Series & Parallel Resistance

Series: one path, sum the squeeze

Resistors in a line carry the same current, one after the other — every one adds its squeeze to the trip:

R_total = R1 + R2 (+ R3 + …)

Sanity checks: add a plain wire (≈ 0 Ω) in series and nothing changes. Two equal resistors double. And you've already measured this law — it's why the voltage drops in a series loop split in proportion to resistance.

Parallel: more paths, less resistance

Side by side, resistors share the work: the current splits between the branches, so the pair passes more current than either alone — which is exactly what "less resistance" means. The two-resistor formula:

R_total = R1 × R2 ÷ (R1 + R2)

Run the worked cases:

Sanity checks: parallel with a plain wire is a short — everything takes the free lane, total ≈ 0. Parallel with nothing (air, ∞ Ω) changes nothing. Two equal branches halve; three equal branches make a third.

You've already built both

Lesson 3's two-LEDs-in-series build and its parallel-branches build are these two pictures with LEDs in the branches. And lesson 1's answer to "why didn't my finger light the LED" is a parallel fact: your ≈ 100 kΩ body in parallel with the circuit takes almost none of the traffic.

Loading — the divider mystery, solved

The Voltage Divider Formula sheet ends with a mystery: hang a load across the tap and the voltage sags. Now you can compute it. The load sits in parallel with R2, making the effective bottom resistor smaller — a smaller fraction, a lower tap.

1. Build the 1 kΩ / 1 kΩ divider: tap = 2.50 V.
2. Hang a 1 kΩ load across the tap. The bottom is now 1 kΩ || 1 kΩ = 500 Ω.
3. Recompute the divider: 5 × 500 ÷ 1500 ≈ 1.7 V. The tap sagged by a third — nothing broke; the fraction changed.

And here's why your multimeter gets away with probing taps all day: its voltage input is roughly 10 MΩ. In parallel with a 1 kΩ bottom resistor that's 1 kΩ || 10 MΩ ≈ 999.9 Ω — the meter is a load so light the circuit barely notices. Meters are built to be ignorable.

N resistors in parallel: sum the reciprocals

The product-over-sum formula only handles two. The real law handles any number of branches:

1 ÷ R_total = 1÷R1 + 1÷R2 + … + 1÷R_N

In words: take each resistance's reciprocal — 1 divided by it — add the reciprocals, then take the reciprocal of the sum. Worked, for 220 Ω || 470 Ω || 1 kΩ:

1. Reciprocal of each: 1÷220 ≈ 0.00455 · 1÷470 ≈ 0.00213 · 1÷1000 = 0.00100
2. Add them: 0.00455 + 0.00213 + 0.00100 = 0.00768
3. Reciprocal of the sum: 1 ÷ 0.00768 ≈ 130 Ω — smaller than the smallest branch ✓

Shortcut worth memorizing: N equal resistors make R ÷ N. Four 1 kΩ in parallel: 250 Ω. (The two-resistor product-over-sum is just this law solved for two branches.)

Practice

Answers at the bottom — and every pure-resistor answer is checkable with the meter (parts out of circuit, ohms mode).

1. 220 Ω and 220 Ω in series? In parallel?
2. 10 kΩ || 10 kΩ || 10 kΩ — all three in parallel?
3. A 10 kΩ / 10 kΩ divider has a 10 kΩ load hung on its tap. What does the tap read?
4. You need ≈ 110 Ω but the kit's smallest resistor is 220 Ω. How do you make it?
5. What's 1 kΩ in parallel with your ≈ 100 kΩ hand-to-hand body? (Estimate before you compute.)

Answers

1. Series 440 Ω; parallel 110 Ω. 2. 3.33 kΩ — three equal branches make a third. 3. Bottom becomes 10 k || 10 k = 5 kΩ; tap = 5 × 5 ÷ 15 ≈ 1.67 V, down from 2.50 V. 4. Two 220 Ω in parallel — this is a real engineer's trick, not a workaround. 5. ≈ 990 Ω — barely below 1 kΩ. Your body is a lane so narrow the traffic doesn't notice it; that's lesson 1's safety fact, computed.