Two Resistors in Parallel

Also known as parallel resistance · product over sum

R_{t} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}

R_t =

R_1 =

R_2 =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

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Wired side by side, two resistors give the current two paths at once, so more current flows for the same voltage and the combination resists less than either branch alone. The tidy product-over-sum form is just 1/R_t = 1/R₁ + 1/R₂ rearranged — conductances, not resistances, are what add in parallel. Worked example: 100 Ω in parallel with 25 Ω gives (100 × 25)/(100 + 25) = 2500/125 = 20 Ω, comfortably below the smaller branch.

Two handy special cases: equal resistors in parallel halve (two 100 Ω resistors make 50 Ω), and a much smaller resistor dominates — 10 Ω in parallel with 10 kΩ is essentially 10 Ω. Household outlets are wired in parallel so every appliance sees full mains voltage. When solving for a branch, the other resistance must exceed the total, since the total is always the smallest value in the circuit.

Two Resistors in Parallel

R_{t} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}

Where

$R_{t}$ = Total resistance
$R_{1}$ = Resistance 1
$R_{2}$ = Resistance 2

Missing one of these? Work it out first, then come back

Total resistance — Ohm's Law, Electrical Power (P = I²R)
Resistance 1 — Ohm's Law, Electrical Power (P = I²R)
Resistance 2 — Ohm's Law, Electrical Power (P = I²R)

Part of this formula set

Series and parallel networks

4 formulasCombining resistors and capacitors two at a time — and the reason the two components follow exactly opposite rules.