Series Circuits - Physics for Pilot Assessments

A series circuit is one in which every component sits on the same single path, so that all the charge that goes through one must go through the next. By the end of this lesson you should be able to say what is the same everywhere in such a circuit and what is not, add resistances to get the total, work out the current from that total and then the voltage across each individual component, run the calculation backwards to find an unknown resistance, and say why one failure stops the lot.

One path, one current

The defining feature is that there is nowhere for the charge to go but forward. If two lamps are in series there is a single loop, and every coulomb that leaves the supply passes through the first lamp, then the second, then returns — there is no junction anywhere for it to take a different route. Combine that with the result from the first lesson, that charge is not consumed on the way round, and the current has to be identical at every point in the loop. Not approximately, and not almost: an ammeter placed anywhere in a series circuit reads the same value as an ammeter placed anywhere else in it.

Current in a series circuit

I_1 = I_2 = I_3 = I_\text{supply}

$I_1, I_2, I_3$: the current through each component in the loop, in amperes (A)
$I_\text{supply}$: the current drawn from the supply, in amperes (A) — the same value

That is a genuinely useful piece of information rather than a triviality, because it means you can work out the current once, from the whole circuit, and then use that same number for every component in it.

Resistances add

If the charge has to fight its way through the first resistance and then through the second, the two oppositions are met one after the other and the total opposition is their sum. There is no subtlety here and no formula to reach for beyond the addition itself: two resistors in series behave exactly like one resistor whose value is the two added together, and a circuit with three or more works the same way.

Total resistance in series

R_\text{total} = R_1 + R_2 + R_3 + \dots

$R_\text{total}$: the resistance of the whole chain, in ohms (Ω)
$R_1, R_2, R_3$: the resistance of each component in the chain, in ohms (Ω)

One consequence is worth stating plainly, because it is a fact about a single path rather than a general truth about circuits: adding a component in series always raises the total resistance, and therefore always lowers the current the supply delivers. Wire three identical lamps in series instead of one and each is markedly dimmer than the single lamp was, because the total resistance has trebled and the current through all of them has fallen to a third.

The supply voltage shares out

The current is shared in the sense of being identical everywhere; the voltage is shared in the ordinary sense of being divided up. Follow one coulomb round the loop. It leaves the supply carrying a certain amount of energy — twelve joules for each coulomb, from a twelve-volt supply. It gives up some of that energy in the first component and the rest in the second, and it arrives back at the supply with none of it left, because if it arrived with energy to spare that energy would have to accumulate somewhere, and it does not. So the energies given up per coulomb must add up to the energy supplied per coulomb, and since energy per coulomb is exactly what voltage means, the voltages across the components add up to the supply voltage.

Voltages in series add up to the supply

V_\text{supply} = V_1 + V_2 + V_3 + \dots

$V_\text{supply}$: the voltage of the supply, in volts (V)
$V_1, V_2, V_3$: the potential difference across each component, in volts (V)

How the sharing works out is decided by the resistances. Since the current through every component is the same, and the voltage across each one is that current multiplied by its own resistance, the component with the larger resistance takes the larger share of the voltage — in direct proportion. A 4 Ω resistor in series with a 2 Ω one takes twice as much of the supply voltage, because twice the resistance at the same current needs twice the push. That is the whole of what is meant when a question talks about the voltage dividing across a series chain, and it is worth reading it as energy sharing rather than as a formula to memorise: the bigger obstacle takes more of the energy each coulomb was carrying.

The 4 Ω and 2 Ω resistors share one path, so both carry the same 2 A. They do not share the voltage equally: the larger resistance takes 8 V and the smaller 4 V, and the two add back to the 12 V the supply provided.

Example 1 — the standard series calculation, end to end

A 12 V supply is connected to a 4 Ω resistor in series with a 2 Ω resistor. Find the total resistance, the current, and the voltage across each resistor.

Add the resistances: $R_\text{total} = 4 + 2 = 6\ \Omega$
Now treat the whole circuit as one 6 Ω resistance across 12 V, and use $\text{current} = \frac{\text{voltage}}{\text{resistance}}$ : $I = \frac{12}{6} = 2\ \text{A}$
That 2 A flows through both resistors, because there is only one path.
Across the 4 Ω resistor: $\text{voltage} = \text{current} \times \text{resistance} = 2 \times 4 = 8\ \text{V}$
Across the 2 Ω resistor: $\text{voltage} = 2 \times 2 = 4\ \text{V}$
Check: $8 + 4 = 12\ \text{V}$ That is the supply. If the two voltages had not added back to the supply, one of the steps would have been wrong.

