Series and parallel circuits
What you’ll learn
- The difference between series and parallel connections
- How to calculate total resistance in each
- How this applies to batteries and sensors in your robot
Series: one path
Parts in series are connected end-to-end, so there’s only one path for current.
(+) ── R1 ── R2 ── R3 ── (−)| Rule | Series |
|---|---|
| Current | Same through every part |
| Voltage | Shared: the voltages across each part add up to the supply |
| Total resistance | Add them: R = R1 + R2 + R3 |
Example: 100 Ω + 220 Ω = 320 Ω total.
Weakness: if one part breaks, the whole chain stops (like old holiday lights).
Parallel: several paths
Parts in parallel are connected across the same two points, so current splits between paths.
(+) ──┬── R1 ──┬── (−) └── R2 ──┘| Rule | Parallel |
|---|---|
| Voltage | Same across every branch |
| Current | Splits: branch currents add up to the total |
| Total resistance | Less than the smallest one: 1/R = 1/R1 + 1/R2 |
Example: two 100 Ω resistors in parallel: 1/R = 1/100 + 1/100 = 2/100 → R = 50 Ω.
Shortcut for two resistors: R = (R1 × R2) ÷ (R1 + R2).
Strength: one branch can fail and the others keep working. That’s why your house is wired in parallel.
Why this matters for robots
- Batteries in series add voltage. Four 1.5 V AA cells in series = 6 V. This is how battery holders work.
- Identical batteries in parallel add capacity (they last longer) at the same voltage. Only do this with matched cells; never casually parallel lithium batteries.
- All sensors connect in parallel to the 5 V and GND rails, so each gets the full 5 V.
- Voltage dividers (two resistors in series) are used to read sensors and to step 5 V signals down to 3.3 V.
Example: two resistors in series
100 Ω and 220 Ω in series on 5 V. Total R = 320 Ω, so I = 5 ÷ 320 ≈ 15.6 mA through both. Voltage across the 220 Ω: 0.0156 × 220 ≈ 3.4 V; across the 100 Ω: ≈ 1.6 V. Together: 5 V.
Example: a voltage divider
Two equal 10 kΩ resistors in series on 5 V make 2.5 V at the middle point. The general formula:
Vout = Vin × R2 ÷ (R1 + R2)
Light sensors (LDRs) use this trick: as light changes the LDR’s resistance, the middle voltage changes, and the Arduino reads it on an analog pin.
Example: battery packs
3 × AA in series = 4.5 V. 2 packs of 4 × AA in parallel = still 6 V, but roughly twice the run time.
Try it
The left circuit has 100 Ω and 220 Ω in series. The right circuit has two 100 Ω in parallel. Both use 5 V. Start the simulator and hover over each resistor:
- Left: is the current the same through both resistors? Do their voltages add up to 5 V?
- Right: is the voltage the same across both? Does each branch carry half the total current?
Runs on Falstad Circuit Simulator, free, no account needed.
Series totalEasy
What’s the total of 1 kΩ, 2.2 kΩ and 470 Ω in series? 1000 + 2200 + 470 = 3670 Ω.
Parallel pairMedium
What’s the total of 220 Ω and 330 Ω in parallel? (220 × 330) ÷ (220 + 330) = 72,600 ÷ 550 = 132 Ω, less than the smaller one, as expected.
5 V to 3.3 VHard
Design a voltage divider that turns a 5 V signal into about 3.3 V for an ESP32 input. Use Vout = 5 × R2 ÷ (R1 + R2) and try R2 = 2 × R1. R1 = 1 kΩ (top), R2 = 2 kΩ (bottom): 5 × 2000 ÷ 3000 = 3.33 V.
Quick check
Finished this lesson?