Parallel Resistor Calculator
Equivalent resistance of resistors in parallel or series
Last updated September 6, 2026
Method: The standard network laws for ideal resistors: 1/Req = sum of 1/Ri for parallel branches and Req = sum of Ri for a series chain, combined with Ohm's law (I = V ÷ R) and the power equation (P = V² ÷ R) to split current and heat across the network.
Included: Up to 10 resistors, parallel or series, three unit scales, the equivalent resistance, the same parts wired the other way, per-resistor current or voltage drop, per-resistor power, the current or voltage share, the nearest E12 stock value, a two-resistor pairing table and a missing-resistor solver for a target value.
Not included: Reactance and frequency effects (capacitors, inductors, AC impedance), resistor tolerance stack-up, temperature coefficients, wiring and contact resistance, and delta-wye network conversions.
🔗 Equivalent resistance of 3 resistors in parallel
🔎 Each resistor in the network
| Resistor | Value | Current share | Current | Power |
|---|---|---|---|---|
| R1 | 100 Ω | 56.90% | 120 mA | 1.44 W |
| R2 | 220 Ω | 25.86% | 54.545 mA | 654.545 mW |
| R3 | 330 Ω | 17.24% | 36.364 mA | 436.364 mW |
| Equivalent | 56.897 Ω | 100.00% | 210.909 mA | 2.531 W |
In parallel every resistor sees the full supply voltage, so the smallest resistor carries the most current and burns the most power.
📊 R1 = 100 Ω combined with one stock resistor
| Second resistor | In parallel | In series |
|---|---|---|
| 10 Ω | 9.091 Ω | 110 Ω |
| 22 Ω | 18.033 Ω | 122 Ω |
| 47 Ω | 31.973 Ω | 147 Ω |
| 100 Ω | 50 Ω | 200 Ω |
| 220 Ω | 68.75 Ω | 320 Ω |
| 470 Ω | 82.456 Ω | 570 Ω |
| 1 kΩ | 90.909 Ω | 1.1 kΩ |
| 2.2 kΩ | 95.652 Ω | 2.3 kΩ |
| 4.7 kΩ | 97.917 Ω | 4.8 kΩ |
| 10 kΩ | 99.01 Ω | 10.1 kΩ |
Two resistors in parallel always land below the smaller of the two; in series they always land above the larger.
Ideal resistors assumed. Real parts carry a tolerance (typically 5% or 1%), drift with temperature, and sit on top of lead, trace and contact resistance. Size each part for at least twice the power shown.
Parallel resistor calculator: everything you need to know
A parallel resistor calculator finds the single equivalent resistance that a group of resistors presents to the rest of the circuit. Parallel branches add reciprocals, so the total always falls below the smallest part: 100 Ω, 220 Ω and 330 Ω wired in parallel measure 56.897 Ω, while the same three parts wired in series measure 650 Ω.
This page covers both wiring styles because the arithmetic is the mirror image of the same idea, and because most real boards mix them. Enter up to ten values, pick parallel or series, add a supply voltage if you want the current and heat in each branch, and switch on the missing-resistor option when you know the total you want and need the one part that gets you there.
How resistors in parallel combine
When resistors sit side by side between the same two nodes, every one of them carries its own current and all of those currents come back together. More paths means less opposition, so the network conducts better than any single branch. Expressed as a formula:
1 ÷ Req = 1 ÷ R1 + 1 ÷ R2 + ... + 1 ÷ Rn Add the reciprocals, then take the reciprocal of the answer. For exactly two resistors that reduces to the familiar product-over-sum shortcut:
Req = (R1 × R2) ÷ (R1 + R2) Series is the easy direction. Resistors in a single chain all carry the same current and their voltage drops stack, so the values simply add:
Req = R1 + R2 + ... + Rn The reciprocal of a resistance is its conductance, measured in siemens. Seen that way the parallel rule stops looking strange: parallel branches add conductance the same way series resistors add resistance. That is exactly what this calculator sums internally before inverting the result.
Worked example: 100 Ω, 220 Ω and 330 Ω on a 12 V supply
Take the three resistors the calculator loads by default. First the reciprocals:
- 1 ÷ 100 = 0.0100000 S
- 1 ÷ 220 = 0.0045455 S
- 1 ÷ 330 = 0.0030303 S
Those add to 0.0175758 S, and the reciprocal of that sum is 56.897 Ω. Notice the answer sits below 100 Ω, the smallest part, exactly as the rule promises. Wired in series instead, the same three parts total 100 + 220 + 330 = 650 Ω, comfortably above the largest.
