Full chapter
Electric charge and current of electricity
All 4 topics and the revision summary on one page.
01
Electric charge and interactions
Make a guess. You are not marked.
A plastic rod rubbed with a cloth becomes negatively charged. What has moved?
Show the answer
Electrons, which have moved from the cloth onto the rod.
Rubbing transfers electrons. The rod gains electrons and becomes negative; the cloth loses them and is left equally positive.
In short
- There are two kinds of charge, positive and negative.
- Like charges repel; unlike charges attract.
- Rubbing moves electrons from one object to the other.
Electric charge has two signs: positive and negative. Charged objects can exert forces on one another without touching.
Charge is represented by q or Q and is measured in coulombs, C. A positive sign means a net positive charge; a negative sign means a net negative charge.
Neutral does not mean no charged particles
Atoms contain positively charged protons in their nuclei and negatively charged electrons. A neutral object has equal total positive and negative charge, so its net charge is zero.
An object with an excess of negative charge has a negative net charge. An object with an excess of positive charge has a positive net charge. Positive and negative describe the charge balance, not whether an object is made only of protons or only of electrons.
Predict attraction and repulsion
Predict the electric force on each charged object
These small objects carry the shown net charges. Only their electric forces on one another are shown; orange arrows are forces, not motion.
What the observation establishes
Two charged balls repel
They have the same sign. They could both be positive or both be negative; repulsion alone does not distinguish those possibilities.
If a ball repels a known positive charge in this model, that additional comparison identifies the ball as positive.
Attraction has more than one possible explanation. A charged object can attract an oppositely charged object, but it can also attract a neutral object. Attraction alone does not prove that both objects carry opposite net charges.
Check your understanding
Two small charged balls, separately suspended and isolated from other charged objects, repel. What can be concluded about their charge signs?
Show the answer
Same sign, but this test cannot tell positive from negative.
Like charges repel. Both positive and both negative fit the observation. A further comparison with a known charge would be needed to identify the sign.
When charge moves through a wire, the rate at which it passes a point is the electric current.
02
Current, charge and electron flow
In short
- Current is the rate of flow of charge: I = Q / t, in A.
- Measure current with an ammeter connected in series.
- Charge is measured in coulombs, C.
Current tells you how quickly charge passes a point in a circuit. A larger current means more charge passes each second.
Charge is measured in coulombs, C. Electrons carry negative charge. In a metal wire, mobile electrons can drift through the material while the positive ions remain in its structure.
Electric current is the rate of flow of charge. It is measured in amperes, A. One ampere means one coulomb passes a cross-section each second: 1 A = 1 C/s. Equivalently, 1 C = 1 A s: a coulomb is an ampere-second. Current is represented by I; charge by Q or q.
Imagine marking one cross-section of a wire and counting the total charge that passes it during an interval. A steady 0.30 A means 0.30 C passes each second. It does not mean the whole wire contains only 0.30 C of charge.
Follow charge through the external metal circuit
The cell is discharging into the resistor. Green arrows show conventional current; blue arrows show electron drift. Arrows give direction, not speed.
Steady current: 0.30 A. Time: 2.0 min = 120 s.
Charge passing X: Q = It = 0.30 × 120 = 36 C.
The resistor transfers energy. It does not use up the charge passing through it.
Conventional current and electron drift
Conventional current is defined in the direction in which positive charge would move. In the external circuit of a discharging cell, it goes from the positive terminal, through the components, towards the negative terminal.
Electrons are negative, so their drift in the metal wire is in the opposite direction: from the negative terminal towards the positive terminal through the external circuit. The cell symbol's long line marks its positive terminal and its short line the negative terminal.
A lamp transfers energy as charge passes through it; it does not use up that charge. Electrons are already present throughout the metal circuit. Do not picture the lamp waiting for one electron to leave the cell and complete a whole journey before it responds.
Charge passing a point
A current flows for two minutes
A steady current of 0.30 A flows for 2.0 min. First convert the time: 2.0 x 60 = 120 s.
Q = It = 0.30 x 120 = 36 C.
In a separate example, 18 C passes in 45 s. Its average current is I = Q/t = 18/45 = 0.40 A.
If the current is given as 300 mA, use 300/1000 = 0.300 A. The prefix milli means one thousandth.
Try it step by step
Work it out in steps: Time in s, then Charge Q = It in C.
Show the answer
Problem 1.
Time in s: t = 3.0 x 60 = 180 s
Charge Q = It in C: Q = It = 0.50 x 180 = 90 C (2 sf)
Problem 2.
Time in s: t = 1.5 x 60 = 90 s
Charge Q = It in C: Q = 2.0 x 90 = 180 C (2 sf)
Problem 3.
