Lesson 10: DC Circuits

Electric circuits (Theme B.5) are where IB candidates gain or lose easy marks: the physics is Kirchhoff's two laws plus Ohm's law and resistivity, but the questions layer them with internal resistance, potential dividers and graph analysis. Work methodically and every circuit question becomes bookkeeping.

The fundamentals

Current I = Q/t, the rate of flow of charge. Potential difference V = W/Q, energy transferred per unit charge. Resistance R = V/I; a component is ohmic if V/I is constant (straight line through the origin on a V–I graph) — a filament lamp is not, because heating raises its resistance. Resistivity: R = ρL/A — doubling length doubles R; doubling diameter quarters R (area ∝ diameter²). Power: P = VI = I²R = V²/R — choose the form using the quantity shared by the components you are comparing (same I in series → I²R; same V in parallel → V²/R).

Kirchhoff's laws: currents at a junction sum to zero (charge conservation); around any closed loop, the sum of emf's equals the sum of pd's (energy conservation).

Worked problem 1 — series–parallel network

A 12 V supply (negligible internal resistance) drives a 6.0 Ω resistor in series with a parallel pair of 4.0 Ω and 12 Ω. Find the supply current and the power dissipated in the 6.0 Ω resistor.

Parallel pair: 1/Rp = 1/4 + 1/12 = 1/3 → Rp = 3.0 Ω
Total resistance: 6.0 + 3.0 = 9.0 Ω ; current from supply: I = 12/9.0 = 1.33 A
Power in 6.0 Ω (the full supply current passes through it): P = I²R = 1.33² × 6.0 = 10.7 W

Check the voltage budget: 1.33 A × 6.0 Ω = 8.0 V across the series resistor; 4.0 V across the parallel pair; 8.0 + 4.0 = 12 V ✓.

Worked problem 2 — internal resistance

A battery of emf 9.0 V drives 1.5 A through a 5.0 Ω load. Find the internal resistance r and the terminal pd.

Terminal pd = IR = 1.5 × 5.0 = 7.5 V
"Lost volts" = emf − terminal pd = 1.5 V = Ir → r = 1.5/1.5 = 1.0 Ω

The governing equation is ε = I(R + r). Internal resistance is why a car battery's voltage sags when the starter motor draws 200 A, and why a nearly-flat battery can show a healthy 1.5 V on a voltmeter (which draws almost no current) yet fail to light a lamp.

Potential dividers and sensing circuits

A potential divider splits a supply in the ratio of resistances: Vout = Vin × R₂/(R₁ + R₂). With a thermistor or LDR as one resistor, the output voltage varies with temperature or light — the basis of sensing circuits in Paper 2 experimental questions. Remember the output is taken across one resistor; if a load is then connected in parallel with it, the divider ratio changes.

Practical worked example — measuring emf and internal resistance

A variable resistor is connected across a cell and V and I recorded. The V–I graph is a straight line with intercept 6.02 V and gradient −0.85 Ω; steepest and shallowest acceptable lines give gradients −0.92 and −0.78 Ω.

ε = intercept = 6.02 V (to ±0.03 V from intercept spread)
r = |gradient| = 0.85 Ω, half-range 0.07 Ω → r = 0.85 ± 0.07 Ω

Because V = ε − Ir, the graph tests the model as well as measuring the parameters: any curvature would signal r changing (the cell warming up during the run — take readings quickly and open the switch between them).

Nature of Science

Kirchhoff's laws are conservation laws in disguise — charge and energy — applied to a network. The modelling choice in circuit theory is to treat wires as resistance-free and components as "lumped"; at very high frequencies (radio engineering) that model fails and the same components behave quite differently, a reminder that models have ranges of validity.

Real-world application

Household wiring uses parallel circuits so every appliance receives the full mains voltage, and fuse ratings are chosen from P = VI: a 2.3 kW kettle on 230 V draws 10 A, needing a 13 A fuse. Series wiring would share the voltage and make every device dim when another is switched on — the fairy-light effect.

Common mistakes (examiner style)

• Using R = ρL/A with diameter as the area or area in mm² unconverted (1 mm² = 10⁻⁶ m²).
• Assuming the current is the same everywhere in a parallel network.
• For potential dividers, taking the output across the wrong resistor (re-read which two points the voltmeter joins).
• Treating emf as the voltage measured across a battery under load — that is the terminal pd, always less by the lost volts.
• For a lamp whose resistance rises with temperature, applying V = IR with the cold resistance throughout.

