Series and Parallel Resistance Calculator
Combine resistors in series or parallel, with the current and power dissipated in each one when a voltage is applied.
How to use this calculator
- 1Choose series or parallel, then enter the resistances separated by commas.
- 2Add an applied voltage to see the current and power in each resistor.
- 3Check the hottest resistor against its power rating — that is what actually fails.
How the calculation works
Series: R = R₁ + R₂ + … + Rₙ Parallel: 1 ÷ R = 1÷R₁ + 1÷R₂ + … + 1÷Rₙ- R
- Equivalent resistance of the whole network, in ohms
- R₁ … Rₙ
- The individual resistances being combined
- series
- Components end to end, so the same current passes through each in turn
- parallel
- Components side by side across the same two nodes, so each sees the same voltage
Series resistance always exceeds the largest single resistor, because the current must pass through every one in turn. Parallel resistance is always below the smallest, because each added path gives the current somewhere else to go.
The parallel formula adds conductances — the reciprocals of resistance — which is why it looks awkward. Conductance is the quantity that genuinely adds when you add paths.
For exactly two resistors in parallel the shortcut R = R₁R₂ ÷ (R₁ + R₂) is equivalent and easier by hand. It does not generalise to three or more without repeated application.
For n identical resistors in parallel, R = Rₙ ÷ n. Two equal resistors in parallel give half the value, which is the most common use of the rule in practice.
Worked example
100 Ω, 220 Ω and 470 Ω in parallel at 12 V
- 1.Add the reciprocals: 1÷100 + 1÷220 + 1÷470 = 0.01 + 0.004545 + 0.002128 = 0.016673.
- 2.Invert: R = 1 ÷ 0.016673 = 59.976 Ω.
- 3.The result is smaller than the smallest resistor, 100 Ω, as parallel combinations always are.
- 4.Total current at 12 V: 12 ÷ 59.976 = 200.1 mA.
- 5.The 100 Ω resistor takes 12 ÷ 100 = 120 mA and dissipates 1.44 W — well over a quarter-watt part's rating.
Result: 59.976 Ω, 200.1 mA
The same three in series
- 1.Series resistances simply add: 100 + 220 + 470 = 790 Ω.
- 2.The result exceeds the largest resistor, 470 Ω, as series combinations always do.
- 3.Current is the same everywhere: 12 ÷ 790 = 15.19 mA.
- 4.Voltage divides in proportion: the 470 Ω resistor drops 15.19 mA × 470 = 7.14 V.
- 5.Total power is only 182 mW, against 2.4 W for the same resistors in parallel.
Result: 790 Ω, 15.19 mA
Why parallel resistance is always smaller
The parallel result surprises people the first time: put a 100 Ω and a 220 Ω resistor side by side and you get 68.75 Ω, less than either. The intuition is that you have not added resistance, you have added a route.
Think of it as conductance rather than resistance. Conductance is the reciprocal of resistance and measures how readily current flows. Adding a parallel path adds conductance, and conductances simply add. Converting back at the end is what produces the awkward reciprocal formula.
Series is the reverse. The current has no choice but to pass through every resistor in turn, so the obstacles accumulate and the total always exceeds the largest single one.
Which resistor gets hot
Power dissipation is not shared evenly, and which resistor runs hottest flips between the two arrangements.
In series, the current is common, so P = I²R makes power proportional to resistance — the largest resistor gets hottest. In parallel, the voltage is common, so P = V²/R makes power inversely proportional to resistance, and the smallest resistor gets hottest.
This matters because power rating is what actually destroys resistors. A standard through-hole resistor is rated at a quarter of a watt. In the worked example above, three resistors in parallel at 12 V put 1.44 W through the 100 Ω part — nearly six times its rating, and it would fail quickly. The same three in series dissipate 182 mW between them and are entirely safe.
The shortcuts worth remembering
Two resistors in parallel: R = R₁R₂ ÷ (R₁ + R₂), often called "product over sum". It is exact for two and only two, though you can apply it repeatedly for more.
n identical resistors in parallel: R = R ÷ n. Two equal resistors halve, four quarter. This is the everyday case, used to get a value you do not have or to share power across several parts.
A very large resistor in parallel barely matters, and a very small one in series barely matters. If one value is more than about twenty times the others, you can usually ignore it and be within a few percent — useful for sanity-checking a result before trusting the arithmetic.
What this assumes, and where it stops
Assumptions
- Ideal resistors with no tolerance, temperature coefficient or parasitic inductance and capacitance.
- A purely resistive network at DC or low frequency.
- An ideal voltage source with no internal resistance.
- All resistors are strictly greater than zero.
Limitations
- Handles a single series chain or a single parallel bank. Mixed series-parallel networks must be reduced in stages, working from the innermost combination outward.
- Real resistors carry a tolerance, typically 1% or 5%, so a computed equivalent is more precise than any network you actually build.
- Ignores temperature coefficient. Resistance drifts as parts heat, which matters most for the resistor dissipating the most power.
- A DC model. At high frequencies, lead inductance and stray capacitance stop the network behaving as pure resistance.
- Does not check power ratings automatically — it reports dissipation, but choosing an adequately rated part is up to you.
Common questions
How do you calculate resistors in parallel?
Add the reciprocals and invert: 1/R = 1/R₁ + 1/R₂ + … For exactly two, the shortcut R = R₁R₂/(R₁+R₂) is easier. For n identical resistors, R = R/n. The answer is always smaller than the smallest resistor in the group.
Why is parallel resistance less than the smallest resistor?
Because each parallel branch is an additional route for current, not an additional obstacle. What adds is conductance — the reciprocal of resistance — so total conductance rises and total resistance falls. Adding any path, however high its resistance, can only make the combination easier for current to cross.
Which resistor dissipates the most power?
In series, the largest, because current is common and P = I²R. In parallel, the smallest, because voltage is common and P = V²/R. This catches people out, since the intuition that bigger resistance means more heat is only true for series.
How do I work out a mixed series-parallel circuit?
Reduce it in stages. Find the innermost group that is purely series or purely parallel, replace it with its equivalent single resistance, and repeat until one value remains. This calculator handles one stage at a time, so feed each result back in as an input to the next.
Sources
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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