K326 / K327 / 2027
Electric charge and current of electricity overview

Full chapter

Electric charge and current of electricity

All 4 topics and the revision summary on one page.

01

Electric charge and interactions

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

Like charges repel. Unlike charges attract.Two positive charges repel, two negative charges repel, and a positive and a negative charge attract.

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.

Like charges repel and unlike charges attractThree rows show separately charged objects A and B. Two positive objects have force arrows pointing apart, as do two negative objects. Opposite-sign objects have force arrows pointing towards one another. Each object has its own force arrow, acting along the line joining the objects. Force direction alone does not specify velocity.Two positive charges repel++ABForce on AForce on BTwo negative charges repel--ABForce on AForce on BUnlike charges attract+-ABForce on AForce on B
Each force arrow acts on the object beside it. Like-sign objects are pushed apart; opposite-sign objects are pulled towards each other in this simple isolated charged-object model.

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.

Optional check Two small charged balls, separately suspended and isolated from other charged objects, repel. What can be concluded about their charge signs?
Two small charged balls, separately suspended and isolated from other charged objects, repel. What can be concluded about their charge signs?

When charge moves through a wire, the rate at which it passes a point is the electric current.

02

Current, charge and electron flow

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.

Current = charge passing / time takenI = Q/t; Q = It; t = Q/I. For steady current, use Q in C, I in A and t in s.

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.

Conventional current and electron drift have opposite directionsA closed rectangular circuit contains a cell on the left and a resistor in the top wire. The cell's long upper plate is positive and its short lower plate is negative. In the external circuit, conventional current goes clockwise from the positive terminal, through the resistor, and back to the negative terminal. Electron drift in the metal wire goes the opposite way. One cross-section X is marked in the bottom wire. At a steady current of 0.30 ampere, 36 coulombs pass this cross-section in 120 seconds. No electron path through the cell's electrolyte is drawn.+-CellResistorConventionalcurrentElectrondriftCross-section X

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.

In the external metal wire, conventional current and electron drift have opposite directions. The marked cross-section carries 0.30 C each second, so 36 C passes it in 120 s. The arrows show direction, not speed.

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.

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.

Optional check A steady current of 0.30 A flows through a wire for 2.0 min. How much charge passes a marked cross-section?
A steady current of 0.30 A flows through a wire for 2.0 min. How much charge passes a marked cross-section?

03

E.m.f. and potential difference

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.
1 volt = 1 joule per coulomb1 V = 1 J/C. Potential difference = work done / charge; work done = potential difference x charge.

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.

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.

Optional check A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?
A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?

04

Resistance and wire dimensions

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.

R = V/IResistance = potential difference / current. Rearrangements: V = IR and I = V/R. The unit is the ohm, Ω; 1 Ω = 1 V/A.

This defines resistance at the stated operating point: the voltage and current under the conditions of that reading. It does not by itself establish that a component's resistance stays the same when conditions change.

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.

An ammeter in series and a voltmeter in parallel measure the resistorThe cell has its positive plate above its negative plate on the left. The top branch runs through ammeter A and then a resistor before returning to the source along the right and bottom wires. The ammeter's left terminal is positive. Two separate connections descend from the exact left and right resistor terminals to voltmeter V; its left terminal is positive. The supplied ammeter reading is 0.20 ampere and the p.d. across the resistor is 6.0 volts. With ideal meters, the ammeter gives the resistor current, so its resistance is 30 ohm. No wire bypasses the resistor or either meter symbol.+-SourceA+-0.20 AMeasured resistorV+-6.0 VBoth readings concern this resistor

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.

The ammeter measures the current in the resistor's path, and the voltmeter connects across its two terminals. With ideal meters, the voltmeter draws negligible current and the ammeter adds negligible resistance. The supplied readings are 6.0 V and 0.20 A.

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.

The prediction uses the stated constant-resistance assumption. The first measurement alone cannot show that heating or other changes will leave the resistance unchanged.

Optional check 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?
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?

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

Original wire: 8.0 ohmThe original wire has length L, circular diameter d and cross-sectional area A. Its supplied resistance is 8.0 ohm. A side view and an end-on circular section show the dimensions. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side viewLdEnd-on cross-sectionAd

Changed wire: 4.0 ohm

Changed wire: 4.0 ohmThe changed wire is twice as long and twice the diameter. At the same drawing scale, its circular section has twice the radius and four times the area. It is the same material at the same temperature, so its resistance is two divided by four, or one half, of 8.0 ohm: 4.0 ohm. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side view2L2dEnd-on cross-section4A2d

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.

The second wire has twice the length and twice the diameter. Its end-on area is four times the original area. Material and temperature are kept the same, so the two changes together halve the resistance.

Compare one change at a time

Twice the length and twice the diameter

The original wire has resistance 8.0 Ω.

  1. Doubling its length gives a resistance factor of 2.
  2. Doubling its diameter gives an area factor of 22 = 4, so the resistance factor from area is 1/4.
  3. 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.

  1. Keep the wire's material and cross-sectional area fixed. Use several measured active lengths.
  2. At each length, record the current through and p.d. across that section. Calculate its resistance from the paired readings.
  3. 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.
  4. 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.

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.

Optional check 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?
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?

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.
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