Now put 12 V across the parallel group. Each branch sees the full 12 V, so Ohm's law gives each current directly, and P = V² ÷ R gives each dissipation:
| Branch | Current at 12 V | Share of current | Power |
|---|---|---|---|
| 100 Ω | 120.000 mA | 56.90% | 1.440 W |
| 220 Ω | 54.545 mA | 25.86% | 0.655 W |
| 330 Ω | 36.364 mA | 17.24% | 0.436 W |
| Equivalent 56.897 Ω | 210.909 mA | 100.00% | 2.531 W |
Two things stand out. The branch currents add up to the total, which is a useful check on any parallel calculation. And the 100 Ω resistor takes more than half the current and burns 1.440 W on its own, so a quarter-watt part in that position would fail within seconds. In a parallel bank, the smallest resistor is always the one to size first.
The same parts in series, for comparison
Rewire those three resistors end to end on the same 12 V supply and the picture flips. The total is 650 Ω, so the chain draws only 12 ÷ 650 = 18.462 mA, and that one current runs through all three. The voltage now splits in proportion to resistance: 1.846 V across the 100 Ω, 4.062 V across the 220 Ω and 6.092 V across the 330 Ω, adding back to 12 V. Dissipation follows P = I² × R, giving 34.083 mW, 74.982 mW and 112.473 mW for a total of just 221.538 mW. The same three components draw eleven times less power in series than in parallel, and now the largest resistor is the hot one instead of the smallest.
Two-resistor parallel combinations
Most bench work is two parts at a time, so this grid is worth keeping close. Each cell is the product-over-sum result for the row value in parallel with the column value, in ohms. Multiply every number by 1,000 to read it as kilohms.
| Parallel with | 100 Ω | 220 Ω | 330 Ω | 470 Ω | 1,000 Ω |
|---|---|---|---|---|---|
| 100 Ω | 50.00 | 68.75 | 76.74 | 82.46 | 90.91 |
| 220 Ω | 68.75 | 110.00 | 132.00 | 149.86 | 180.33 |
| 330 Ω | 76.74 | 132.00 | 165.00 | 193.88 | 248.12 |
| 470 Ω | 82.46 | 149.86 | 193.88 | 235.00 | 319.73 |
| 1,000 Ω | 90.91 | 180.33 | 248.12 | 319.73 | 500.00 |
The diagonal is the equal-pair case, always exactly half the value. Read across any row and you can see how quickly a large partner stops mattering: a 100 Ω resistor drops to 50.00 Ω next to another 100 Ω, but only to 90.91 Ω next to a 1,000 Ω.
Identical resistors: parallel versus series
Stacking copies of one value is the most common trick in the book, whether you are hunting a value you do not stock or spreading heat across packages. Using 1,000 Ω parts:
| Count | Parallel (R ÷ n) | Series (R × n) | Combined rating, 1/4 W parts |
|---|---|---|---|
| 2 | 500.00 Ω | 2,000 Ω | 0.50 W |
| 3 | 333.33 Ω | 3,000 Ω | 0.75 W |
| 4 | 250.00 Ω | 4,000 Ω | 1.00 W |
| 5 | 200.00 Ω | 5,000 Ω | 1.25 W |
| 8 | 125.00 Ω | 8,000 Ω | 2.00 W |
| 10 | 100.00 Ω | 10,000 Ω | 2.50 W |
The power column is the practical payoff: identical resistors share the load evenly, so four 1/4 W parts in parallel handle a full watt at a quarter of the original resistance, and four in series handle the same watt at four times the resistance. This is how a bench dummy load or a high-wattage bleeder is usually built.
Solving for a missing resistor
The most useful mode on this page runs the formula backwards. You know what total you want and what you already have soldered in, and you need the one part that closes the gap. In parallel:
1 ÷ Rx = 1 ÷ Rtarget − (1 ÷ R1 + 1 ÷ R2 + ...) The table below starts from a single 100 Ω resistor and lists the second resistor needed to pull the pair down to each target. Every value comes straight from that rearrangement.
