Resistors in series and parallel: formulas and worked examples

Series resistors: add them up

When resistors are connected in series – one after another, forming a single path for current – the total resistance is simply the sum of the individual values: R_total = R1 + R2 + R3 + …. This is the simpler of the two formulas, and the intuition matches it: current has to push through every resistor in turn, so each one adds its own opposition on top of the others.

In a series circuit, the current is identical at every point of the loop – there's only one path, so the same amount of current flows through every component. What differs is the voltage: each resistor drops a share of the total voltage in proportion to its own resistance (a voltage divider), while the current through all of them stays the same.

Parallel resistors: reciprocal sum

When resistors are connected in parallel – each one bridging the same two nodes, giving current multiple paths to choose from – the combination formula is different: the reciprocals add, not the resistances themselves. 1/R_total = 1/R1 + 1/R2 + 1/R3 + …, so R_total has to be calculated by summing the reciprocals and then taking the reciprocal of that sum.

In a parallel circuit, it's the voltage that's identical across every resistor – each one is connected directly across the same two points – while the current splits between the branches, with more current flowing through the lower-resistance branches. Adding another parallel path always gives current an additional route, which can only reduce the total resistance, never increase it – the parallel total is always lower than the smallest individual resistor in the group.

Worked example: three resistors, both ways

Take three resistors – 220 Ω, 470 Ω and 1,000 Ω – and combine them both ways to see how differently the two configurations behave.

The series total (1,690 Ω) is higher than any individual resistor; the parallel total (130.33 Ω) is lower than even the smallest one (220 Ω). This is the defining difference between the two configurations, and it's true for any set of resistors, not just this particular example.

Worked example: current in each configuration

Take a 9 V supply connected to 220 Ω and 470 Ω. In series, the total resistance is 690 Ω (220 + 470), so the current through both resistors is the same: I = U ÷ R = 9 ÷ 690 = 13.04 mA (0.01304 A).

Wired in parallel instead, both resistors see the full 9 V, but each draws its own current: the 220 Ω branch carries 9 ÷ 220 = 40.91 mA, and the 470 Ω branch carries 9 ÷ 470 = 19.15 mA. The total current drawn from the supply is the sum of the two branches, 60.06 mA – more than four times the series case, because parallel resistors present a much lower total resistance (149.86 Ω here) to the same 9 V source.

Equal resistors: a quick mental shortcut

When all the resistors in a parallel group are equal, the total is simply that value divided by how many there are. Two 100 Ω resistors in parallel give exactly 50 Ω (100 ÷ 2); three 100 Ω resistors in parallel give 33.33 Ω (100 ÷ 3). This shortcut only works when every resistor in the group has the same value – as soon as the values differ, the full reciprocal-sum calculation is needed.

Mixed series-parallel networks

Real circuits often combine both configurations in the same network – for example, two resistors in parallel, with that combination then in series with a third resistor. The method is to work from the innermost combination outward: calculate the parallel (or series) sub-group first, treat its result as a single equivalent resistor, then combine that with whatever it's in series or parallel with next. There's no single formula for a mixed network – it's solved step by step, one sub-combination at a time, using the series and parallel formulas above in sequence.

1. 1

   Identify the innermost group

   Find the smallest self-contained series or parallel group in the circuit – one that doesn't depend on any other resistor's value first.

2. 2

   Reduce it to a single equivalent value

   Apply the series formula (sum) or the parallel formula (reciprocal sum) to that group, replacing it conceptually with one resistor of the resulting value.

3. 3

   Repeat with the simplified circuit

   Treat that equivalent resistor as a normal component and look for the next group to reduce – series or parallel, whichever applies at that stage.

4. 4

   Continue until one value remains

   Keep reducing sub-groups until the whole network collapses to a single total resistance value between the two original terminals.