Calcylator
Resistor series network

Series resistors:
one path, one current, simple addition

Add series resistance in one step, then use the single shared current to find each resistor's voltage and heat.

Calcylator Editorial Team

Updated · 5 min read

One path, so one current

Resistors connected end to end, with no branches between them, form a series network. All the charge that leaves the supply has only a single route to follow, so the same current passes through each resistor in turn. What differs from one resistor to the next is the voltage across it.

That one fact explains everything useful about series circuits. The supply voltage is shared among the components, each taking a slice proportional to its resistance, and the total opposition to current is just the sum of the parts.

An analogy with a single hose helps. The same volume of water per second passes every point along the hose, but the pressure falls steadily from one end to the other. Each narrow section takes a bigger share of the pressure, just as each larger resistor takes a bigger share of the voltage.

It also means that a measurement of current taken anywhere in the chain tells you the current everywhere in it, and a single meter reading is enough to check the whole circuit.

The series rule

Total series resistance =R(total) = R1 + R2 + R3 + …
R1, R2, R3:
Individual resistances in ohms
R(total):
Equivalent single resistance
Always at least as large as the largest single resistor.

Adding another resistor in series can only increase total resistance, and so reduce current for a fixed supply. Contrast this with parallel arrangements, where extra branches reduce the total.

Mixed units must be unified before adding. A 1 kΩ and a 220 Ω resistor give 1,000 + 220 = 1,220 Ω, or 1.22 kΩ.

The rule comes straight from Ohm's law. The voltages across each resistor add to the supply voltage, and the current is common, so V = I × R1 + I × R2 + I × R3 = I × (R1 + R2 + R3). The bracket is the equivalent resistance.

A handy check: the total must be larger than the biggest resistor in the chain. If your answer is smaller, you have probably applied the parallel rule by mistake.

Worked example: 100 Ω, 220 Ω and 330 Ω on 12 V

  • R1

    100 Ω

  • R2

    220 Ω

  • R3

    330 Ω

  • Supply

    12 V

Total resistance

650 Ω (current 18.5 mA)

R(total) = 100 + 220 + 330 = 650 Ω. Current I = 12 ÷ 650 = 0.01846 A ≈ 18.5 mA.

With the common current known, find each voltage drop from V = I × R. Across 100 Ω: about 1.85 V. Across 220 Ω: about 4.06 V. Across 330 Ω: about 6.09 V. They add to 12.0 V, which confirms the sums.

Heat follows from P = I² × R. The 330 Ω part dissipates about 0.112 W, the 220 Ω about 0.075 W and the 100 Ω about 0.034 W; the total of 0.22 W equals 12² ÷ 650.

Suppose the target is exactly 20 mA instead. The resistance needed is 12 ÷ 0.02 = 600 Ω, so you would remove 50 Ω from the chain, for example by replacing the 330 Ω part with a 270 Ω one. Series arithmetic makes this kind of adjustment easy because every change is simply added or subtracted.

How the supply is shared

ResistorValueShare of totalVoltage on 12 VPower
R1100 Ω15.4%1.85 V0.034 W
R2220 Ω33.8%4.06 V0.075 W
R3330 Ω50.8%6.09 V0.112 W
Total650 Ω100%12.00 V0.221 W

The percentage column is the voltage divider rule in disguise: each resistor takes its fraction of the supply equal to its value divided by the total. That is how potentiometers and sensor circuits produce a chosen fraction of a voltage.

Tolerance and power ratings

Real resistors are not exact. A 5% tolerance on each part means 650 Ω nominal could land anywhere between 617.5 Ω and 682.5 Ω in the worst case. For current-setting jobs, calculate the extreme values and check that the circuit still works.

  • Check the power rating of each resistor against I² × R, and add margin. Half-watt and quarter-watt parts are common, but a design running near the limit gets hot.
  • Use a higher-tolerance part where the exact value matters, such as 1% metal film.
  • Combine standard values to reach a target. A 330 Ω and 100 Ω pair gives 430 Ω, and adding a third part opens up many more reachable values.
  • Remember that resistance drifts with temperature, though usually only slightly for these types.

Where one resistor of a larger power rating is unavailable, a series string spreads the heat across several parts, each carrying the same current but dissipating its own share.

Temperature rise matters when several resistors sit close together on a board, because their heat adds up in the same space. Spacing them or choosing a larger body improves the margin and keeps their values closer to nominal.

Where a series chain shows up

  • Current limiting for LEDs and sensors, where a resistor sets the current from a fixed supply.
  • Voltage dividers for scaling a high voltage down to a range a microcontroller or meter can measure.
  • String of heaters or lamps, where one open element breaks the entire loop.
  • Extending a resistor's value or power rating with parts you already have.

A single failed component in a series loop stops the current completely, which is both the weakness and, in some safety devices, the point. A series resistance tool will add a long list of values for you and is helpful when the string is long or when units are mixed.

When designing, pick the part with the highest power dissipation first and check its rating, since it is the most likely to overheat. Then confirm that the sum of the voltages matches the supply so that you have not mis-keyed a value.

Common questions

What is the formula for resistors in series?

Add the values directly: R(total) = R1 + R2 + R3 and so on. With 100 Ω, 220 Ω and 330 Ω the total is 650 Ω, and the total is always larger than any single resistor.

What current flows through resistors in series?

The same current passes through each one. Divide the supply voltage by the total resistance: 12 V ÷ 650 Ω is about 18.5 mA, and that value applies to all three resistors simultaneously.

How do I find the voltage across each series resistor?

Multiply the common current by the resistor's value. At 18.46 mA, 100 Ω drops 1.85 V, 220 Ω drops 4.06 V and 330 Ω drops 6.09 V. The three drops add up to the supply voltage.

Does adding a resistor in series increase or decrease current?

It decreases current for a fixed voltage, because total resistance rises. Adding 100 Ω to the 650 Ω chain makes it 750 Ω, and 12 V then drives only 16 mA instead of 18.5 mA.

What happens if one resistor in series fails open?

The whole circuit stops, because there is only one path. This is why series strings are unreliable for loads that must stay on, and why a single open part can be found by checking voltage across each in turn.

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