| Target with 100 Ω | Second resistor Rx | Nearest E12 stock part |
|---|---|---|
| 90 Ω | 900.000 Ω | 820 Ω |
| 80 Ω | 400.000 Ω | 390 Ω |
| 75 Ω | 300.000 Ω | 270 Ω |
| 60 Ω | 150.000 Ω | 150 Ω |
| 50 Ω | 100.000 Ω | 100 Ω |
| 40 Ω | 66.667 Ω | 68 Ω |
| 25 Ω | 33.333 Ω | 33 Ω |
| 20 Ω | 25.000 Ω | 27 Ω |
Two of those targets land exactly on a stock part, which is a nice reminder that a 150 Ω across a 100 Ω gives a clean 60 Ω. In series the same idea is plain subtraction: Rx = Rtarget − sum(Ri), so a 650 Ω chain that already holds 100 Ω and 220 Ω needs a 330 Ω to finish. If the calculator tells you the target is not reachable, the direction is wrong: a parallel target must sit below the equivalent resistance of the resistors already listed, and a series target must sit above their sum.
How to use this calculator
- Pick the wiring: parallel for branches between the same two nodes, series for a single chain.
- Choose the unit: ohms, kilohms or megohms. All values on the page use the same scale, so convert before you mix.
- Type the values: start with the three defaults, edit them, add rows up to ten, or remove any row you do not need. Blank and zero rows are simply ignored.
- Add a supply voltage if you want current and power. Each branch then shows its own current in parallel, or its own voltage drop in series.
- Switch on missing-resistor solving when you know the total you want, and type that target. The calculator returns the exact Rx and the nearest E12 stock part.
- Read the pairing table at the bottom to see what one more standard resistor would do to your first value in either wiring.
Everything recalculates as you type, so it is quick to sweep a value and watch where the equivalent resistance lands.
Who this calculator is for
- Hobbyists and makers building a value that is not in the parts drawer out of two that are.
- Students checking homework on network reduction, current dividers and power sharing.
- Electronics technicians reading an unfamiliar board and needing to know what a resistor bank presents to the supply.
- Anyone building a load bank or a bleeder resistor where the wattage matters more than the exact value.
- Automotive and LED tinkerers adding a load resistor across an existing circuit and needing to know the new total and its heat.
Key terms
- Equivalent resistance (Req): the single resistance that would behave identically to the whole network, seen from its two terminals.
- Conductance (G): the reciprocal of resistance, in siemens. Parallel branches add conductance.
- Current divider: the parallel counterpart of the voltage divider. Branch current is proportional to conductance, so the smallest resistor takes the largest share.
- E12 series: the standard 12-value-per-decade resistor ladder (10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82) behind most 5% and 10% stock.
- Tolerance: how far a real part may sit from its printed value, commonly 5% or 1%.
- Power rating: the heat a resistor can shed continuously, typically 1/8 W, 1/4 W, 1/2 W or 1 W in through-hole parts.
What changes the result the most
- The smallest resistor in a parallel group dominates everything. It sets most of the equivalent value, carries most of the current and takes most of the heat.
- How many branches you add. Each new parallel branch lowers the total, but with rapidly diminishing returns once its value is far above the smallest.
- The wiring choice itself. Our three example parts read 56.897 Ω in parallel and 650 Ω in series, a factor of more than eleven from the same components.
- The supply voltage, which does not touch the resistance at all but scales current linearly and power with the square. Doubling 12 V to 24 V quadruples the 2.531 W total to 10.124 W.
- Unit slips. Mixing a kilohm value into an ohm field is off by a thousand and is the single most common wrong answer.
Tips from the bench
- Sanity-check the direction first. A parallel answer below the smallest part and a series answer above the largest are free checks that catch most arithmetic slips.
- Use equal pairs where you can. Two identical parts halve the resistance, double the power handling and keep the tolerance behavior simple.
- Reduce big networks in stages. Collapse each parallel cluster, then treat the results as a series chain, and run this calculator once per stage.
- Derate for heat. Pick parts rated for at least twice the calculated dissipation, and more if they sit in a sealed enclosure or against other hot components.
- Measure after you build. With the power off, an ohmmeter across the network should land within the combined tolerance of the calculated value. A wide miss usually means a solder bridge or an open joint.
Limitations and assumptions
- Ideal resistors only. Tolerance, temperature coefficient and aging are not modeled; two 5% 100 Ω parts nominally give 50.00 Ω but can legitimately measure anywhere from 47.50 Ω to 52.50 Ω.
- DC or purely resistive AC. Capacitance and inductance make the combination frequency dependent, which needs impedance math rather than plain resistance.
- No wiring resistance. Leads, traces, connectors and solder joints add their own milliohms, which matter once the network itself is only a few ohms.