Example 2 — working backwards to an unknown resistor

A 12 V supply drives 1.5 A through a series circuit containing a 4 Ω resistor and one unknown resistor. What is the unknown resistance, and what is the voltage across each?

The current and the supply voltage give the total resistance: $\text{resistance} = \frac{\text{voltage}}{\text{current}} = \frac{12}{1.5}$ Multiply both numbers by 10 to clear the decimal: $R_\text{total} = \frac{120}{15} = 8\ \Omega$
Series resistances add, so the unknown is the total minus the known one: $8 - 4 = 4\ \Omega$
Across the known 4 Ω resistor: $V = 1.5 \times 4 = 6\ \text{V}$
Across the unknown, which is also 4 Ω: $V = 1.5 \times 4 = 6\ \text{V}$
Check: $6 + 6 = 12\ \text{V}$ The two resistances came out equal, so the supply voltage split equally between them — which is the proportional sharing rule with a ratio of one to one.

Example 3 — why three lamps in series are dim

A single lamp of resistance 4 Ω is connected to a 12 V supply. Two more identical lamps are then added, so that all three sit in series in one loop. Find the current in each arrangement, and the voltage across each lamp once there are three.

The single lamp alone: $\text{current} = \frac{\text{voltage}}{\text{resistance}} = \frac{12}{4} = 3\ \text{A}$ The lamp has the whole 12 V across it.
With three in series, the resistances add: $R_\text{total} = 4 + 4 + 4 = 12\ \Omega$
The current — the same at every point of the loop — is now $\frac{12}{12} = 1$ ampere, a third of what the single lamp drew.
Each lamp takes voltage = current × resistance = 1 × 4 = 4 volts, and the check works: $4 + 4 + 4 = 12\ \text{V}$ That is the supply.
Each filament now carries a third of the current with a third of the voltage across it, which is why all three are markedly dimmer than the single lamp was. Adding the lamps did not share the original 3 A out between them — it reduced the current for everything in the loop.

Why a break stops everything

Because there is only one path, anything that interrupts that path stops the current in all of it. That is what makes a switch in a series circuit able to control everything on the loop from wherever it happens to be fitted, and it is also the well-known weakness of the arrangement: in a string of lamps wired in series, one lamp with a broken filament leaves the whole string dark, and there is nothing in the behaviour of the others to tell you which one failed. Note carefully what has happened in that case, because items probe it: the filament breaking has not lowered the circuit's resistance, it has made it effectively infinite, so the current falls to zero rather than redistributing itself. There is nowhere for it to redistribute to.

Why it matters in aviation

Series connection appears in an aircraft mainly where a single break is supposed to stop everything, which is precisely the job of a protective device fitted in line with a load: it sits on the one path the current has, so opening it removes that load and nothing else. The same reasoning explains the battery. Each cell adds its own energy to every coulomb passing through, so cells are stacked one after another until their EMFs add to the system voltage, and one failed or reversed cell in that stack affects the whole battery rather than a part of it.

Elsewhere series wiring is avoided deliberately, and the two rules of this lesson say why. Components in series share the supply voltage between them, so each receives less than the full busbar voltage and its share depends on what else sits in the chain. And because there is only one path, a break anywhere opens all of it. Neither property is wanted from equipment that has to keep working when something else has failed.

What the test asks

The distractor to expect is a current that changes as it goes round — an item showing a series circuit with two unequal resistors and offering a larger current before the bigger resistor than after it, or a current that splits between the two. Neither happens in a single loop; it is the voltage that is unequal, and reading which of the two quantities the question is about settles most items on the topic. Expect a total-resistance item where the trap is to average the resistances instead of adding them, and expect one where you are given the voltage across one component and asked for the other, which is nothing more than a subtraction from the supply voltage. Finally, expect a lamp-brightness item: adding lamps in series makes all of them dimmer, and the tempting wrong answer has each one unaffected because the supply has not changed.