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B.5

Current and circuits

Theme B · The particulate nature of matter · SL and HL

Circuits are where Theme B stops being about molecules in a box and starts being about charges in a wire — but it is the same physics. A current is a flow of particles; a potential difference is energy per unit charge; resistance is what happens when those particles keep colliding with the lattice they are moving through. Get those three definitions exactly right and every circuit question becomes bookkeeping.

🎯What you need to be able to do

  • Use \( I = \Delta q/\Delta t \) and \( Q = Ne \), and distinguish conventional current from electron flow.
  • Use \( I = nAvq \) and explain why drift speed is so small.
  • Define potential difference and emf as energy per unit charge, and say how they differ.
  • Use \( R = V/I \), and sketch and interpret the \(I\)–\(V\) characteristics of an ohmic conductor, a filament lamp and a diode.
  • Use \( R = \rho L/A \) and predict how resistance changes with the dimensions of a wire.
  • Use \( P = VI = I^{2}R = V^{2}/R \).
  • Combine resistors in series and in parallel, and place ideal ammeters and voltmeters correctly.
  • Use \( \varepsilon = I(R + r) \) for a cell with internal resistance, and find the terminal potential difference.
  • Analyse a potential divider.

⚡Charge and current

Charge is quantised: it comes in whole multiples of the elementary charge \( e = 1.60 \times 10^{-19} \) C. So any charge \(Q\) is \( Q = Ne \) for some whole number \(N\). It is also conserved — charge is never created or destroyed, only moved, which is why current is the same all the way round a series circuit.

Electric current is the rate of flow of charge:

\[ I = \frac{\Delta q}{\Delta t} \qquad \text{1 ampere} = \text{1 coulomb per second} \]
A wire connected to a cell, with free electrons drifting from the negative terminal towards the positive one. An arrow beneath the wire shows conventional current defined in the opposite direction, from positive to negative, because the convention was fixed before anyone knew the carriers were negative. A note states that this makes no difference to any calculation, and that current is the same at every point in a series circuit because charge is conserved. Beside it, charge is shown as quantised in multiples of the elementary charge.
Conventional current runs from + to −; the electrons actually go the other way. The convention was fixed before anyone knew what the carriers were, and nothing in any calculation depends on it.

🐌Drift velocity: how slowly a current actually moves

Consider a conductor of cross-sectional area \(A\) containing \(n\) free charge carriers per unit volume, each of charge \(q\), drifting along at average speed \(v\):

\[ I = nAvq \]
A cylindrical section of wire of cross-sectional area A. In a time delta t, every carrier within a distance v delta t of one end passes through it, so the volume that empties is A v delta t, containing n A v delta t carriers, each with charge q. Dividing the total charge by delta t gives the current I equals n A v q. A worked figure beneath shows that for a copper wire of one square millimetre carrying one amp, with n equal to 8.5 times ten to the twenty-eight per cubic metre, the drift speed is only about 0.07 millimetres per second.
The derivation is a volume count. The answer it gives is startling — and it is the reason the next paragraph exists.

For a typical copper wire the drift speed works out at well under a millimetre per second: a given electron would take hours to travel the length of a room. Yet the lamp lights instantly. There is no contradiction — the wire is already full of free electrons everywhere along its length, and the electric field that sets them all moving is established at close to the speed of light. Nothing has to travel from the switch to the lamp; everything is already in place, waiting to be pushed.

🔋Potential difference, emf, and resistance

Both of the first two are energy per unit charge, and the distinction between them is a favourite exam question.

A simple circuit with a cell and a resistor. Inside the cell, chemical energy is converted to electrical energy, which is the electromotive force: the energy given to each coulomb by the source. Across the resistor, electrical energy is converted to thermal energy, which is the potential difference: the energy transferred out of the circuit by each coulomb. Both are measured in volts, both are joules per coulomb, and the difference is only the direction of the conversion. Resistance is defined beside them as the potential difference divided by the current.
Same unit, same definition, opposite direction of conversion. Emf puts energy in; potential difference takes it out.
emf\( \varepsilon = \dfrac{\text{energy supplied}}{\text{charge}} \)
p.d.\( V = \dfrac{\text{energy transferred}}{\text{charge}} \)
resistance\( R = \dfrac{V}{I} \)

Resistance is defined by that last equation for every component, ohmic or not. What varies is whether \(R\) is constant.

