Series and parallel resistor combinations, Kirchhoff's two laws, EMF and internal resistance, and the relationship $\varepsilon = I(R + r)$ with terminal potential difference.

What this dot point is asking

Edexcel wants you to combine resistors in series and parallel, apply Kirchhoff's two laws to multi-loop circuits, and account for the internal resistance of a real source using $\varepsilon = I(R + r)$ and the terminal potential difference $V = \varepsilon - Ir$. You should be able to explain why a battery's terminal voltage drops as it delivers more current, and design the experiment that measures EMF and internal resistance from a graph.

The answer

Combining resistors

In series the current has only one path, so charge conservation forces the same $I$ everywhere; the potential differences add to the supply voltage. In parallel each branch sees the full potential difference across the combination, and the branch currents add at the junctions to give the total current. Two equal resistors in parallel give half the single value, and $n$ equal resistors in parallel give $R/n$.

Kirchhoff's two laws

To solve a network, label every branch current, apply the junction rule to reduce the number of unknowns, then write a loop equation for each independent loop. Be consistent with sign conventions: choose a direction to traverse each loop, count an EMF as positive when you pass from the negative to the positive terminal, and count an $IR$ drop as positive when you travel with the current.

EMF and internal resistance

The electromotive force $\varepsilon$ is the energy transferred to each coulomb of charge by the source, measured in volts (joules per coulomb). A real source is not ideal: chemical reactions and the resistance of the electrolyte mean some energy per coulomb is dissipated inside the cell across its internal resistance $r$.

Rearranging, $V = \varepsilon - Ir$ is a straight line of $V$ against $I$ with $y$-intercept $\varepsilon$ and gradient $-r$. As the load current rises, the lost volts rise and the terminal voltage falls. At open circuit ($I = 0$) the terminal voltage equals the EMF; under short circuit the current is limited to $\varepsilon / r$.

Required practical: measuring EMF and internal resistance

The standard Edexcel core practical connects a cell to a variable resistor (rheostat), with a voltmeter across the terminals and an ammeter in series. Vary the load, record matched $V$ and $I$ pairs, and plot $V$ against $I$. The intercept gives $\varepsilon$ and the magnitude of the gradient gives $r$. Take readings quickly so the cell does not warm up and change $r$, and use a high-resistance voltmeter so it draws negligible current.

Examples in context

In a car, the starter motor draws a huge current (hundreds of amps), so the lost volts $Ir$ across the battery's small internal resistance become large and the terminal voltage sags, which is why the headlights dim while cranking. In a laboratory, a high-quality power supply has a very low internal resistance so its terminal voltage barely changes with load. A nearly flat battery has a high internal resistance: its open-circuit voltage can still read near normal, but under load the lost volts dominate and the terminal voltage collapses, which is how battery testers diagnose a tired cell.

Try this

Q1. State Kirchhoff's first law. [1 mark]

• Cue. The total current into a junction equals the total current out (conservation of charge).

Q2. Two $6.0$ ohm resistors are connected in parallel. Find the combined resistance. [2 marks]

• Cue. $\frac{1}{R} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0}$, so $R = 3.0$ ohm.

Q3. A cell of EMF $1.5$ V has internal resistance $0.30$ ohm. It delivers $2.0$ A. Calculate the terminal potential difference. [2 marks]

• Cue. $V = \varepsilon - Ir = 1.5 - 2.0 \times 0.30 = 0.90$ V.