HomeLearning HubIB DP PhysicsB.5 Current and circuits
B.5

Current and circuits

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

If it seems odd to find circuits filed under “the particulate nature of matter”, that is deliberate: current is the drift of charged particles through matter, and the syllabus wants it understood that way rather than as an isolated topic.

🎯What you need to be able to do

  • Define current as rate of flow of charge, and relate it to drift speed.
  • Distinguish emf from potential difference, and use internal resistance.
  • Use \( R = V/I \), and identify ohmic and non-ohmic behaviour from an \(I\)–\(V\) graph.
  • Use resistivity, and combine resistors in series and parallel.
  • Calculate electrical power and energy.
  • Analyse potential divider circuits.

📚The physics

Current is the rate of flow of charge, \( I = \Delta q/\Delta t \), measured in amperes. Conventional current is defined as flowing from positive to negative, which is opposite to the actual motion of the electrons — an accident of history that has never been corrected and never causes a problem provided you stay consistent.

emf and potential difference are not the same quantity. Electromotive force is the energy given to each coulomb by the source; potential difference is the energy delivered by each coulomb to a component. The difference between them is the energy wasted inside the source itself, in its internal resistance \(r\).

\( \varepsilon = I(R + r) \)
\( V = \varepsilon - Ir \)

This is why a car’s headlights dim when the starter motor turns: the huge current through the starter causes a large \(Ir\) drop inside the battery, so less is left for everything else.

Resistance is \( R = V/I \), always. Ohm’s law is the stronger claim that \(R\) stays constant as \(V\) changes, and it only holds for an ohmic conductor at constant temperature. Three \(I\)–\(V\) characteristics are worth recognising instantly: an ohmic conductor gives a straight line through the origin; a filament lamp gives a curve that flattens, because heating raises the resistance; a diode conducts sharply above about 0.6 V in forward bias and essentially not at all in reverse.

Resistivity \(\rho\) is the material property behind resistance: \( R = \rho L/A \). Longer wire means more resistance, thicker wire means less. Note that \(A\) is the cross-sectional area, so doubling the diameter quarters the resistance, not halves it.

Combining resistors. In series the current is the same everywhere and \( R_{\text{total}} = R_1 + R_2 + \ldots \) In parallel the potential difference is the same across each branch and \( 1/R_{\text{total}} = 1/R_1 + 1/R_2 + \ldots \) Two sanity checks that catch most arithmetic errors: a parallel combination is always smaller than the smallest individual resistor, and a series combination is always larger than the largest.

Power comes in three forms of the same statement:

\( P = VI \)
\( P = I^{2}R \)
\( P = V^{2}/R \)

Choose whichever matches what you already know. The \(I^{2}\) version explains why transmission lines run at very high voltage: for a given power, higher voltage means lower current, and losses scale as the square of the current.

Potential dividers split a supply voltage in the ratio of the resistances: \( V_{\text{out}} = V_{\text{in}} \times R_2/(R_1 + R_2) \). Replace one resistor with a thermistor or an LDR and the output voltage becomes a temperature or light sensor, which is the basis of most simple electronic control circuits.

✏️Worked example

A battery of emf 9.0 V and internal resistance 1.5 Ω is connected to a 12 Ω resistor in parallel with a 6.0 Ω resistor.

Reduce the external circuit first. \( 1/R = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12 \), so \( R = 4.0 \) Ω. Sanity check: 4.0 Ω is smaller than 6.0 Ω, as a parallel combination must be.

Find the current from the battery. \( I = \varepsilon/(R + r) = 9.0/(4.0 + 1.5) = 1.64 \) A.

Terminal potential difference. \( V = \varepsilon - Ir = 9.0 - 1.64 \times 1.5 = 6.5 \) V. Note that this, not 9.0 V, is what appears across the parallel combination.

Branch currents. Through the 12 Ω: \( 6.5/12 = 0.55 \) A. Through the 6.0 Ω: \( 6.5/6.0 = 1.09 \) A. They sum to 1.64 A, matching the battery current — always check this.

Power wasted inside the battery. \( P = I^{2}r = 1.64^{2} \times 1.5 = 4.0 \) W, out of a total 14.8 W.

The trap. Over a quarter of the battery’s output is heating the battery itself. Ignoring internal resistance and using 9.0 V throughout would have given branch currents about 40% too high.

🔭See it happen

PhET, Circuit Construction Kit: DC. Build the circuit above, switch on the battery’s internal resistance, and watch the voltmeter reading fall from 9.0 V to 6.5 V as you close the switch. The gap you are watching appear is \(Ir\).

📝Practise

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

1. A current of 2.5 A flows for 3.0 minutes. Find the charge transferred.
\( q = I\Delta t = 2.5 \times 180 = 450 \) C. The time must be in seconds.
2. A copper wire has resistivity \( 1.7 \times 10^{-8} \) Ω m, length 2.5 m and cross-sectional area \( 1.2 \times 10^{-6} \) m\(^{2}\). Find its resistance.
\( R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 2.5}{1.2 \times 10^{-6}} = 3.5 \times 10^{-2} \) Ω.
3. Find the combined resistance of a 4.0 Ω and a 12 Ω resistor in parallel.
\( 1/R = 1/4.0 + 1/12 = 3/12 + 1/12 = 4/12 \), so \( R = 3.0 \) Ω. Sanity check: it is smaller than the smaller resistor, as a parallel combination must be.
4. A cell of emf 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current and the terminal potential difference.
\( I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{3.0} = 0.50 \) A. Then \( V = \varepsilon - Ir = 1.5 - 0.50 \times 0.50 = 1.25 \) V.
5. A 60 W lamp operates on a 240 V supply. Find the current it draws and its resistance.
\( I = P/V = 60/240 = 0.25 \) A. Then \( R = V/I = 240/0.25 = 960 \) Ω (or directly, \( R = V^{2}/P = 57600/60 = 960 \) Ω).
6. A 12 V supply is connected across a 3.0 kΩ resistor in series with a 6.0 kΩ resistor. Find the potential difference across the 6.0 kΩ resistor.
\( V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2} = 12 \times \dfrac{6.0}{9.0} = 8.0 \) V. The supply divides in the ratio of the resistances.

🔗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.

  • The Physics Classroom — Electric Circuits
  • The Physics Hypertextbook — electric current and resistance