Three current against voltage graphs. For an ohmic conductor at constant temperature the line is straight and passes through the origin, so resistance is constant and Ohm's law holds. For a filament lamp the curve bends over towards the voltage axis as the voltage rises, because the filament heats up and its resistance increases. For a diode the current is essentially zero until a forward voltage of about 0.6 volts, after which it rises very steeply, and in reverse bias the current stays near zero, so the component conducts in only one direction.
Only the first is ohmic. On any \(I\)–\(V\) graph, resistance at a point is \(V/I\) for that point — not the gradient, unless the line happens to pass through the origin.
The gradient of an \(I\)–\(V\) graph is not \(1/R\) unless the graph is a straight line through the origin. For a filament lamp, the resistance at any point is \(V/I\) at that point — the ratio of the coordinates, which you can get by drawing a line from the origin. Taking the tangent instead gives a different number that is not the resistance. This costs marks every year.

Ohm’s law is not the definition of resistance; it is the observed fact that for some materials, at constant temperature, \(V\) is proportional to \(I\). A filament lamp disobeys it, and does not thereby stop having a resistance.

📏Resistivity: resistance from the material and the shape

The resistance of a wire depends on what it is made of and on its dimensions:

\[ R = \frac{\rho L}{A} \]
Three wires compared. Doubling the length doubles the resistance, because the charges must travel through twice as much lattice. Doubling the cross-sectional area halves the resistance, because there are twice as many parallel paths. Doubling the diameter quadruples the area and so quarters the resistance, which is the step most often missed. The resistivity rho is identified as the property of the material alone, measured in ohm metres, with copper at 1.7 times ten to the minus eight and a typical insulator many orders of magnitude higher.
\( \rho \) belongs to the material; \(R\) belongs to the object. It is the same distinction as specific heat capacity against thermal capacity in B.1.

Watch the area term when a question gives you a diameter: \( A = \pi r^{2} = \pi d^{2}/4 \), so doubling the diameter quarters the resistance, not halves it.

✏️Worked example 1 — charge, electrons, and a wire’s resistance

(a) A current of 2.0 A flows for 5.0 minutes. Find the charge transferred and the number of electrons involved. (b) Find the resistance of a copper wire of length 2.0 m and diameter 0.50 mm. Take \( \rho = 1.7\times10^{-8} \) Ω m.

(a) Convert the time first — seconds, always:

\[ Q = I\Delta t = 2.0 \times 300 = 600\ \text{C} \qquad N = \frac{Q}{e} = \frac{600}{1.60\times10^{-19}} = 3.8\times10^{21} \]

(b) The radius is half the diameter, and the area goes as the radius squared:

\[ A = \pi r^{2} = \pi\,(0.25\times10^{-3})^{2} = 1.96\times10^{-7}\ \text{m}^{2} \]
\[ R = \frac{\rho L}{A} = \frac{1.7\times10^{-8} \times 2.0}{1.96\times10^{-7}} = 0.17\ \Omega \]
Both answers pass a plausibility test. \( 10^{21} \) electrons is an enormous number, which it should be — each one carries only \( 10^{-19} \) C. And a fifth of an ohm for two metres of thin copper is about right: connecting wires are supposed to have negligible resistance compared with the components they connect. If (b) had come out in the hundreds of ohms, the likely error is using the diameter as the radius, which would be four times too small an area.

🔌Power in a circuit

Power is energy per unit time, and in a circuit that gives one equation with two useful rearrangements:

\( P = VI \)
\( P = I^{2}R \)
\( P = \dfrac{V^{2}}{R} \)
The three power equations shown as one. Power equals potential difference times current, because potential difference is energy per coulomb and current is coulombs per second, so their product is joules per second. Substituting V equals I R gives P equals I squared R, and substituting I equals V over R gives P equals V squared over R. A note advises which form to use: the I squared R form when the current is the quantity shared between components, as in a series circuit, and the V squared over R form when the potential difference is shared, as in a parallel circuit.
Not three formulas — one formula and two substitutions of \(V = IR\). Choosing the right one is just choosing the quantity you already know.