- Simple topologies. Purely parallel and purely series groups are covered; bridge, delta and wye networks need a conversion step before you can reduce them.
- Steady state. Power figures assume continuous operation, not pulses, where a resistor can briefly absorb far more energy than its rating.
Sources
- National Institute of Standards and Technology (NIST) - Special Publication 811, Guide for the Use of the International System of Units (SI): the ohm, the siemens, the volt, the ampere and the watt, and the prefixes used on this page.
- International Electrotechnical Commission - IEC 60063, preferred number series for resistors and capacitors: the source of the E6, E12 and E24 ladders, including the twelve E12 values listed above.
- The parallel rule (1/Req = sum of 1/Ri), the series rule (Req = sum of Ri), Ohm's law (V = I x R) and the power equation (P = V² / R) are exact algebraic identities for ideal resistors, so every figure on this page is reproducible by hand and needs no external data table.
How it compares to related calculators
This page answers "what does this resistor network measure, and what does each part carry?" Sister tools take neighboring questions. Use the Ohm's Law Calculator when you have a single element and need to move between voltage, current, resistance and power. Use the Voltage Divider Calculator when two resistors in series are there to produce a specific output voltage rather than a specific resistance. Use the Resistor Color Code Calculator when you need to read the bands on a part before you can enter it here. For the drop along a run of wire rather than through components, the Voltage Drop Calculator is the right tool, and the Watts to Amps Calculator converts a power figure into the current a supply has to deliver. If the heat you calculate here runs continuously, the Electricity Cost Calculator turns those watts into dollars per month.
⚠️ Common mistakes & edge cases
Forgetting the final reciprocal
Adding 1/100 + 1/220 + 1/330 gives 0.0175758, and it is tempting to stop there. That number is conductance in siemens, not resistance. Invert it to get the 56.897 Ω answer.
Using product-over-sum on three resistors
R1 × R2 ÷ (R1 + R2) is a two-resistor shortcut only. For three or more, use the full reciprocal sum, or apply the shortcut in pairs: combine two, then combine that result with the next.
Mixing ohms and kilohms
Typing 4.7 when you mean 4.7 kΩ into an ohms field is off by a factor of a thousand. Set the unit switch first, then convert every value to that scale before you enter it.
Sizing the wrong resistor for power
In parallel the smallest resistor runs hottest, not the largest. At 12 V the 100 Ω part burns 1.440 W while the 330 Ω part burns only 0.436 W, so a quarter-watt part in the low position would cook.
Expecting parallel to raise resistance
Adding a branch can never increase the total. If your calculated parallel value came out above the smallest resistor, an entry is wrong or the network is really in series.
Assuming tolerance averages out
Combining parts does not cancel their error. Two 5% resistors in parallel still carry a 5% window, so 50.00 Ω nominal can measure 47.50 Ω or 52.50 Ω. Use 1% parts when the value has to hold.
❓ Frequently asked questions
What is the formula for resistors in parallel?
The reciprocal of the equivalent resistance equals the sum of the reciprocals of every branch: 1/Req = 1/R1 + 1/R2 + ... + 1/Rn. Take the reciprocal of that sum to get Req. With 100 Ω, 220 Ω and 330 Ω the reciprocals are 0.010000, 0.0045455 and 0.0030303 siemens, summing to 0.0175758, so Req = 1 ÷ 0.0175758 = 56.897 Ω.
How do I calculate two resistors in parallel?
For exactly two resistors the reciprocal formula simplifies to the product-over-sum shortcut: Req = R1 × R2 ÷ (R1 + R2). A 100 Ω and a 220 Ω resistor give 100 × 220 ÷ 320 = 68.75 Ω. The shortcut only works for two resistors at a time, but you can apply it repeatedly: combine two, then combine the answer with the third.
Is the parallel resistance always smaller than the smallest resistor?
Yes, always. Every extra branch gives the current another path, so the total resistance can only fall. Adding a 1,000 Ω resistor across an existing 100 Ω resistor drops the pair to 90.91 Ω, and adding a 10,000 Ω resistor across the same 100 Ω still lowers it, to 99.01 Ω. That one-way rule is a fast sanity check on any parallel answer.
What is the resistance of two equal resistors in parallel?
Half of one of them. Two 1,000 Ω resistors in parallel give 500 Ω, two 470 Ω resistors give 235 Ω, and two 4.7 kΩ resistors give 2.35 kΩ. More generally, n identical resistors of value R in parallel come to R ÷ n, so four 1,000 Ω resistors give 250 Ω and ten give 100 Ω.
