Current, Resistance & Ohm's Law
Welcome to DC Circuits. The last two topics studied charges that stay where they are: the forces between them, and the fields and potentials they set up. This topic lets charges move. Connect a wire across the two terminals of a battery and the battery keeps the ends of the wire at different potentials, so the free charges inside are pushed along it, and they keep flowing round the circuit for as long as it is closed. "DC" stands for direct current, a current that keeps flowing the same way round. This page sets up the three quantities that describe every circuit, current, voltage and resistance, and the law that ties them together, Ohm's law. The second page combines resistors in series and in parallel, the third finds the power and the heat in a circuit, and the fourth adds the battery's own EMF and internal resistance.
The topic has had two or three questions in every paper so far, 11 of the 192. Two of them belong to this page, and both treat current as a flow of charge: the charge that a current of 1.6 A carries through a wire in 0.5 s (January 2026, Q27), and what the slope of a charge–time graph stands for (March 2026, Q26). Both are worked through below. The rest of the page builds what the other three pages lean on. Ohm's law is a step in most of their questions, and the official sample paper asks students to compare resistances on an I–U graph, a kind of graph no real paper has used yet. This page teaches how to read one, and the Series & Parallel Circuits page finishes the sample's question type. Most marks lost here come from two habits, which the page names as they come up: copying a number from the stem across as the answer, and reading a graph's slope the wrong way up.
Current: charge per second
A metal conducts because some of its electrons are free: they are not tied to any one atom and can wander through the whole piece of metal, as the Charge & Coulomb's Law topic showed. In a wire on its own they wander at random, as many one way as the other, so on average no charge is carried anywhere. Connect the wire to a battery and its two ends are held at different potentials. The electric field this sets up along the wire pushes the free electrons towards the end at the higher potential, and a steady drift of charge along the wire begins. That drift is an electric current.
To say how large a current is, pick one cross-section of the wire, an imaginary slice straight across it, and measure the charge that passes through it. A large current carries a lot of charge through the slice each second, and a small one carries a little. So current is defined as the charge that passes a cross-section per unit time. If a charge q passes in a time t,
I = q/t.
The unit of current is the ampere, A, one of the SI base units from the Quantities & Units topic. A current of 1 A carries 1 coulomb through the slice every second. Rearranged, the definition gives the charge carried in a time t, q = It, and it shows where the coulomb comes from: 1 C = 1 A·s, the charge that a current of 1 A carries in one second. Small currents are often given in milliamperes: 1 mA = 10⁻³ A.
A current also has a direction. By convention it is the direction in which positive charge would move: out of the positive terminal of a battery, round the circuit and back into the negative terminal. In a metal the moving charges are electrons, which are negative, so they drift the opposite way to the current. The convention was fixed before the electron was discovered, and it changes none of the arithmetic, because electrons drifting to the left carry the same current as positive charge drifting to the right. The direction of a current is a direction along the wire, though, not a direction in space: where wires meet, the currents add as plain numbers, not by the parallelogram rule. So current is a scalar, even though it has a direction.
If the current passing through a wire is 1.6 A, the amount of charge passing through the cross-section of the wire within 0.5 s is
A. 1.6C B. 1.2C C. 2.0C D. 0.8C
A current of 1.6 A carries 1.6 C through the cross-section every second, so half a second carries half of that: q = It = 1.6 × 0.5 = 0.8 C. The answer is D.
A, 1.6 C, copies the current across as the charge. That is the charge carried in one second, but the question asks about half a second. Dividing instead of multiplying, 1.6 ÷ 0.5, gives 3.2 C, which is not offered. B, 1.2 C, and C, 2.0 C, do not follow from any common slip.
A question can go one step further and ask for electrons instead of coulombs. Each electron carries e = 1.6 × 10⁻¹⁹ C, so 0.8 C is 0.8 ÷ (1.6 × 10⁻¹⁹) = 5 × 10¹⁸ electrons through the slice in half a second.
