Apply Kirchhoff's current and voltage laws, combine resistors in series and parallel, and analyse potential dividers and the effect of internal resistance

What this dot point is asking

SEAB wants you to apply Kirchhoff's two circuit laws, combine resistors in series and parallel, analyse a potential divider, and account for the internal resistance of a source. Kirchhoff's laws are simply conservation of charge and energy applied to circuits, and they solve any d.c. network.

The answer

Kirchhoff's current law (charge conservation)

The sum of currents entering a junction equals the sum leaving it:

This expresses conservation of charge: charge does not accumulate at a junction.

Kirchhoff's voltage law (energy conservation)

Around any closed loop, the sum of the e.m.f.s equals the sum of the potential differences across the components:

This expresses conservation of energy: a charge returning to its start has gained and lost equal amounts of energy.

Combining resistors

• Series: the same current flows through each, and resistances add: $R_{\text{total}} = R_1 + R_2 + \dots$
• Parallel: the same voltage is across each, and reciprocals add: $\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$

A parallel combination always has a smaller resistance than the smallest individual resistor.

The potential divider

Two resistors in series split the supply voltage in proportion to their resistances. The voltage across $R_2$ is:

A potential divider provides a chosen fraction of the supply, and with a variable resistor (or a sensor such as a thermistor or LDR) it produces an output that responds to a physical quantity.

Electromotive force and internal resistance

The electromotive force (e.m.f.) $\mathcal{E}$ of a source is the energy it gives per unit charge. A real source has internal resistance $r$, so some energy is dissipated inside it. The terminal potential difference is:

The term $Ir$ is the "lost volts". With external resistance $R$, the current is $I = \dfrac{\mathcal{E}}{R + r}$. The terminal voltage falls as more current is drawn, which is why a battery's voltage drops under heavy load.

Examples in context

Example 1. A thermistor temperature sensor. Placing a thermistor as $R_1$ in a potential divider produces an output voltage that changes with temperature, because the thermistor's resistance falls as it warms. This output can switch a transistor or feed a microcontroller, the basis of simple temperature-controlled circuits.

Example 2. Why a car's headlights dim when starting. The starter motor draws a very large current from the battery, so the lost volts $Ir$ across the internal resistance become significant and the terminal voltage drops. The headlights, supplied by this lower terminal voltage, momentarily dim, a direct demonstration of internal resistance.

Try this

Q1. State Kirchhoff's two laws and the conservation principle each expresses. [2 marks]

• Cue. Current law: currents in equal currents out at a junction (charge conservation). Voltage law: sum of e.m.f.s equals sum of p.d.s round a loop (energy conservation).

Q2. Three resistors of $3.0\ \Omega$, $6.0\ \Omega$ and $6.0\ \Omega$ are all in parallel. Find the combined resistance. [2 marks]

• Cue. $\dfrac{1}{R} = \dfrac{1}{3.0} + \dfrac{1}{6.0} + \dfrac{1}{6.0} = \dfrac{2 + 1 + 1}{6.0} = \dfrac{4}{6.0}$, so $R = 1.5\ \Omega$.

Q3. A cell of e.m.f. $1.5\ \text{V}$ and internal resistance $0.30\ \Omega$ supplies $0.50\ \text{A}$. Find the terminal potential difference. [2 marks]

• Cue. $V = \mathcal{E} - Ir = 1.5 - (0.50)(0.30) = 1.35\ \text{V}$.