Pull an old string of fairy lights out of a box in the loft, plug it in — and half the time nothing lights. Not one bulb. You jiggle the wire, you squint at the little glass beads, and eventually you find it: a single bulb, somewhere in the middle, quietly burnt out. That one dead bulb has switched off the whole string.

That is a series circuit giving away its deepest secret. Everything is threaded onto one single loop, one component after another, so the charge has only one road to travel. On this page we'll follow that single loop all the way round and pin down its three golden rules: the current is the same everywhere, the supply's push is shared out between the components, and the resistances simply add up.

Beads on one string

A series circuit is the simplest way to wire things up: take the battery, run a wire to the first component, run a wire from that to the second component, and a wire from the last one back to the battery. No branches, no side-roads, no forks. Every component sits in the same loop, lined up in a row like beads threaded on a single string.

Because there is exactly one path, the charge that leaves the battery has no choice at all: it must go through every component, in order, before it can get home. This one fact is where all three rules come from. Let's take them one at a time.

Rule 1: the current is the same everywhere

Current is how fast charge flows past a point — how many charges stream by each second. In a series loop there is only one path, so the charge cannot pile up anywhere and it cannot leak away anywhere. Whatever flows out of the battery flows through the first resistor, then straight through the second, and back to the battery, all in one file. So the current has the same value at every single point in the loop:

Put an ammeter anywhere you like — just after the battery, between the two bulbs, just before the battery — and it reads the same number. Think of the bicycle chain again: every link moves at once, at the same speed, all the way round. The charge does exactly that.

Rule 2: the supply voltage is shared

The battery gives the charge a total push — its potential difference (pd, or voltage), say 12\ \text{V}. As the charge squeezes through each component it "spends" some of that push, and by the time it gets back to the battery it has spent it all. So the pd across the components must add up to the pd of the supply:

A bigger resistor grabs a bigger share of the push; a smaller one takes a smaller share — but the slices always add back up to the whole. With a 12\ \text{V} battery and two equal bulbs, each gets 6\ \text{V}. Make one bulb a much bigger resistor and it might hog 9\ \text{V}, leaving just 3\ \text{V} for the other.

Rule 3: the resistances add up

Every extra component the charge has to force its way through is one more obstacle in the single loop — so the resistances stack up, end to end, into one big total:

This is beautifully simple: to find the total resistance of a series circuit, you just add the numbers. And it has a striking consequence. Because the same battery now pushes against a larger total resistance, the current drops (from I = V / R_\text{total}). So adding another bulb in series makes every bulb dimmer, not brighter — more resistance in the loop chokes the current for all of them at once.

Build a series circuit

Here is a single loop: a battery at the bottom and two resistors, R_1 and R_2, sitting one after the other along the top. Drag the sliders to change the supply pd and each resistance, and watch all three rules work together.

Notice as you play: the current I printed at the left and right of the loop is always the same number — that's Rule 1. The two pd readings V_1 and V_2 always add up to the supply — that's Rule 2. And the total resistance in the middle is just R_1 + R_2 — that's Rule 3. Crank up a resistance and see the shared current fall for the whole loop.

Worked example: total resistance, current, and each pd

A 12\ \text{V} battery is connected in series with two resistors, R_1 = 4\ \Omega and R_2 = 6\ \Omega. Find the total resistance, the current in the circuit, and the pd across each resistor.

Step 1 — total resistance (Rule 3, just add).

Step 2 — the current (same everywhere, from I = V / R_\text{total}).

This 1.2\ \text{A} flows through both resistors — the current is the same all round the loop.

Step 3 — the pd across each resistor (use V = IR on each, with the shared current).

Step 4 — check Rule 2. The two pds should add up to the supply:

Perfect — the whole push is accounted for. Notice the bigger resistor (R_2 = 6\ \Omega) took the bigger share of the voltage (7.2\ \text{V}). The share of voltage follows the share of resistance.

Worked example: adding a bulb dims them all

Start with a 6\ \text{V} battery and a single bulb of resistance 3\ \Omega. The current is

Now add a second identical 3\ \Omega bulb in series. The total resistance is now R_\text{total} = 3 + 3 = 6\ \Omega, so

The current has halved — from 2\ \text{A} down to 1\ \text{A} — even though the battery is unchanged. Less current through each bulb means each one glows more dimly. That is the price of a series circuit: the more you add, the dimmer they all become, because they must share the same choked-down current.