The slope of a q–t graph
Suppose a meter could keep a running total of the charge that has passed the slice since a clock was started, and you plotted that total against time. For a steady current I, the total after a time t is q = It. The charge grows in proportion to the time, so the q–t graph is a straight line through the origin. Its slope is the rise in charge divided by the time it took, Δq/Δt, and for a steady current that is exactly the current. The larger the current, the faster the total climbs and the steeper the line.
This is the reading you learned on the Position–Time Graphs page of Describing Motion, where the slope of a displacement–time graph is the velocity. There the slope says how fast the position changes; here it says how fast charge passes, which is the current. A q–t graph that curves is read the same way too. Where it gets steeper the current is growing, where it flattens the current is falling, and where it is level no charge is passing at all. The current at any one moment is the slope of the tangent there. Read the other way round, the area under a current–time graph is the charge that has passed, just as the area under a velocity–time graph is the displacement.
Inside the wire: I = nqSv
The definition I = q/t counts charge at a slice. A current can also be built up from the moving charges themselves, and that shows what it depends on. Call the moving charges the carriers. Suppose a wire of cross-sectional area S contains n carriers in each cubic metre, each with charge q, all drifting along the wire at a speed v.
In a time t each carrier moves a distance vt. So every carrier that starts within a distance vt behind the slice crosses it in that time, and no other carrier does. Those carriers fill a cylinder of length vt and cross-section S, whose volume is Svt. It holds nSvt carriers, which carry a charge nSvt × q between them. Dividing by the time gives the current:
I = nqSvt/t = nqSv.
For a metal the carriers are electrons, and q = e = 1.6 × 10⁻¹⁹ C. The formula says that a current is larger when there are more carriers, when each carries more charge, when the wire is thicker or when the carriers drift faster. It also explains a surprise. Copper has about 8.5 × 10²⁸ free electrons per cubic metre, so a current of 1 A in a copper wire with a cross-section of 1 mm² needs a drift speed of only v = I/(nqS) ≈ 7 × 10⁻⁵ m/s, less than a tenth of a millimetre per second. A lamp still lights the instant its switch closes, because the field that pushes the electrons spreads along the whole circuit at nearly the speed of light, and electrons everywhere in the wire start drifting at once.
One more consequence is used on the next page. In a steady current, charge cannot pile up anywhere along a wire, because whatever flows into a stretch of wire each second must flow out of it. So the same current crosses every slice of the wire, thick or thin.
Resistance and Ohm's law
A current needs a voltage to drive it. The voltage U across a piece of wire is the potential difference between its two ends, from the Electric Field & Potential topic: the work the field does on each coulomb of charge that passes through, measured in volts. How the current depends on the voltage was found by experiment. Connect a length of metal wire to a supply whose voltage can be changed, with an ammeter to measure the current through the wire and a voltmeter to measure the voltage across it. Keep the wire at a steady temperature, set several voltages and record the current each time:
| Voltage U | Current I | U/I |
|---|---|---|
| 1.0 V | 0.20 A | 5.0 V/A |
| 2.0 V | 0.40 A | 5.0 V/A |
| 3.0 V | 0.60 A | 5.0 V/A |
| 4.0 V | 0.80 A | 5.0 V/A |
Double the voltage and the current doubles; triple it and the current triples. The current is proportional to the voltage, I ∝ U, so the ratio U/I is the same for every reading, 5.0 volts per ampere. Repeat the experiment with a different wire and the current is again proportional to the voltage, but the ratio is different: a longer or a thinner wire needs more voltage for the same current. The ratio measures how strongly the wire opposes a current through it, and it is called the wire's resistance:
R = U/I.
The unit of resistance is the ohm, Ω. A resistance of 1 Ω needs a voltage of 1 V to drive a current of 1 A, so 1 Ω = 1 V/A, and the wire in the table has R = 5.0 Ω. A part made to have a definite resistance is a resistor, and one whose resistance does not change is a fixed resistor (定值电阻).
The finding that U/I stays constant is Ohm's law, after Georg Ohm, who published it in 1827. It is usually written with the current as the subject, because the voltage is what you set and the current is what follows:
I = U/R.
