the left, which way is conventional current?
I = Q/t = 12/4.0 = 3.0 A. Conventional current is to the right, opposite electron motion in the metal.
Independent check: a steady 0.25 A flows through one lamp for 2.0 minutes. Calculate the charge that enters the lamp. A student says only half that charge leaves because the lamp uses it up. Repair the account.
Q = It = 0.25 × 120 = 30 C. The same 30 C leaves during this steady interval. The lamp transfers electrical energy; it does not consume charge.
E.m.f., potential difference and energy
A source supplies energy to the charge
The electromotive force (e.m.f.) of an electrical source is measured in volts (V). It describes the energy supplied per coulomb by the source. Despite its name, e.m.f. is not a mechanical force: a force is measured in newtons, whereas one volt is one joule per coulomb. In a cell, chemical energy is transferred to electrical energy; the cell does not manufacture the circulating charge.
A component transfers energy from the charge
The potential difference (p.d.) across a component is the work done per unit charge in driving charge through the component. It is measured in volts:
V = W / Q
A p.d. of 6.0 V means 6.0 J of energy is transferred for each coulomb that passes through the component. Equivalently, the energy transferred is E = QV (W is another symbol for this work done). Match the voltage to the component whose energy transfer you are calculating; do not automatically substitute the source voltage.
Worked energy account: an ideal 6.0 V source supplies 6.0 J per coulomb. In a series circuit with a 4.0 V p.d. across a motor and 2.0 V across a lamp, each coulomb transfers 4.0 J in the motor and 2.0 J in the lamp. For 3.0 C, those transfers are 12 J and 6 J, adding to the source's 18 J. Charge is conserved while energy is transferred.
Try first: a component transfers 20 J when 5.0 C passes through it. Find its p.d. and explain what the result means.
V = E/Q = 20/5.0 = 4.0 V. Each coulomb transfers 4.0 J in that component.
Independent check: a 9.0 V source drives a circuit in which the p.d. across a heater is 6.0 V. When 8.0 C passes, a learner calculates the heater's energy as 8.0 × 9.0 = 72 J. Identify the first error and calculate the heater's transfer.
The learner used the source voltage for a different component. The heater transfers E = QV = 8.0 × 6.0 = 48 J. The remaining source energy is transferred elsewhere in the stated circuit; it is not missing charge.
Resistance and wire dimensions
Use the voltage and current of the same component
R = V / I
Resistance, R, is measured in ohms (Ω). A component has a p.d. of 9.0 V and a current of 0.30 A. Its resistance isR = 9.0 / 0.30 = 30 Ω.
The voltmeter reading must be across this component and the current must pass through it. Dividing the supply voltage by a branch current generally does not give an individual series component's resistance. One ohm means one volt per ampere.
Compare wires at the same material and temperature
A longer wire gives moving electrons a longer path through the material. A larger cross-sectional area provides more paths alongside one another. Compare one dimension at a time: material and temperature must stay the same, and a length comparison also keeps area fixed.
Length
Resistance is proportional to length: doubling length doubles resistance.
Cross-sectional area
Resistance is inversely proportional to area: doubling area halves resistance.
If diameter doubles, cross-sectional area becomes four times as large, so resistance becomes one quarter when material and length remain unchanged.
Try first: two wires have the same material, temperature and length, but B has twice A's cross-sectional area. Compare their resistances. Would the conclusion still follow if B were also twice as long?
With length fixed, B has half the resistance. Doubling its length as well cancels that reduction, giving the same resistance. A fair one-variable comparison must hold the other quantities constant.
Independent check: a wire has 3.0 V across it and carries 0.20 A. Find its resistance. A second wire of the same material, temperature and area is three times as long; predict its resistance and state the control conditions.
R = V/I = 3.0/0.20 = 15 Ω. The longer wire has resistance 45 Ω. This uses fixed material, temperature and cross-sectional area; a different material or temperature would not justify the length-only prediction.
Common mistakes
“Electrons follow conventional current.”
Remember: in metals, the two directions are opposite.
“A lamp uses up current.”
Remember: the lamp transfers energy; charge continues around the circuit.
“E.m.f. is a force.”
Remember: it is a source quantity measured in volts.
“A thicker wire has more resistance.”
Remember: larger cross-sectional area gives lower resistance.
Check your understanding
Worked example: connect current, p.d. and resistance
A charge of 24 C passes through a component in 8.0 s. The component transfers 48 J of energy. Find the current, potential difference and resistance.
- I = Q/t = 24/8.0 = 3.0 A.
- V = W/Q = 48/24 = 2.0 V. Each coulomb transfers 2.0 J.
- R = V/I = 2.0/3.0 = 0.67 Ω to two significant figures.
The important check is that each relationship uses quantities from the same time interval and the same component.
Guided practice
What charge passes when 2.0 A flows for 15 s?
Q = It = 2.0 × 15 = 30 C.
Find R when V = 12 V and I = 0.50 A.
R = V/I = 12/0.50 = 24 Ω.
A wire’s length triples at fixed area. What happens to R?
Resistance triples because R is proportional to length.
Practise this independently
- Spheres A and C are positively charged; sphere B is negatively charged. State the interaction between A and B, the interaction between A and C, and the unit of charge.
- 18 C passes a point in 6.0 s. Calculate the current, then state the directions of conventional current and electron flow in a metal connected to a cell.
- A current of 0.25 A flows for 8.0 min. Calculate the charge transferred.
- A source is labelled 9.0 V. State the unit of its e.m.f. A motor transfers 24 J when 4.0 C passes through it; find the p.d. across the motor.
- A wire carries 0.20 A when the p.d. across it is 6.0 V. Find its resistance. A second wire of the same material and temperature is twice as long and has twice the diameter. Compare its resistance with the first wire.
Check your answers
- A and B attract because their charges are unlike. A and C repel because their charges are alike. Charge is measured in coulombs (C).
- I = Q/t = 18/6.0 = 3.0 A. Conventional current is from the positive terminal towards the negative terminal; electrons in the metal move the opposite way.
- 8.0 min = 480 s, so Q = It = 0.25 × 480 = 120 C.
- E.m.f. is measured in volts (V). The motor p.d. is work done per unit charge: V = W/Q = 24/4.0 = 6.0 V.
- R = V/I = 6.0/0.20 = 30 Ω. Doubling the length multiplies R by 2; doubling diameter makes area four times as large and divides R by 4. The combined change gives half the original resistance, or 15 Ω.
Try this next
Close the answers and redo questions 3–5 with your own sensible values. If one step is wrong, return to that relationship, explain what each symbol means, and try again.
Practise this topic
The topic check covers charge flow, current and voltage, and resistance. Use the feedback to revisit the matching explanation before trying a fresh question.
Continue with the next resource in this course.
Course and syllabus information
- Course
- SEC G3 Combined Science Physics component
- Edition
- SEC G3 Combined Science Physics component 2027