How is series resistance different from parallel resistance?
In series the resistors sit end to end on one path, so the values simply add: Req = R1 + R2 + ... The same 100 Ω, 220 Ω and 330 Ω resistors that give 56.897 Ω in parallel give 650 Ω in series. Series raises the total above the largest part; parallel lowers it below the smallest.
How do I find a missing resistor to hit a target resistance?
Rearrange the formula. In parallel, 1/Rx = 1/Rtarget − sum of the reciprocals you already have, then invert. To pull a 100 Ω resistor down to 75 Ω you need 1/75 − 1/100 = 0.0033333, so Rx = 300 Ω. In series it is simple subtraction: Rx = Rtarget − sum(Ri). Switch on the missing-resistor option in the calculator and it does both.
How much current flows through each resistor in parallel?
Every parallel branch sees the full supply voltage, so each branch current is V ÷ R by Ohm's law. On a 12 V supply, the 100 Ω branch carries 120.000 mA, the 220 Ω branch 54.545 mA and the 330 Ω branch 36.364 mA, for a total of 210.909 mA. The smallest resistor always carries the largest share, here 56.90% of the total.
How much power does each parallel resistor dissipate?
Because the voltage is shared, use P = V² ÷ R for each branch. At 12 V, a 100 Ω resistor dissipates 1.440 W, a 220 Ω resistor 0.655 W and a 330 Ω resistor 0.436 W, totaling 2.531 W. The lowest-value resistor runs hottest, which is why it is the one that usually fails first in a parallel bank.
Why would I put resistors in parallel on purpose?
Three common reasons: to reach a value that is not a stock part, to spread heat across several packages, and to lower the resistance of an existing circuit. Four quarter-watt resistors in parallel share the load and handle a full watt together, and two stock parts often land closer to an odd target than any single E12 value does.
Does the order of the resistors matter?
No. Both the parallel sum and the series sum are commutative, so 100 Ω, 220 Ω and 330 Ω give the same 56.897 Ω in parallel and the same 650 Ω in series no matter which one you list first. Physical placement matters only for heat spreading and wiring practicality, not for the arithmetic.
How do I calculate a series-parallel network?
Work from the inside out. Collapse each purely parallel group into a single equivalent value, collapse each purely series chain into a single value, and repeat until one number remains. Run the calculator once per group, note each equivalent value, then feed those numbers back in with the mode switched.
How does resistor tolerance affect the combined value?
The tolerance carries through roughly unchanged. Two nominally 100 Ω resistors with 5% tolerance combine to 50.00 Ω nominal, but the real value can land anywhere from 47.50 Ω (both parts 5% low) to 52.50 Ω (both 5% high), which is still 5% of the nominal. Combining parts does not average the error away unless the deviations happen to run in opposite directions.
Can I put resistors of different values in parallel?
Yes, and the calculator handles up to ten mixed values. Be aware that the smallest resistor dominates: it sets most of the equivalent value, carries most of the current and takes most of the heat. A 100 Ω resistor paralleled with a 10,000 Ω resistor still measures 99.01 Ω, so the large one contributes almost nothing.
What is conductance and why does the parallel formula use it?
Conductance G is the reciprocal of resistance, measured in siemens (S). Parallel branches add in conductance because they add paths for current, so Gtotal = G1 + G2 + ... and Req = 1 ÷ Gtotal. Thinking in conductance makes the parallel formula feel natural instead of arbitrary, and it is exactly what this calculator sums internally.
Is this parallel resistor calculator free?
Yes. It runs entirely in your browser with no sign-up, no fee and no limit on how many networks you solve. Nothing you type is sent anywhere, and it works the same on a phone at the bench as it does on a desktop.
💡 Good to know
Parallel is a current divider
A voltage divider splits voltage between series resistors; a parallel group splits current between branches. The share is proportional to conductance, so the 100 Ω branch in our example takes 56.90% of the current while the 330 Ω branch takes only 17.24%.
Parallel resistors multiply the power rating
Four 1/4 W resistors in parallel share the heat evenly and handle a full watt together, at a quarter of the single-part resistance. Series stacking gives the same wattage boost while multiplying the resistance instead.
Odd values are usually cheaper as two parts
A 75 Ω value is not in the E12 ladder, but a 100 Ω across a 300 Ω lands on it exactly, and a stock 330 Ω gets within a couple of percent. Combining two common parts is almost always faster than ordering one precise one.