In words, the current through a conductor is proportional to the voltage across it and inversely proportional to its resistance. The law holds for metals and for conducting solutions (electrolytes), as long as their temperature stays steady. It does not hold for everything: the gas in a neon lamp, and many semiconductor devices, carry currents that are not proportional to the voltage. The last section of this page shows what a filament lamp does.
A 24 Ω resistor is connected across a 12 V battery. What current flows through it, and how much charge passes through it in 5 minutes?
Ohm's law: I = U/R = 12 ÷ 24 = 0.5 A.
The charge: 5 minutes is 300 s, so q = It = 0.5 × 300 = 150 C.
Check the units: volts divided by ohms are amperes, and amperes times seconds are coulombs.
The March 2026 paper joined the two ideas of this page so far, the slope of a q–t graph and Ohm's law, in a single question.
A resistor of resistance R has voltage U applied across it. The graph of charge q passing through its cross-section versus time has a slope representing ( ).
The paper's figure shows a straight line through the origin, with q on the vertical axis, t along the horizontal one and no values.
A. R B. U/R C. 1/R D. U
Step 1, the graph: the slope of a q–t graph is Δq/Δt, the charge passing per second, which is the current I.
Step 2, Ohm's law: the current through a resistance R with a voltage U across it is I = U/R. So the slope represents U/R. The answer is B.
Units settle it too. The slope is in coulombs per second, which are amperes, and of the four options only U/R, volts divided by ohms, is in amperes. A, R, is the slope of a graph of U against I, a different graph. C, 1/R, is the slope of an I–U graph, which a section below reads; it is the current per volt, not the current. D, U, is the voltage, which drives the current but is not the current. The stem names U and R, and three of the options are built from them one at a time.
What sets a resistance: R = ρL/S
The table above came from one wire. Measure wires of one material but different sizes, all at the same temperature, and two patterns appear. Double the length and the resistance doubles: a wire twice as long is like two equal halves one after the other, and the same current needs a voltage to push it through each half in turn, so twice the voltage in all. Double the cross-sectional area and the resistance halves: a wire twice as thick is like two equal wires side by side, and the same voltage drives the same current through each, so twice the current in all. So R ∝ L/S. Wires of copper and of nichrome with the same size, though, have very different resistances, so the constant of proportionality depends on the material. It is called the resistivity, ρ:
R = ρL/S.
Rearranged, ρ = RS/L, so its unit is Ω·m² divided by m, which is Ω·m. Resistivity belongs to a material, the way density does, while resistance belongs to one particular piece of it. Good conductors have tiny resistivities: copper about 1.7 × 10⁻⁸ Ω·m and aluminium about 2.7 × 10⁻⁸ Ω·m. Nichrome, used in heating elements, has about 1.1 × 10⁻⁶ Ω·m, some 65 times copper's. The resistivity of a metal also grows as the metal gets hotter, which matters for the lamp at the end of this page. This formula has not been asked in any paper so far, so the section stays short.
A copper wire is 100 m long and has a cross-sectional area of 1 mm². The resistivity of copper is 1.7 × 10⁻⁸ Ω·m. What is the wire's resistance?
Convert the area first: 1 mm² = (10⁻³ m)² = 10⁻⁶ m², not 10⁻³ m².
Then R = ρL/S = 1.7 × 10⁻⁸ × 100 ÷ 10⁻⁶ = 1.7 Ω.
I–U graphs and U–I graphs
Ohm's law can be drawn. Plot the current through a fixed resistor against the voltage across it, with I up the vertical axis and U along the horizontal one, as the papers do. Every reading has I = U/R with the same R, so the points fall on a straight line through the origin. This is the resistor's I–U graph, called in Chinese its 伏安特性曲线, its volt–ampere characteristic. Its slope is the rise in current over the run in voltage:
slope = ΔI/ΔU = I/U = 1/R.
So the slope of an I–U line is not the resistance but its reciprocal. A steeper line gives more current for each volt, and that means a smaller resistance. Now swap the axes and plot U up the vertical axis against I. The slope becomes ΔU/ΔI = R, so on this U–I graph a steeper line means a larger resistance. The same two resistors swap their order of steepness between the two graphs.
The safest way to compare two lines is to read one point on each, then divide.
