Free Parallel Resistance Calculator

Compute the combined resistance of up to four resistors connected in parallel. Enter any one to four values in ohms (Ω) and get the total parallel resistance instantly, plus the series sum for comparison.

Enter one to four resistor values in ohms (Ω). Blank or zero fields are ignored.

Result
Total parallel resistance (Rt)50 Ω
Series sum (ΣRi)200 Ω
Resistors used2

Quick answer

For resistors in parallel, the reciprocal of the total resistance equals the sum of the reciprocals of each resistor: 1/Rt = 1/R1 + 1/R2 + ... + 1/Rn. For two equal 100 Ω resistors this gives Rt = 50 Ω. The total parallel resistance is always smaller than the smallest individual resistor.

Formula & method

1/Rt = 1/R1 + 1/R2 + 1/R3 + 1/R4
  • Rt Total equivalent resistance of the parallel network (Ω)
  • R1…R4 Resistance of each individual resistor (Ω)

The conductances (1/R) of resistors in parallel add together. Invert the sum to get the total resistance Rt. Blank or zero entries are skipped.

Rt = (R1 · R2) / (R1 + R2)
  • R1, R2 The two parallel resistor values (Ω)

Convenient product-over-sum shortcut for exactly two resistors in parallel.

Examples

Example 1: Two equal 100 Ω resistors
Input
R1 = 100 Ω, R2 = 100 Ω
Result
Rt = 50 Ω (series sum = 200 Ω)
Why
1/Rt = 1/100 + 1/100 = 0.02, so Rt = 1 / 0.02 = 50 Ω. Two identical resistors in parallel always halve the resistance. The series sum (100 + 100) is 200 Ω for reference.
Example 2: Three resistors: 100, 200, 300 Ω
Input
R1 = 100 Ω, R2 = 200 Ω, R3 = 300 Ω
Result
Rt ≈ 54.5455 Ω (series sum = 600 Ω)
Why
1/Rt = 1/100 + 1/200 + 1/300 = (6 + 3 + 2)/600 = 11/600 ≈ 0.018333. Rt = 600/11 ≈ 54.5455 Ω, which is below the smallest resistor (100 Ω), as expected for parallel.
Example 3: Four equal 10 Ω resistors
Input
R1 = R2 = R3 = R4 = 10 Ω
Result
Rt = 2.5 Ω (series sum = 40 Ω)
Why
1/Rt = 1/10 + 1/10 + 1/10 + 1/10 = 0.4, so Rt = 1/0.4 = 2.5 Ω. For n identical resistors of value R in parallel, Rt = R/n = 10/4 = 2.5 Ω.
Example 4: Two unequal resistors: 60 Ω and 40 Ω
Input
R1 = 60 Ω, R2 = 40 Ω
Result
Rt = 24 Ω (series sum = 100 Ω)
Why
Using the product-over-sum shortcut: Rt = (60 · 40)/(60 + 40) = 2400/100 = 24 Ω. The same result comes from 1/60 + 1/40 = 5/120, so Rt = 120/5 = 24 Ω.

When to use this tool

  • Designing or analyzing circuits where two or more resistors share the same two nodes (a parallel branch).
  • Choosing resistor combinations to hit a target resistance you cannot buy as a single standard value.
  • Calculating equivalent load resistance for current-divider, LED, or speaker wiring problems.
  • Checking homework or lab results in physics and electronics courses involving parallel networks.

Common mistakes

  • Adding resistances directly (R1 + R2) — that is the series formula. Parallel resistance requires adding reciprocals, then inverting the sum.
  • Expecting the total to be larger than an individual resistor. The parallel total is always smaller than the smallest resistor in the group.
  • Forgetting to invert at the end. 1/R1 + 1/R2 gives the total conductance, not the resistance; you must take the reciprocal of that sum to get Rt.
  • Entering 0 Ω expecting it to be ignored as 'no resistor'. A true 0 Ω path (a short) would force Rt to 0; this tool treats 0 and blanks as 'not connected' so only real resistor values are summed.

Frequently asked questions

What is the formula for resistors in parallel?

The reciprocal of the total resistance equals the sum of the reciprocals of each resistor: 1/Rt = 1/R1 + 1/R2 + ... + 1/Rn. After summing the reciprocals, take the reciprocal of that result to get the total resistance Rt.

Why is parallel resistance always less than the smallest resistor?

Adding a parallel path gives current another route, which increases the total conductance and therefore lowers resistance. Because every reciprocal you add makes the sum 1/Rt larger, inverting it always yields a value below the smallest individual resistor.

How do I calculate two resistors in parallel quickly?

For exactly two resistors, use the product-over-sum shortcut: Rt = (R1 × R2) / (R1 + R2). For example, 60 Ω and 40 Ω give (60 × 40)/(60 + 40) = 2400/100 = 24 Ω.

What happens if all the resistors are equal?

For n identical resistors of value R in parallel, the total is simply Rt = R / n. Four 10 Ω resistors in parallel give 10/4 = 2.5 Ω, and two 100 Ω resistors give 100/2 = 50 Ω.

Do I have to fill in all four resistor fields?

No. Leave any field blank to use fewer resistors — the calculator only sums the values you enter. Blank and zero entries are ignored so they do not distort the parallel result.

What is the series sum shown next to the result?

The series sum (R1 + R2 + ...) is provided for comparison, showing what the same resistors would total if wired in series instead of parallel. It is always larger than the parallel total and equals the simple arithmetic sum of your inputs.

Sources & references

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