Time in s: t = 4.0 x 60 = 240 s
Charge Q = It in C: I = 250 mA = 0.25 A, so Q = 0.25 x 240 = 60 C (2 sf)
More detail (optional)
Q = It applies directly when I is constant over the interval. For a changing current, use the average current for that interval, or find the charge in each stated constant-current part and add the amounts. An instantaneous reading does not establish the current throughout a long interval.
Measure current through a component
An ammeter, shown as a circle containing A, connects in series with the branch being measured. Open the circuit at the chosen position and include the meter in the path, so the branch's charge passes through it.
For a d.c. measurement, connect its positive terminal on the side from which conventional current enters. Choose a suitable range; if the expected value is uncertain, start with a sufficiently high range and reduce it when appropriate for a clearer reading. Keep the reading within the selected range.
An ammeter has very low resistance. Connecting it directly across a cell would provide a low-resistance path, rather than measure the current through the intended component. The resistance arrangement shows the correct current and voltage connections together.
An ammeter reads amperes, not coulombs. To determine charge for a steady current, also measure the time interval, for example with a stopwatch, then use Q = It.
Watch out for this
A student says: "Current is used up as it flows through a lamp, so less current returns to the cell." What is wrong with this?
Show the answer
Charge is not used up. In a series circuit the current is the same at every point. What the lamp takes is energy, which the charge carries from the cell.
Check your understanding
A steady current of 0.30 A flows through a wire for 2.0 min. How much charge passes a marked cross-section?
Show the answer
36 C
The current carries 0.30 C past the cross-section each second. In 120 s, Q = 0.30 x 120 = 36 C.
03
E.m.f. and potential difference
In short
- Potential difference is the energy transferred per unit charge between two points, in V.
- Measure it with a voltmeter connected in parallel across the component.
- E.m.f. is the energy the source gives to each coulomb of charge.
Voltage describes energy transferred per coulomb of charge. The source and the component have different roles in that energy transfer.
Work done is an energy transfer, measured in joules, J. Charge is measured in coulombs, C. Current measures coulombs per second; voltage measures joules per coulomb.
- Electromotive force: e.m.f.
- The e.m.f. of a source is the work done per unit charge by the source in driving charge around a complete circuit. It is measured in volts. A 1.5 V source supplies 1.5 J per coulomb in this energy account.
- Potential difference: p.d.
- The p.d. across a component is the work done per unit charge in driving charge through the component. It is measured in volts. A p.d. of 3.0 V means 3.0 J is transferred in that component for each coulomb passing through it.
Small voltages may be given in millivolts: 1 mV = 0.001 V. Divide a reading in millivolts by 1000 to convert it to volts.
A cell transfers energy from its chemical store into the electrical pathway. In a resistor, energy is transferred to internal stores; in a motor, part can be transferred mechanically to a load. Charge carries on through the circuit while energy is transferred. The energy stores and pathways account helps keep these two ideas separate.
The name electromotive force does not mean a force measured in newtons. E.m.f. is an energy-per-charge quantity, just as p.d. is. Neither quantity measures how many coulombs pass each second.
Energy per coulomb
Find a component's potential difference
A component transfers 60 J when 12 C passes through it.
P.d. = 60/12 = 5.0 J/C = 5.0 V.
For a separate component with p.d. 3.0 V, passing 8.0 C transfers work of 3.0 x 8.0 = 24 J.
These are supplied energy and charge amounts. A temperature rise alone would not measure every energy transfer unless the heated mass, its properties and other transfers were also accounted for.
Try it step by step
Work it out in steps: Charge Q = It in C, then Energy transferred E = VQ in J.
Show the answer
Problem 1.
Charge Q = It in C: Q = It = 0.20 x 50 = 10 C
Energy transferred E = VQ in J: E = VQ = 6.0 x 10 = 60 J (2 sf)
Problem 2.
Charge Q = It in C: Q = 1.5 x 40 = 60 C
Energy transferred E = VQ in J: E = 12 x 60 = 720 J (2 sf)
Problem 3.
Charge Q = It in C: Q = 0.40 x 25 = 10 C
Energy transferred E = VQ in J: E = 3.0 x 10 = 30 J (2 sf)
Measure voltage across two points
A voltmeter, shown as a circle containing V, connects in parallel across the component: one lead at each of its terminals. This measures the p.d. between those two points. It is not inserted into the main current path as an ammeter is.
For a positive d.c. reading, connect the meter's positive lead to the higher-potential side and its negative lead to the lower-potential side. Reversing the leads reverses the sign shown by a suitable digital meter; it does not mean the component has changed its resistance.
