Resistors in series add: 220, 330 and 470 ohms make 1,020. Resistors in parallel follow a reciprocal sum, so 1 kΩ, 2.2 kΩ and 4.7 kΩ together give 599.8 Ω — always less than the smallest resistor in the set. Those two rules produce everything else in this area, including both dividers.
The parallel result being smaller than any individual resistor is the part that feels wrong until you see why: adding a path for current to take can only make it easier for current to flow.
What are the shortcuts worth knowing?
Three, and they cover most bench work without a calculator.
- Equal resistors in parallel: the value divided by the count. Ten 1 kΩ resistors in parallel are 100 Ω.
- Two resistors in parallel: R₁R₂ ÷ (R₁ + R₂), which is quicker by hand than the reciprocal sum.
- Series: straight addition, and the same current flows through every resistor.
The two-resistor shortcut is the one to memorise, because pairing is the standard workshop trick for hitting a value you do not have. A 2.2 kΩ with a 4.7 kΩ gives 1,498.55 Ω — closer to 1.5 kΩ than a 5 per cent 1.5 kΩ resistor is guaranteed to be.
How does the voltage divider follow?
From series behaviour: the same current flows through both resistors, so the voltage across each is proportional to its resistance. The output is Vin × R2 ÷ (R1 + R2), and two equal 10 kΩ resistors on 12 V give 6 V at 0.6 mA.
The critical condition is that the load must draw far less current than the divider itself — a factor of ten is the usual minimum. A divider is for sensing, not for supplying: scaling a battery voltage into an ADC input works because an ADC draws almost nothing, and the same divider powering an LED collapses immediately.
Choosing the values is a trade. High resistances waste less quiescent power and are more susceptible to noise and to the input impedance of whatever is measuring them; low resistances are stiff and burn current continuously.
Why does the current divider swap ends?
Because current takes the easier path, so the branch with less resistance carries more. The formula is I₁ = I × R2 ÷ (R1 + R2) — note that the branch current uses the opposite resistor in the numerator, which is the exact reverse of the voltage divider.
One amp through a 100 Ω and a 300 Ω branch gives 0.75 A in the 100 Ω branch and 0.25 A in the 300 Ω one. The two formulas look identical with the labels swapped, and that swap is where almost every mistake in this area happens.
The practical case is a shunt: a very low resistance in parallel with a meter, sized so a known small fraction of the current goes through the meter and the rest bypasses it.
Which resistor gets hot?
The opposite one in each arrangement, which is worth internalising before choosing wattages.
| Arrangement | Same across all | Largest dissipation |
|---|---|---|
| Series | Current | The largest resistance |
| Parallel | Voltage | The smallest resistance |
In series, P = I²R with a common current, so the biggest resistor dissipates the most. In parallel, P = V²/R with a common voltage, so the smallest resistor does. Mixing wattages in a parallel set means the small-value part needs the high-wattage rating, which is the opposite of the intuition that a bigger number needs a bigger part.
Do tolerances add up?
Better than intuition suggests, in a long chain. Random errors in many resistors partially cancel rather than accumulating linearly, so a string of ten 1 per cent resistors is considerably better than 10 per cent overall.
Pairing for accuracy works for the same reason. Two resistors from the same batch and the same value tend to err in the same direction, which is why a matched pair gives a better ratio than the individual tolerances imply — and it is the ratio, not the absolute value, that a divider depends on.
That is also the argument for a resistor ladder: a string of equal resistors tapped at each junction gives evenly spaced references, which is exactly how a flash ADC and a simple resistor DAC work.
Questions people ask
Can I power something from a divider? Only if it draws under about a tenth of the divider current. Otherwise the output sags, and the sag is not constant — it varies with the load, which makes it worse than a wrong fixed voltage.
How do I choose a dropping resistor? From Ohm’s law on the difference. A 5 V supply feeding something wanting 3.3 V at 20 mA needs to drop 1.7 V at 20 mA, which is 85 Ω. The catch is that the drop depends on the current, so it only holds at that current.
Can I use the divider formula for more than two branches? For the voltage divider, yes — any tap on a series string. For the current divider, work out the parallel combination of the other branches first and treat it as a two-branch problem.
Do parallel LEDs share current equally? No, and this is the classic misuse. LEDs are not resistors; small differences in forward voltage mean one takes most of the current. Each parallel LED needs its own resistor.
Two rules, two dividers, and one formula that swaps ends. The parallel resistor calculator and resistors in parallel handle the reciprocal sum, series resistor and resistors in series the addition, and the voltage divider and current divider the two that follow from them.