🔗Series and parallel

Two circuits compared. In series, the same current passes through every component because there is only one path and charge is conserved, the potential differences add up to the supply voltage, and the total resistance is the sum of the individual resistances. In parallel, every branch has the same potential difference across it because each is connected to the same two points, the branch currents add up to the total current, and the reciprocal of the total resistance is the sum of the reciprocals, so the combined resistance is always smaller than the smallest branch. Ideal meters are shown: an ammeter in series with zero resistance, a voltmeter in parallel with infinite resistance.
Everything follows from two conservation laws: charge is conserved at a junction, and energy is conserved round a loop. The formulas are consequences, not extra facts.
series\( R_{\text{total}} = R_1 + R_2 + \ldots \)
parallel\( \dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots \)

Two sanity checks worth applying to every answer: a series combination is always larger than the biggest resistor in it, and a parallel combination is always smaller than the smallest. If your parallel answer is bigger than one of the branches, you have forgotten to invert at the end.

Ideal meters are defined by not disturbing what they measure. An ammeter goes in series and ideally has zero resistance, so it drops no voltage. A voltmeter goes in parallel and ideally has infinite resistance, so it draws no current.

🔋Internal resistance: why a battery’s voltage sags

A real cell is not just a source of emf. The chemicals inside it have resistance too, and that internal resistance \(r\) is in series with everything else in the circuit:

\[ \varepsilon = I(R + r) \qquad\text{so}\qquad V_{\text{terminal}} = \varepsilon - Ir \]
A cell drawn as an ideal source of emf in series with a small internal resistance r, inside a dashed box representing the real battery, connected to an external resistor R. The terminal potential difference is what a voltmeter across the battery reads, and it is the emf minus the voltage lost across the internal resistance. Beside it, a graph of terminal potential difference against current is a straight line with intercept equal to the emf and gradient equal to minus the internal resistance, which is the standard experiment for measuring both.
The emf is what the cell would deliver at zero current. Draw any current and some of it is spent inside the cell itself — which is why a car’s headlights dim when the starter motor turns.

✏️Worked example 2 — a cell under load

A cell of emf 6.0 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current, the terminal potential difference, the power delivered to the resistor and the power wasted inside the cell.
\[ I = \frac{\varepsilon}{R + r} = \frac{6.0}{2.5 + 0.50} = 2.0\ \text{A} \]
\[ V_{\text{terminal}} = \varepsilon - Ir = 6.0 - 2.0\times0.50 = 5.0\ \text{V} \]
\[ P_R = I^{2}R = 2.0^{2}\times2.5 = 10\ \text{W} \qquad P_r = I^{2}r = 2.0^{2}\times0.50 = 2.0\ \text{W} \]
Check the total. The cell supplies \( \varepsilon I = 6.0\times2.0 = 12 \) W, and \( 10 + 2.0 = 12 \) W is accounted for. Note that a sixth of the energy is being wasted heating the cell itself, and that the terminal voltage has sagged from 6.0 V to 5.0 V. Draw a bigger current and it sags further — which is exactly what a voltmeter shows when a battery is described as “flat”: the emf is often nearly unchanged, but \(r\) has risen.

⚖️The potential divider

Two resistors in series across a supply split the voltage between them in the ratio of their resistances, because the same current passes through both:

\[ V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2} \]
A potential divider: two resistors in series across a supply, with the output taken across the lower one. Because the same current flows through both, the supply voltage divides in the ratio of the resistances. A worked case shows 12 volts across a 4 kilohm and an 8 kilohm resistor giving 8 volts across the larger one. Beside it, the same circuit with the lower resistor replaced by a light-dependent resistor or a thermistor, so that the output voltage responds to light or temperature, which is how a sensing circuit is built.
The formula is worth deriving rather than memorising: the current is \( V_{\text{in}}/(R_1+R_2) \), and \( V_{\text{out}} \) is that current times \(R_2\).

✏️Worked example 3 — dividing a supply

A 12 V supply is connected across a 4.0 kΩ resistor in series with an 8.0 kΩ resistor. Find the potential difference across the 8.0 kΩ resistor, and the power dissipated in it.
\[ V_{\text{out}} = 12 \times \frac{8.0}{4.0 + 8.0} = 8.0\ \text{V} \]
\[ P = \frac{V^{2}}{R} = \frac{8.0^{2}}{8.0\times10^{3}} = 8.0\times10^{-3}\ \text{W} = 8.0\ \text{mW} \]
The bigger resistor gets the bigger share. Two thirds of the resistance, two thirds of the voltage — and the ratio is all that matters, so 4 Ω and 8 Ω would divide it identically. Note also that the kilohms had to become ohms before the power calculation: using 8.0 rather than 8000 would have given 8 W, a thousand times too large.

