Series and Parallel Calculator
List resistor or capacitor values separated by commas or new lines to get the series and parallel equivalents together.
Formulas
Resistors in series add: R = R1 + R2 + …. In parallel: (1)/(R) = (1)/(R1) + (1)/(R2) + …. Capacitors work the other way around: they add in parallel and combine as reciprocals in series.
Worked example: 10 kΩ, 4.7 kΩ and 2.2 kΩ
- Series: 10 kΩ + 4.7 kΩ + 2.2 kΩ = 16.9 kΩ
- Two in parallel: product over sum, R = (R1·R2)/(R1 + R2) = (10 kΩ × 4.7 kΩ)/(14.7 kΩ) = 3.197 kΩ
- All three in parallel: (1)/(R) = (1)/(10 kΩ) + (1)/(4.7 kΩ) + (1)/(2.2 kΩ) → 1.303 kΩ. A parallel combination is always below the smallest part.
Two common resistors in series and parallel
| Resistor 1 | Resistor 2 | Series | Parallel |
|---|---|---|---|
| 1 kΩ | 1 kΩ | 2 kΩ | 500 Ω |
| 10 kΩ | 10 kΩ | 20 kΩ | 5 kΩ |
| 1 kΩ | 2.2 kΩ | 3.2 kΩ | 687.5 Ω |
| 10 kΩ | 4.7 kΩ | 14.7 kΩ | 3.197 kΩ |
| 10 kΩ | 22 kΩ | 32 kΩ | 6.875 kΩ |
| 100 kΩ | 10 kΩ | 110 kΩ | 9.091 kΩ |
| 4.7 kΩ | 4.7 kΩ | 9.4 kΩ | 2.35 kΩ |
| 330 Ω | 470 Ω | 800 Ω | 193.9 Ω |
Two capacitors in series and parallel
| Capacitor 1 | Capacitor 2 | Series | Parallel |
|---|---|---|---|
| 100 nF | 100 nF | 50 nF | 200 nF |
| 100 nF | 47 nF | 31.97 nF | 147 nF |
| 1 µF | 1 µF | 500 nF | 2 µF |
| 10 µF | 100 nF | 99.01 nF | 10.1 µF |
Two equal parts in parallel give exactly half, in series exactly double. Capacitors work the other way round: they add in parallel and add as reciprocals in series.
Checked on our bench: the voltage a series loop lost
The numbers below are computed from values the site owner recorded in lab classes. Measured values are left exactly as recorded.
In a series circuit the part voltages should add up to the supply (Kirchhoff's voltage law). Measured on a 1 kΩ + 3 kΩ series circuit, the sum came up slightly short.
| Supply | Supply measured | VR1 + VR2 | Missing voltage | Missing ÷ current |
|---|---|---|---|---|
| Supply10 V | Supply measured10.023 V | VR1 + VR29.983 V | Missing voltage40 mV | Missing ÷ current16 Ω |
| Supply16 V | Supply measured16.015 V | VR1 + VR215.819 V | Missing voltage196 mV | Missing ÷ current50 Ω |
The missing voltage sat across the ammeter and leads that were also in series. The last column shows how many ohms that part of the loop behaved like. This calculator only knows the parts you type in, so when a meter sits in the loop, count it as one more resistor.
The same lab's measured resistors also show how part tolerance carries into a combination:
| Combination | From nominal values | From the measured parts | Difference |
|---|---|---|---|
| CombinationThree 1 kΩ in series | From nominal values3 kΩ | From the measured parts3.046 kΩ | Difference+1.5% |
| CombinationThree 10 kΩ in parallel | From nominal values3.333 kΩ | From the measured parts3.359 kΩ | Difference+0.8% |
In series and in parallel alike, the combined error ends up around the average of the individual errors. Errors can cancel, but parts from one batch often lean the same way (all three 1 kΩ parts here read high), so they shrink less than you might hope.
Combining parts in practice
Entering a network
- Choose resistors or capacitors, then list the values separated by commas or spaces, with prefixes if you like (10k, 4.7k, 2k2).
- Read both the series and the parallel equivalent; you usually need only one, but the other is a quick sanity check.
- If you are combining parts to hit an odd value, compare the result with the nearest single standard value first.
Series or parallel: the usual mix-ups
- Swapping the rules for capacitors. Capacitors add in parallel and shrink in series, the opposite of resistors.
- Expecting the parallel value to land between the parts. A parallel combination is always smaller than its smallest member. Two equal resistors in parallel give exactly half.
- Forgetting power sharing. In series the largest resistor dissipates the most power; in parallel the smallest one does. Check each part's rating, not just the total.
Tricks with series parts
Two resistors in series give a precise value cheaply: a 10 kΩ and a 2.2 kΩ make 12.2 kΩ with no special part. Series capacitors are used to raise the voltage rating, but leakage differences make the voltage split unevenly, so add equal balancing resistors across each one in high-voltage work. On a breadboard, measure the combination with the parts out of the rest of the circuit, since anything else connected in parallel changes the reading.
Worked example: a mixed series-parallel network
A 1 kΩ resistor feeds 2.2 kΩ and 3.3 kΩ in parallel, all from 5 V. Reduce the parallel pair first: (2.2 kΩ × 3.3 kΩ)/(2.2 kΩ + 3.3 kΩ) = 1.32 kΩ. Add the series part: 1 kΩ + 1.32 kΩ = 2.32 kΩ, so the supply delivers 5 V ÷ 2.32 kΩ = 2.155 mA.
Now work back out. The parallel pair has 2.155 mA × 1.32 kΩ = 2.845 V across it, so 1.293 mA flows in the 2.2 kΩ and 0.862 mA in the 3.3 kΩ. They add to 2.155 mA, which is the check that nothing went wrong. Enter the pair here to get 1.32 kΩ, then use the Ohm's law calculator for the currents.
Making an odd value from standard parts
- Series for a sum. 2.5 kΩ is not an E24 value, but 2.2 kΩ + 300 Ω is exactly 2.5 kΩ, and both are E24.
- Parallel to trim down. 10 kΩ in parallel with 3.3 kΩ gives 2.481 kΩ. Putting a large resistor across a smaller one is a fine adjustment: the result moves only a little below the smaller part.
- Capacitors add in parallel. 3.3 µF and 1 µF side by side make 4.3 µF.
Each part keeps its own tolerance, so two 5% resistors still give a combination that is good to about 5%, not better. Look up the nearest single value with the resistor color code page before building a combination you may not need.
Frequently asked questions
Is there a quick way to combine two resistors in parallel?
Use (R1 × R2)/(R1 + R2). Two equal resistors in parallel give half the value, so two 10 kΩ resistors make 5 kΩ.
Why put capacitors in series?
The total capacitance drops, but the combination can withstand a higher voltage. Capacitance tolerances can split the voltage unevenly, so at high voltage add equal balancing resistors across each capacitor.