How do you calculate resistors in parallel?

Add the reciprocals and invert the result: 1/Req = 1/R1 + 1/R2 + … + 1/Rn. For exactly two resistors you can use Req = R1×R2/(R1+R2) instead, which avoids the reciprocals entirely.

Why is parallel resistance less than the smallest resistor?

Each parallel branch is an additional path for current. More paths mean less total opposition, so adding any branch - however large - can only reduce the total. The answer is therefore always below the smallest single branch.

What is the formula for two resistors in parallel?

Req = (R1 × R2) / (R1 + R2), the product over the sum. It is exact for two resistors and wrong for three or more, where the reciprocal formula is needed.

What happens to two equal resistors in parallel?

The total is exactly half of one of them, and each carries half the current. n equal resistors in parallel give R/n, which is the quickest way to check a parallel result.

Is the voltage the same across resistors in parallel?

Yes. Parallel branches share the same two nodes, so they all see the same voltage. What differs is the current in each branch, which is inversely proportional to its resistance.

Parallel Resistance Calculator

Resistors in parallel combine reciprocally, and the equivalent resistance is always smaller than the smallest resistor in the group. Each branch sees the same voltage and carries its own share of the current.

Parallel Resistance Formula

Every resistor in a parallel group is connected across the same two nodes, so they all have the same voltage V across them. That is the mirror image of the series case, and it is where this formula starts: parallel means one voltage, shared.

Kirchhoff's current law says the current arriving at a node equals the sum of the currents leaving it, so the total current splits between the branches:

I = I_{1} + I_{2} + I_{3}

Ohm's law gives each branch current as V divided by that branch's resistance, and because the voltage is common the same V appears in every term:

\frac{V}{R_{e q}} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}

Divide through by V and it cancels, leaving the reciprocal formula for equivalent parallel resistance:

\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \dots + \frac{1}{R_{n}}

The reciprocal of the total is the sum of the reciprocals. Remember the final flip.

The reciprocals are what makes this formula easy to get wrong: the sum gives you 1/Req, not Req. The last step is always to invert.

Two resistors: the product-over-sum shortcut

For exactly two resistors the reciprocals can be cleared algebraically, which gives the product-over-sum form most people memorise:

R_{e q} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}

This is a shortcut for two resistors only. Applied to three it silently gives the wrong answer, which is one of the most common errors in circuit arithmetic. For three or more, use the reciprocal form, or apply product-over-sum twice - combine two, then combine that result with the third.

Equal resistors in parallel

If all n resistors are identical the formula collapses to a division, which is worth knowing as a sanity check on any parallel answer:

R_{e q} = \frac{R}{n}

Two 10 kΩ in parallel are 5 kΩ; four are 2.5 kΩ. If a calculation of equal resistors does not match this, the arithmetic went wrong somewhere.

Worked example

Given

R1 = 4.7 kΩ
R2 = 10 kΩ

Working

Req = (R1 × R2) / (R1 + R2)
Req = (4700 × 10000) / (4700 + 10000)
Req = 47,000,000 / 14,700

Answer3.1973 kΩ

The result is below 4.7 kΩ, the smaller of the two, as it must be. The nearest E24 value is 3.3 kΩ, about 3.2 % high — the calculator will show you that.

Why the Total Is Always Smaller

Adding a parallel branch gives the current somewhere else to go, and more paths means less total opposition. So the equivalent parallel resistance is always below the smallest branch - not between the two values, and never above them.

That gives you a free check on every parallel calculation, and it is the fastest way to catch the reciprocal mistake. If 4.7 kΩ and 10 kΩ came out as 14.7 kΩ, the series rule got used. If they came out as 7.35 kΩ, the two values were averaged. The answer has to be under 4.7 kΩ, and it is 3.1973 kΩ.

The bounds worth remembering

Two equal resistors in parallel: exactly half the value of one.
One resistor much larger than the other: the answer is just under the smaller one.
Any number of branches: the total is always less than the smallest branch.

Reaching Awkward Values With Parallel Resistors

Standard values leave gaps, and parallel combinations fill them. Two E24 parts in parallel reach values no single E24 part can: if you need 3.2 kΩ and only have E24 stock, 4.7 kΩ with 10 kΩ gets you to 3.1973 kΩ — an error of about half a percent, which is inside the tolerance of the parts themselves.

This is also how a value is trimmed after the fact. Soldering a large resistor across an existing one pulls the total down by a known amount, which is a common repair and calibration technique. Use the find-missing-value mode above to work out what to add: give it the target and the resistor you already have.

Parallel Resistance in Practice

Sharing current and power

Two resistors in parallel each carry part of the current, so each dissipates part of the power. Two equal parts in parallel handle twice the power of one, which is the standard way to build a high-power load or a current-sense shunt from ordinary parts.

Loading you did not intend

Anything connected across a divider is in parallel with the lower resistor, and it changes the ratio. A 10 kΩ input impedance across a 10 kΩ divider leg halves that leg. This is the most common reason a divider measures lower than the calculation predicted, and the reason high-impedance dividers need a buffer.

Meter loading

A multimeter has finite input resistance, typically 10 MΩ, and that resistance goes in parallel with whatever you measure. On a 100 kΩ node it is irrelevant; on a 10 MΩ node it halves the reading.

Common mistakes

Forgetting the final reciprocal. The sum of 1/R terms gives 1/Req. Invert it, or the answer is out by a large factor.
Using product-over-sum with three or more resistors. It is valid for exactly two. For three, combine two first and then combine the result with the third.
Averaging the two values. 4.7 kΩ with 10 kΩ is not 7.35 kΩ — it is 3.1973 kΩ.
Accepting an answer larger than the smallest resistor. That is impossible in parallel and always means the wrong rule was applied.
Ignoring the load. A parallel calculation that leaves out the thing connected across the node describes a circuit you did not build.

Frequently asked questions

How do you calculate resistors in parallel?: Add the reciprocals and invert the result: 1/Req = 1/R1 + 1/R2 + … + 1/Rn. For exactly two resistors you can use Req = R1×R2/(R1+R2) instead, which avoids the reciprocals entirely.
Why is parallel resistance less than the smallest resistor?: Each parallel branch is an additional path for current. More paths mean less total opposition, so adding any branch - however large - can only reduce the total. The answer is therefore always below the smallest single branch.
What is the formula for two resistors in parallel?: Req = (R1 × R2) / (R1 + R2), the product over the sum. It is exact for two resistors and wrong for three or more, where the reciprocal formula is needed.
What happens to two equal resistors in parallel?: The total is exactly half of one of them, and each carries half the current. n equal resistors in parallel give R/n, which is the quickest way to check a parallel result.
Is the voltage the same across resistors in parallel?: Yes. Parallel branches share the same two nodes, so they all see the same voltage. What differs is the current in each branch, which is inversely proportional to its resistance.

Assumptions and limitations for Parallel Resistance Calculator are listed on the About page. Every worked example on this site is checked against the same solver the calculator uses.