Resistors in series

One path, one after another

Two parts are in series when they sit one after another on a single path, with nothing branching off between them. Every bit of current that leaves the battery has to pass through the first resistor, then the second, then back to the battery. There is no other way around.

Each resistor "squeezes" the current a little (the water-pipe picture from earlier in the series), so the squeezes stack up.

Total resistance just adds

For resistors in series, the total (or equivalent) resistance is the sum:

R total = R1 + R2 + R3 + ...

So a 100 Ω resistor followed by a 200 Ω resistor acts exactly like one 300 Ω resistor. The total is always bigger than any single resistor in the row.

Same current everywhere

Connect that pair to a 9 V battery. Ohm's law gives the current: I = V ÷ R = 9 V ÷ 300 Ω = 0.03 A, which is 30 mA.

Because there's only one path, that same 30 mA flows through the 100 Ω resistor and the 200 Ω resistor. Put a meter anywhere in the loop and it reads 30 mA.

The voltage gets shared

The voltage is different: each resistor takes a share of it, called a voltage drop. Use Ohm's law on each one with the 30 mA current:

• 100 Ω: 0.03 A × 100 Ω = 3 V
• 200 Ω: 0.03 A × 200 Ω = 6 V

Together, 3 V + 6 V = 9 V, all of the battery's voltage. The bigger resistor takes the bigger share.

Add a third resistor in series and the total resistance climbs, so the current from the same battery drops.

Try it yourself

Pick two resistor values, say 470 Ω and 1,000 Ω, and a 9 V battery. Work out the total resistance, the current, and the voltage across each resistor. Check that your two voltage drops add back up to 9 V. If you have a kit and a multimeter, build it and measure.

Good to know

• The series rule: R total = R1 + R2 + .... The current is the same through every part of a single loop (Kirchhoff's current law, coming later in the series).
• The voltage drops around the loop add up to the source voltage (Kirchhoff's voltage law, also coming later).
• These numbers assume an ideal battery and wires with zero resistance. A real 9 V battery's voltage sags a little under load, so you'd measure slightly less.
• "Squeeze" is the series' water-pipe analogy; it's a simplification.
• 100 Ω and 200 Ω were picked for easy math. 200 Ω isn't a standard E12 value; 220 Ω is common.
• Battery voltages only. Safe at hobby level.

Answer to the comment question

Two ohms and four ohms in series on a twelve-volt battery. What's the current? 2 amps. In series the resistances add, so 2 Ω + 4 Ω = 6 Ω total, and Ohm's law gives 12 V ÷ 6 Ω = 2 A, the same current through both resistors. (A hypothetical practice circuit, assuming an ideal battery and wires.)