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Electrical Circuits · DC Circuits

Series and parallel

Reading1h 2mLesson 2 of 3 in this module

Solve series, parallel and series-parallel resistor networks for total resistance, current and voltage.

Why it matters

Real circuits are rarely one battery and one resistor. Lights on a string, outlets on a branch circuit, sensors on a control panel and the resistors on a circuit board are all connected in series, in parallel or in a mix of both. Reducing a network to a single equivalent resistance lets you find the total current, then work back to the voltage and current at every part. It is the most common hand calculation in circuit work.

Series circuits

Components are in series when they sit one after another on a single path, so the same current flows through each one.

  • Total resistance is the sum: R_total = R1 + R2 + R3 + ...
  • The source voltage is shared out: each resistor's voltage is I × R, and the drops add up to the source voltage.
  • If one part opens (a lamp filament breaks), current stops everywhere on that path.

Worked example 1. A 12 V battery drives 100 Ω, 200 Ω and 300 Ω in series.

R_total = 100 + 200 + 300 = 600 Ω
I = 12 / 600 = 0.02 A (20 mA)
V1 = 0.02 × 100 = 2 V,  V2 = 4 V,  V3 = 6 V   (2 + 4 + 6 = 12 V)

The biggest resistor takes the biggest share of the voltage. This is the idea behind a voltage divider: two resistors in series give an output of Vout = Vin × R2 / (R1 + R2).

Worked example (divider). A sensor input needs about 3 V from a 9 V battery. With R1 = 20 kΩ on top and R2 = 10 kΩ below:

Vout = 9 × 10,000 / (20,000 + 10,000) = 3 V

Dividers are used for reference voltages and for reading sensors, but they are not power supplies: connecting a load across R2 puts it in parallel with R2 and pulls the output voltage down.

Spotting series and parallel on a drawing

Schematics are not always drawn neatly, so look at the nodes (the connection points), not at the shape of the drawing. Trace each wire with a finger or a coloured pen:

  • Two parts are in series if the only thing joined at the node between them is those two parts.
  • Two parts are in parallel if both of their ends land on the same pair of nodes, even if one is drawn at the top of the page and the other at the bottom.

Colouring every node a different colour first is a habit that prevents most reduction mistakes.

Parallel circuits

Components are in parallel when they connect across the same two points, so they all have the same voltage.

  • Currents add: the total current is the sum of the branch currents.
  • The equivalent resistance is found from reciprocals:
1 / R_total = 1/R1 + 1/R2 + 1/R3 + ...
  • For exactly two resistors there is a shortcut, "product over sum": R_total = (R1 × R2) / (R1 + R2).
  • The equivalent resistance is always smaller than the smallest branch. Adding a branch gives current another path, so total current goes up.
  • If one branch opens, the others keep working. That is why outlets and lights in a building are wired in parallel.

Worked example 2. 100 Ω and 300 Ω in parallel across 12 V.

R_total = (100 × 300) / (100 + 300) = 75 Ω
I_total = 12 / 75 = 0.16 A
I1 = 12 / 100 = 0.12 A,  I2 = 12 / 300 = 0.04 A   (0.12 + 0.04 = 0.16 A)

Two equal resistors in parallel give half the value: 220 Ω ∥ 220 Ω = 110 Ω.

Series-parallel circuits

Reduce the network step by step, from the part furthest from the source back toward it:

  1. Find groups that are clearly in series or clearly in parallel.
  2. Replace each group with its equivalent resistance and redraw.
  3. Repeat until one resistance is left.
  4. Find the total current, then expand the drawing again in reverse to find each voltage and current.

Worked example 3. A 10 V source feeds R1 = 40 Ω in series with a parallel pair R2 = 60 Ω and R3 = 30 Ω.

R2 ∥ R3 = (60 × 30) / (60 + 30) = 20 Ω
R_total = 40 + 20 = 60 Ω
I_total = 10 / 60 = 0.1667 A
V across R1 = 0.1667 × 40 = 6.67 V
V across the pair = 0.1667 × 20 = 3.33 V
I2 = 3.33 / 60 = 0.0556 A,  I3 = 3.33 / 30 = 0.111 A

Check: 6.67 + 3.33 = 10 V, and 0.0556 + 0.111 = 0.1667 A. Always close the loop with a check like this.

Common mistakes

  • Adding parallel resistances. If your parallel answer is bigger than the smallest branch, you added instead of using reciprocals.
  • Forgetting the final reciprocal. 1/100 + 1/300 = 0.01333, but that is 1/R. The resistance is 1 / 0.01333 = 75 Ω.
  • Calling resistors "in series" because they are drawn in a line. Series means the same current with no branch point between them; parallel means the same two end nodes.
  • Skipping the expand-back step. The equivalent resistance gives the total current, not the current in each branch.

Recap

  • Series: same current, resistances add, voltages share out.
  • Parallel: same voltage, currents add, 1/R_total = Σ 1/R, result smaller than the smallest branch.
  • Reduce series-parallel networks step by step, then expand back and check with the totals.

Try it

Open the Do tab and write the parallel-resistance formula in a small network calculator.

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