About this calculator
Sometimes the value you need does not exist as a single part, or you want to use up what is already in the drawer. Combining components gets you there: resistors add in series, capacitors add in parallel, and both do something less obvious the other way round.
This calculator takes as many components as you want to add, in either arrangement, and returns the equivalent value.
How it works
For resistors in series, the same current passes through each one and the voltages add, so the resistances simply add. Three 1 kΩ resistors in a row behave as 3 kΩ.
For resistors in parallel, each one offers the current another path, so the total resistance falls below the smallest of them. The reciprocals add: 1/Req = 1/R1 + 1/R2 + …. Two equal resistors in parallel give exactly half their value.
Capacitors behave the opposite way, and this trips people up constantly. In parallel, the plate area effectively grows, so capacitances add. In series the reciprocals add, and the total comes out smaller than the smallest capacitor.
The reason for the inversion is that capacitance is the inverse of the impedance relationship resistance has. In series a capacitor's impedance adds just like a resistor's — but impedance goes as 1/C, so adding impedances means adding reciprocal capacitances.
Worked example
You need 150 Ω but only have 100 Ω and 220 Ω resistors on hand.
- Try them in parallel: 1/Req = 1/100 + 1/220
- 1/Req = 0.01 + 0.004545 = 0.014545
- Req = 1 / 0.014545 = 68.75 Ω — too low
- Try two 100 Ω in series with nothing else: 200 Ω — too high
- Try 100 Ω in series with (220 ∥ 220 = 110 Ω): 210 Ω
- Try 220 ∥ 470 = 149.9 Ω — near enough
220 Ω in parallel with 470 Ω gives 149.9 Ω, within 0.1% of the 150 Ω you wanted — comfortably inside the tolerance of either resistor.
Practical notes
- Parallel resistance is always smaller than the smallest resistor in the group. If your answer is bigger, you have used the series formula by mistake.
- Series capacitance is always smaller than the smallest capacitor. Same sanity check, opposite component.
- Power splits unevenly in parallel. The smallest resistor carries the most current and therefore dissipates the most — check its rating individually rather than assuming the load shares out evenly.
- Putting resistors in series does not improve tolerance, but it does average it out somewhat: random errors partially cancel, so a chain of four 5% resistors is typically closer to nominal than one.
- Electrolytic capacitors in series need balancing resistors across each one. Leakage varies between parts, and without balancing one capacitor can end up with most of the voltage across it.
Frequently asked questions
How do I calculate resistors in parallel?
Add the reciprocals and invert the result: 1/Req = 1/R1 + 1/R2. For exactly two resistors, the shortcut (R1 × R2) / (R1 + R2) is quicker.
Why do capacitors work backwards from resistors?
Because capacitance is inversely related to impedance. Putting capacitors in parallel effectively increases the plate area, so the values add. In series their impedances add, which means the reciprocals of the capacitances add.
Can I make any value by combining resistors?
Very nearly. Two standard E24 values in series or parallel can land within about 1% of almost any target. Bear in mind that component tolerance usually swamps that last fraction of a percent anyway.
Do parallel resistors share power equally?
Only if they are equal values. Otherwise the lower-resistance path carries proportionally more current and dissipates more power. Always check the smallest resistor against its own rating.
Does combining resistors increase the power rating?
Yes. Four 100 Ω ¼ W resistors arranged as two series pairs in parallel still measure 100 Ω but can handle 1 W in total, since the dissipation is shared across four bodies.