🔭See it happen

PhET, Circuit Construction Kit: DC. Build the circuits above, switch on the electron view, and watch how slowly the charges actually crawl while the bulb lights at once. Then give the battery some internal resistance and watch the terminal voltage sag as you lower the external resistance — the whole of worked example 2, live.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A charge of 45 C passes a point in 90 s. Find the current, and the number of electrons involved.
\( I = \Delta q/\Delta t = 45/90 = 0.50 \) A. Number of electrons \( N = Q/e = 45/(1.60\times10^{-19}) = 2.8\times10^{20} \).
2. Explain why a lamp lights almost instantly even though the drift speed of the electrons is less than a millimetre per second.
The wire already contains free electrons along its entire length — nothing has to travel from the switch to the lamp. Closing the switch establishes an electric field throughout the circuit at close to the speed of light, and every electron everywhere begins to drift almost at once, including the ones already inside the lamp. The drift speed governs how far an individual electron gets, not how quickly the current starts.
3. A wire of resistance 12 Ω is stretched uniformly to twice its original length. Find its new resistance.
Stretching does not change the volume, so doubling the length halves the cross-sectional area. In \( R = \rho L/A \), the numerator doubles and the denominator halves, so the resistance goes up by a factor of 4: \( R = 48\ \Omega \). Answering 24 Ω means the area was forgotten.
4. Sketch the \(I\)–\(V\) characteristic of a filament lamp and explain its shape.
The curve starts straight through the origin, then bends over towards the voltage axis: equal increases in \(V\) produce smaller and smaller increases in \(I\). As the current rises the filament gets hotter, so the lattice ions vibrate with greater amplitude, the free electrons collide with them more often, and the resistance rises. Since \( R = V/I \) increases with \(V\), the graph must flatten.
5. Resistors of 6.0 Ω and 3.0 Ω are connected in parallel, and that combination is in series with a 4.0 Ω resistor across a 12 V supply of negligible internal resistance. Find the total resistance and the current from the supply.
Parallel first: \( 1/R = 1/6.0 + 1/3.0 = 0.5 \), so \( R = 2.0\ \Omega \) — and note it is smaller than either branch, as it must be. Total \( = 2.0 + 4.0 = 6.0\ \Omega \). Current \( I = V/R = 12/6.0 = 2.0 \) A.
6. State where an ammeter and a voltmeter are connected, and what resistance each ideally has.
An ammeter is connected in series with the component whose current is being measured, and ideally has zero resistance so that it drops no potential difference and does not reduce the current. A voltmeter is connected in parallel with the component, and ideally has infinite resistance so that it draws no current and does not alter the circuit it is measuring.
7. A cell of emf 1.5 V has an internal resistance of 0.80 Ω. Find the terminal potential difference when it drives a current of 0.50 A.
\( V = \varepsilon - Ir = 1.5 - 0.50\times0.80 = 1.5 - 0.40 = 1.1 \) V. The missing 0.40 V is dissipated inside the cell itself.
8. A student plots terminal potential difference against current for a cell and obtains a straight line with intercept 4.5 V and gradient −1.5 V A\(^{-1}\). State the emf and the internal resistance.
Rearranged, \( V = \varepsilon - Ir \) is the equation of a straight line with \(V\) on the \(y\)-axis and \(I\) on the \(x\)-axis. The intercept is the emf, so \( \varepsilon = 4.5 \) V, and the gradient is \( -r \), so \( r = 1.5\ \Omega \). The intercept is the emf because it is the terminal voltage at zero current, when nothing is lost inside the cell.
9. A 24 V supply is connected across a 3.0 kΩ resistor in series with a 9.0 kΩ resistor. Find the output voltage taken across the 3.0 kΩ resistor.
\( V_{\text{out}} = 24 \times \dfrac{3.0}{3.0 + 9.0} = 24 \times 0.25 = 6.0 \) V. The smaller resistor takes the smaller share — a quarter of the resistance, a quarter of the voltage.
10. A 60 W lamp and a 100 W lamp are both designed for 240 V. Which has the greater resistance, and what current does each draw?
From \( P = V^{2}/R \), \( R = V^{2}/P \), so a lower power at the same voltage means a higher resistance. The 60 W lamp: \( R = 240^{2}/60 = 960\ \Omega \), drawing \( I = P/V = 0.25 \) A. The 100 W lamp: \( R = 240^{2}/100 = 576\ \Omega \), drawing 0.42 A. The brighter lamp has the lower resistance, which surprises most people the first time.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — DC circuits and resistivity
  • The Physics Hypertextbook — electric current and resistance
  • PhET — Circuit Construction Kit: DC