Select a voltage range that includes the expected reading. For example, a 0-5 V range is unsuitable for a 6.0 V p.d. Once a suitable range is established, finer resolution helps distinguish nearby readings. For an analogue scale, read at eye level to reduce parallax.
A voltmeter connected across a source with no other external circuit can give an estimate of its e.m.f. when the meter draws negligible current. For the ideal-source model, the source's terminal p.d. equals its e.m.f. The combined meter diagram instead measures the p.d. across a resistor.
Check your understanding
A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?
Show the answer
5.0 V: each coulomb transfers 5.0 J in the component.
Potential difference is work done per unit charge: V = 60 / 12 = 5.0 J/C = 5.0 V. It is an energy-per-charge quantity, not a rate.
04
Resistance and wire dimensions
Make a guess. You are not marked.
A wire has a resistance of 4 ohm. A second wire of the same material and the same thickness is twice as long. What is its resistance?
Show the answer
8 ohm, twice as much.
Resistance is proportional to length for the same material and cross-section. Charge has twice as much wire to pass through, so the resistance doubles.
In short
- Resistance R = V / I, in ohms.
- A longer wire has more resistance; a thicker wire has less.
- A filament lamp's resistance rises as it gets hotter.
Resistance relates the potential difference across a component to the current through it. At the same p.d., a greater resistance gives a smaller current.
Current I is measured through a component in amperes. Potential difference V is measured across its terminals in volts. Use readings for the same component when finding its resistance.
This defines resistance at the stated operating point: the voltage and current under the conditions of that reading. If the component heats up or its conditions change, its resistance can change, so a new reading may give a different value.
Measure current through and p.d. across the same resistor
A is an ammeter in series. V is a voltmeter connected to the resistor's two terminals. The marked meter polarities suit this d.c. source.
R = V/I = 6.0/0.20 = 30 Ω.
This arrangement uses the ideal meter model: the voltmeter takes negligible current and the ammeter causes negligible p.d. The ammeter therefore measures the resistor current.
Use corresponding readings
Determine resistance, then predict current
The p.d. across a resistor is 6.0 V and the current through it is 0.20 A.
R = V/I = 6.0/0.20 = 30 Ω.
If the same resistor remains at 30 Ω when the p.d. becomes 9.0 V, then I = V/R = 9.0/30 = 0.30 A.
This prediction assumes the resistance stays at 30 Ω. If the resistor heats up, its resistance can change and the current will differ.
Try it step by step
Work it out in steps: Resistance R = V / I in ohm, then Current at the new p.d. in A.
Show the answer
Problem 1.
Resistance R = V / I in ohm: R = 4.0 / 0.25 = 16 ohm
Current at the new p.d. in A: I = V / R = 6.0 / 16 = 0.375 A, which is 0.38 A to 2 sf
Problem 2.
Resistance R = V / I in ohm: R = 9.0 / 0.30 = 30 ohm
Current at the new p.d. in A: I = 12 / 30 = 0.40 A (2 sf)
Problem 3.
Resistance R = V / I in ohm: R = 7.5 / 0.50 = 15 ohm
Current at the new p.d. in A: I = 3.0 / 15 = 0.20 A (2 sf)
Check your understanding
A voltmeter reads 6.0 V across a resistor while an ammeter measures 0.20 A through that resistor. What is its resistance at this operating point?
Show the answer
30 ohm
R = V/I = 6.0 / 0.20 = 30 ohm. The two readings refer to the same resistor. This result alone does not show that its resistance stays constant at other operating points.
How wire dimensions affect resistance
Compare wires of the same material at the same temperature. Changing the material or temperature can change the relationship between their dimensions and resistance.
- Greater length gives greater resistance
- Resistance is proportional to length, R ∝ L. Doubling the length while keeping the cross-section unchanged doubles the resistance. Charge must travel through a longer section of the material.
- Greater cross-sectional area gives smaller resistance
- Resistance is inversely proportional to cross-sectional area, R ∝ 1/A. Doubling that area while keeping length unchanged halves the resistance. The wider cross-section provides more conducting material alongside the path.
The cross-sectional area is the area you see when looking straight at a cut end. It is not the curved outer surface of the wire. For a circular wire, A = πd2/4, so area changes with the square of the diameter.
Double the length and double the diameter
Compare the same material at the same temperature. Both drawings use the same scale; the labels give ratios, not dimensions to measure from your screen. The area is the circular end section.
Original wire: 8.0 ohm
Changed wire: 4.0 ohm
Twice the diameter gives four times the cross-sectional area. The resistance changes by 2/4 = 1/2, so 8.0 Ω becomes 4.0 Ω.
