Electronics & Circuits
Intermediate

Resistors in Parallel

Reciprocals of parallel resistances add.

Formula

1Req=1R1+1R2+⋯\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots

Variables

R_iIndividual resistances

Example

10∥10 = 5Ω

How it works

When resistors are connected in parallel, each branch has the same voltage across it, so the current splits among the branches. Because current adds while voltage stays the same, the total conductance adds, which is why the reciprocals of the resistances are summed. The equivalent resistance is always less than the smallest individual resistor because adding more parallel paths makes it easier for current to flow.

When to use it

Use this formula whenever resistors are wired across the same two nodes, such as in parallel branches of a circuit or when combining multiple loads into one equivalent resistance. It is also useful for checking current division and simplifying complex resistor networks.

Worked example, step by step

Find the equivalent resistance of 6 Ω, 3 Ω, and 2 Ω connected in parallel. 1/R_eq = 1/6 + 1/3 + 1/2 1/R_eq = 1/6 + 2/6 + 3/6 1/R_eq = 6/6 = 1 R_eq = 1/1 = 1 Ω Final answer: the equivalent resistance is 1 Ω.

Common mistakes to avoid

  • Adding the resistor values directly instead of adding their reciprocals
  • Forgetting that parallel branches must share the same two connection points
  • Miscalculating the final inverse step after summing the reciprocals

Did You Know?

Adding a parallel resistor lowers total resistance — more lanes for the current to flow.

Share this formula

Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026