The I–U graphs of two fixed resistors a and b are straight lines through the origin. Line a passes through the point (2 V, 0.4 A) and line b through (6 V, 0.4 A). Which line is the steeper, which resistor is larger, and what is Ra : Rb?
Read the resistances: Ra = 2 ÷ 0.4 = 5 Ω and Rb = 6 ÷ 0.4 = 15 Ω. So Ra : Rb = 1 : 3, and b is the larger resistor.
The steepness: a reaches 0.4 A at only 2 V, so its line climbs three times as fast. The steeper line, a, is the smaller resistance, as the slope 1/R says.
Check with a third reading: with 3 V across each, a carries 3 ÷ 5 = 0.6 A and b carries 3 ÷ 15 = 0.2 A. At the same voltage the smaller resistance carries the larger current.
A lamp's curved I–U line
Not everything obeys Ohm's law. Put a small filament lamp in place of the fixed resistor and repeat the readings. At low voltages the I–U graph is close to a straight line, but as the voltage grows the current grows more and more slowly, and the graph bends towards the U-axis. The reason is the filament's temperature. A larger current heats the thin metal filament until it glows, and, as the last section said, the resistivity of a metal rises as it gets hotter. So the lamp's resistance rises with the voltage, and each extra volt adds less current than the one before. A part whose I–U graph is a straight line through the origin is called a linear element (线性元件); one with a curved graph, like the lamp, is a non-linear element (非线性元件).
The definition R = U/I still holds at every point of the curve: it gives the resistance at that voltage and current. What changes is how it is read off the graph. The resistance at a point P is U/I for that point, the reciprocal of the slope of the straight line from the origin O to P. It is not found from the tangent at P, nor from the slope between two points on the curve. As P moves up the curve, the line OP gets flatter, which on an I–U graph means a larger resistance.
The I–U curve of the lamp in the figure passes through P (2 V, 0.2 A) and Q (8 V, 0.4 A). What is the lamp's resistance at P and at Q?
At P: R = U/I = 2 ÷ 0.2 = 10 Ω. At Q: R = 8 ÷ 0.4 = 20 Ω. The resistance has doubled, although the voltage has risen four times and the current only twice.
The tempting shortcut is the slope between the two points: (8 − 2) ÷ (0.4 − 0.2) = 30 Ω. That is the resistance at neither point. The lamp has no single resistance, only one at each voltage, and each comes from its own point.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Current | I = q/t, the charge through a cross-section per second, so q = It. 1 A = 1 C/s. January 2026 Q27: 1.6 × 0.5 = 0.8 C. |
| Direction | The way positive charge moves; in a metal the electrons drift against it. Current has a direction but is a scalar. |
| q–t graph | The slope Δq/Δt is the current: steeper means a larger current, level means none. The height is the charge passed so far, not the current. |
| Inside a wire | I = nqSv, with q the charge on one carrier. Drift speeds are tiny. The same current crosses every slice of a wire, thick or thin. |
| Resistance | R = U/I, in ohms: 1 Ω = 1 V/A. It is set by the conductor, not by U or I: doubling U doubles I and leaves R alone. |
| Ohm's law | I = U/R, for metals and electrolytes at a steady temperature. March 2026 Q26: the slope of a q–t graph is I = U/R. |
| Notation | U for voltage, with V its unit; E for EMF, r for internal resistance, S for cross-sectional area. I–U graphs put I on the vertical axis. |
| Resistivity | R = ρL/S, with ρ in Ω·m, set by the material and its temperature. 1 mm² = 10⁻⁶ m². Stretching a wire to twice its length gives four times the resistance. |
| I–U vs U–I | I–U graph: slope = 1/R, so the steeper line is the smaller R. U–I graph: slope = R, so the steeper line is the larger R. Read values, not angles. |
| Lamp | Its I–U curve bends towards the U-axis as the hot filament's resistance rises. R at a point is U/I for that point, not the slope between two points. |
| Common slips | The current given as the charge; minutes not turned into seconds; R, 1/R or U for the slope of a q–t graph; steeper read the wrong way round; the slope between two points taken as a lamp's resistance. |