Compare one change at a time
Twice the length and twice the diameter
The original wire has resistance 8.0 Ω.
- Doubling its length gives a resistance factor of 2.
- Doubling its diameter gives an area factor of 22 = 4, so the resistance factor from area is 1/4.
- Combine the factors: new R = 8.0 x 2/4 = 4.0 Ω.
For any comparison at the same material and temperature, Rnew/Rold = (Lnew/Lold) x (Aold/Anew).
Determine resistance and investigate wire length
Use a suitable low-voltage classroom supply, with an ammeter in the current path and a voltmeter across the component. Record the p.d. and current together after the readings settle, and calculate R = V/I. Check meter ranges and polarity before collecting readings.
To investigate length, use the same uniform wire and vary the distance between its electrical contacts. Measure this active length with a ruler. The unused wire beyond the contacts is not part of the measured section.
- Keep the wire's material and cross-sectional area fixed. Use several measured active lengths.
- At each length, record the current through and p.d. across that section. Calculate its resistance from the paired readings.
- Keep the wire temperature as steady as practical: use a suitably low current and switch off between readings where appropriate. Wait for it to return to the comparison conditions if it warms.
- Plot resistance vertically against active length horizontally. Proportionality predicts a straight line through the origin for the wire alone at constant cross-section and temperature.
More detail (optional)
If diameter is needed, use a micrometer or suitable calipers, check for zero error and take readings at several positions and orientations. Avoid squeezing or measuring a damaged part of the wire. Convert the diameter to consistent units before calculating area.
Contact and lead resistance may add to a measured resistance, depending on the voltage lead positions. Wire heating can change readings, and a varying diameter means the cross-section is not uniform. Repeating and averaging readings can reveal scatter, but it does not automatically remove these effects. Explain the specific limitation and how the method addresses it.
Twice the diameter is four times the area. Also keep the voltage and current paired: dividing a voltage across one branch by the current through a different branch does not generally give either branch's resistance.
Check your understanding
A wire has resistance 8.0 ohm. A second wire of the same material and temperature has twice its length and twice its diameter. What is the second resistance?
Show the answer
4.0 ohm: length factor 2, area factor 4.
For the same material and temperature, R is proportional to L/A. The new resistance is 8.0 x 2 / 4 = 4.0 ohm.
Revision summary
Charge and current
Charge has positive and negative signs and is measured in coulombs, C. Like charges repel; unlike charges attract. Neutral means balanced total positive and negative charge. Attraction alone does not prove that two objects have opposite net charges.
- Charge, Q or q
- Unit: coulomb, C; 1 C = 1 A s (one ampere-second). For steady current, Q = It. Convert time to seconds and milliamperes to amperes: 1 mA = 0.001 A.
- Current, I
- Rate of flow of charge; unit: ampere, A = C/s. I = Q/t gives the average over the stated interval. An ammeter connects in series with the measured path.
- Direction in an external metal circuit
- Conventional current goes from the discharging cell's positive terminal towards its negative terminal. Electron drift is opposite. A component transfers energy without consuming the charge.
Energy per charge
- E.m.f., E
- Work done per unit charge by the source in driving charge around the complete circuit; unit: volt, V = J/C.
- Potential difference, V
- Work done per unit charge in driving charge through a component; also measured in volts. 1 mV = 0.001 V. Work done = p.d. x charge. A voltmeter connects across the component's two terminals.
Resistance and wire comparisons
- Resistance, R
- R = V/I; V = IR; I = V/R. Unit: ohm, Ω = V/A. Use the voltage across and current through the same component at the same operating point.
- Wire dimensions
- Same material and temperature: R ∝ L and R ∝ 1/A. For a circular cross-section, A ∝ d2. Twice the length and twice the diameter give half the resistance.
- Measurement
- Use suitable meter ranges and d.c. polarities. Keep wire temperature steady in length comparisons and measure the active length between contacts. Contact resistance, heating and diameter variation need specific attention.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
A cell is connected to a lamp. Distinguish between the electromotive force (e.m.f.) of the cell and the potential difference (p.d.) across the lamp. [2]
Saved on this browser only. Anyone using this browser can see it.
Exam-style question
Define electric current. A current of 0.40 A flows through a lamp for 5.0 minutes. Calculate the charge that passes through the lamp. [3]
Saved on this browser only. Anyone using this browser can see it.
Exam-style question
Define the resistance of a component. When the potential difference across a resistor is 6.0 V, the current in it is 0.25 A. Calculate its resistance. [3]
Saved on this browser only. Anyone using this browser can see it.
Quick recall
Two small charged balls, separately suspended and isolated from other charged objects, repel. What can be concluded about their charge signs?