IB Exam Experience

Paper 1-style questions. 1. Two resistors, 3.0 Ω and 6.0 Ω, are connected in parallel. Their combined resistance is: A. 2.0 Ω B. 4.5 Ω C. 9.0 Ω D. 0.50 Ω. Key: A — 1/R = 1/3 + 1/6 = 1/2. Parallel combinations are always smaller than the smallest resistor. 2. A wire of length L and diameter d has resistance R. A second wire of the same material, length 2L and diameter 2d, has resistance: A. 4R B. 2R C. R D. R/2. Key: D — length doubles (×2) but area quadruples (÷4): net R/2. 3. A 12 V battery with internal resistance 2.0 Ω is short-circuited. The current is: A. zero B. 6.0 A C. 24 A D. infinite. Key: B — I = ε/r = 6.0 A (and a great deal of heating).

Paper 2-style structured question (7 marks). A battery of emf 12 V and internal resistance 1.5 Ω is connected to a 10 Ω resistor in series with a parallel combination of a 15 Ω resistor and a variable resistor set to 30 Ω. (a) Calculate the total resistance of the external circuit. [2] (b) Calculate the current drawn from the battery. [2] (c) Calculate the potential difference across the parallel combination. [2] (d) State the effect on this pd of decreasing the variable resistance, without calculation. [1] *Marking notes:* (a) Parallel: 1/R = 1/15 + 1/30 = 1/10 → 10 Ω; total external = 10 + 10 = 20 Ω. (b) I = 12/(20 + 1.5) = 0.56 A. (c) pd across parallel = I × 10 = 5.6 V (or 12 − 0.56 × 11.5). (d) The pd decreases: the parallel resistance falls, so it takes a smaller share of the total while current rises — accept either correct reasoning route.

IA idea: Investigate how the internal resistance of a fruit/vegetable cell or a commercial cell changes with temperature — or with age — and how that affects the maximum power it can deliver.

Kirchhoff's laws on a two-loop circuit

Consider a 6.0 V cell driving a 2.0 Ω resistor in series with a parallel pair (3.0 Ω and 6.0 Ω). The parallel pair combine to 2.0 Ω, so the total is 4.0 Ω and the main current is 1.5 A. The pd across the pair is 1.5 × 2.0 = 3.0 V; hence the branch currents are 3.0/3.0 = 1.0 A and 3.0/6.0 = 0.50 A, summing to 1.5 A at the junction — Kirchhoff's first law checked explicitly. The second law around the 6.0 Ω branch loop: 6.0 V = (1.5 A × 2.0 Ω) + (0.50 A × 6.0 Ω) = 3.0 + 3.0 ✓. In harder questions the two laws are used simultaneously with unknown currents and solved as simultaneous equations; the checking step (junctions and loops balance) is identical.

Sensing circuits with real numbers

A potential divider pairs a thermistor (resistance 1800 Ω at 20 °C, falling to 600 Ω at 40 °C) with a fixed 1200 Ω resistor across 6.0 V, output taken across the fixed resistor. At 20 °C: Vout = 6.0 × 1200/(1800 + 1200) = 2.4 V. At 40 °C: Vout = 6.0 × 1200/(600 + 1200) = 4.0 V. A 1.6 V swing over 20 K is easily measured; the design question — which resistor the output should span, and its value relative to the sensor's range — decides sensitivity. Examination questions often ask exactly this: choose the fixed resistor near the sensor's mid-range resistance for the widest output swing.

Why transmission uses high voltage (the circuit version)

Delivering 100 kW to a town along cables of total resistance 2.0 Ω: at 1000 V the current is 100 A and the cable loss is I²R = 20 kW (20% wasted, and the delivered voltage sags). At 25 kV the current is 4.0 A and the loss is 32 W (0.03%). This is Lesson 23's transformer argument stated purely in Theme B.5 language — the IB's synoptic questions reward exactly this cross-theme fluency.

The loading effect: voltmeters are part of the circuit

A real voltmeter has a finite resistance (a school digital meter ~10 MΩ; an analogue meter much less). Measuring the output of a potential divider whose resistors are themselves megohms puts the meter in parallel with the lower resistor and drags the reading down: a divider of 1.0 MΩ and 1.0 MΩ fed by 10 V should read 5.0 V, but a 1.0 MΩ analogue meter in parallel with the lower arm makes that arm 0.50 MΩ and the reading becomes 10 × 0.50/1.50 = 3.3 V. The IB's experimental questions use this to test whether you model the instrument as part of the system. Rule of thumb: the meter's resistance should exceed the measured resistance by a factor of 100 for 1% fidelity.