IMAT Physics · Electricity

Resistor Networks, Kirchhoff's Rules and Internal Resistance

Resistors in series add, resistors in parallel give a total smaller than the smallest, current splits at junctions, and a real battery loses some voltage inside itself.

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Why this matters

This is the most asked page of the chapter: past papers from 2011, 2014, 2016, 2022 and 2025 all hinged on combining resistors, either to find a current, the largest and smallest totals, meter readings or the resistance of a big regular network.

Concept 1 of 3: Resistors in series and in parallel

In series the current has to pass through every resistor in turn, so the obstacles add up. In parallel the current has several roads to choose from, so adding another road always makes the whole thing easier to get through. Collapse a network one group at a time, from the inside out.

Definition

Series: the same current flows through each; the total is the sum, $R = R_1 + R_2 + \dots$
Parallel: each has the same p.d. across it; reciprocals add, $1/R = 1/R_1 + 1/R_2 + \dots$
Two in parallel: $R = \dfrac{R_1 R_2}{R_1 + R_2}$ (product over sum).
$n$ identical resistors $R$ : $nR$ in series, $R/n$ in parallel.
A parallel total is always smaller than the smallest branch. Adding a branch in parallel lowers the total and raises the current from the battery.

Series and parallel

R_{\text{series}} = R_1 + R_2 + \dots \qquad \frac{1}{R_{\parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots

$R_1, R_2$ the individual resistances, in Ω
$R_{\parallel}$ the single resistance equal to the parallel group

Worked example

A 22 V battery with no internal resistance is connected to a 5.0 Ω resistor in series with a parallel pair of 10 Ω and 15 Ω. Find the current from the battery and the current in each branch of the pair.

Practice this conceptself-check · 4 quick reps

A parallel total is smaller than every branch

Putting resistors in parallel gives the current extra paths, so the total is less than the smallest single resistor. If your parallel answer is bigger than one of the branches, you have added the resistances instead of their reciprocals.

Concept 2 of 3: Kirchhoff's rules, the potential divider and meters

Charge does not pile up anywhere in a circuit, so whatever flows into a junction must flow out. Energy is not made or lost either: going all the way round a loop, the voltage the battery gives is used up exactly by the components. These two bookkeeping rules solve every simple circuit.

Definition

Junction rule (charge conservation): the total current into a junction equals the total current out.
Loop rule (energy conservation): round any closed loop, the e.m.f.s add up to the sum of the p.d.s ( $IR$ values).
Potential divider: resistors in series share the supply voltage in proportion to their resistances, $V_1 = V\,R_1/(R_1 + R_2)$ .
An ammeter goes in series and ideally has zero resistance. A voltmeter goes in parallel and ideally has infinite resistance (it takes no current).

Potential divider

V_1 = V \, \frac{R_1}{R_1 + R_2}

$V$ total p.d. across the series chain, in V
$V_1$ p.d. across $R_1$

Worked example

A 12 V supply with no internal resistance is connected across a 2.0 kΩ and a 4.0 kΩ resistor in series. Find the current and the p.d. across each resistor.

Practice this conceptself-check · 3 quick reps

Meters in the wrong place

An ammeter in parallel with a component short-circuits it, because the ammeter has almost no resistance. A voltmeter placed in series stops the current almost completely and reads close to the full supply voltage. Read each meter's position before using its reading.

Concept 3 of 3: EMF and internal resistance of a battery

A battery is not a perfect voltage source. Its chemicals have some resistance of their own, so when current flows, some of the battery's energy is spent inside it. The more current you draw, the more voltage is lost inside and the less reaches the outside circuit.

Definition

The e.m.f. $\varepsilon$ is the energy the battery gives each coulomb, in volts. Despite its name it is not a force.
The internal resistance $r$ acts like a small resistor in series inside the battery.
The terminal p.d. (what reaches the circuit) is $V = \varepsilon - Ir$ . The $Ir$ part is the "lost volts".
With no current (an open switch, or an ideal voltmeter alone), $V = \varepsilon$ .
A short circuit ( $R = 0$ ) draws the largest possible current, $I = \varepsilon / r$ .

EMF and internal resistance

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

$\varepsilon$ e.m.f. of the battery, in V
$R$ external (load) resistance, in Ω
$r$ internal resistance, in Ω
$V$ terminal p.d., in V

Worked example

A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Find the current, the terminal p.d. and the lost volts.

Practice this conceptself-check · 3 quick reps

Terminal p.d. is less than the e.m.f. when current flows

The e.m.f. is reached only when no current flows. Draw a current and the battery loses

Ir

inside itself, so the voltage across the circuit falls. The bigger the current (the smaller the load resistance), the bigger the drop.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Resistors in series and in parallel
Series and parallel
$R_{\text{series}} = R_1 + R_2 + \dots \qquad \frac{1}{R_{\parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$
Kirchhoff's rules, the potential divider and meters
Potential divider
$V_1 = V \, \frac{R_1}{R_1 + R_2}$
EMF and internal resistance of a battery
EMF and internal resistance
$\varepsilon = I(R + r) \qquad V = \varepsilon - I r$

Watch out for (3)

A parallel total is smaller than every branch
→ Resistors in series and in parallel
Meters in the wrong place
→ Kirchhoff's rules, the potential divider and meters
Terminal p.d. is less than the e.m.f. when current flows
→ EMF and internal resistance of a battery

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