How Do You Combine Three Resistors in Parallel?

To combine three resistors in parallel, replace them with a single equivalent resistance that draws the same total current at the same voltage. Because every parallel branch has the same voltage, the branch currents add. This gives a reciprocal-sum formula: add the reciprocals of the three resistances, then take the reciprocal of that sum.

The formula for three parallel resistors

For three finite, positive resistances $R_1$, $R_2$, and $R_3$, their equivalent resistance $R_{\mathrm{eq}}$ is

$$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}.$$

Equivalently,

$$R_{\mathrm{eq}}=\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)^{-1}.$$

The resistors are parallel when each one connects between the same pair of circuit nodes. Their physical placement on a drawing does not matter; what matters is which nodes their ends connect to.

Why the reciprocals add

Each resistor in parallel has the same voltage $V$ across it. By Ohm’s law, the current in each branch is $I_k=V/R_k$. The total current is the sum of the branch currents:

$$I_{\mathrm{total}}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}.$$

For a nonzero applied voltage, a single equivalent resistor obeys $I_{\mathrm{total}}=V/R_{\mathrm{eq}}$. Equating these expressions and dividing by $V$ gives the reciprocal-sum formula. This reasoning also shows why you do not add the resistances directly: direct addition is the rule for resistors in series, where the same current passes through each resistor.

Worked example

Suppose three resistors of $6\,\Omega$, $12\,\Omega$, and $4\,\Omega$ are connected in parallel. Substitute their values:

$$\begin{aligned} \frac{1}{R_{\mathrm{eq}}} &=\frac{1}{6\,\Omega}+\frac{1}{12\,\Omega}+\frac{1}{4\,\Omega}\\ &=\frac{2}{12\,\Omega}+\frac{1}{12\,\Omega}+\frac{3}{12\,\Omega}\\ &=\frac{6}{12\,\Omega}=\frac{1}{2\,\Omega}. \end{aligned}$$

Taking the reciprocal gives

$$R_{\mathrm{eq}}=2\,\Omega.$$

A current check confirms the result. If the parallel network is connected across $12\,\mathrm{V}$, the three branch currents are $12/6=2\,\mathrm{A}$, $12/12=1\,\mathrm{A}$, and $12/4=3\,\mathrm{A}$. Their total is $6\,\mathrm{A}$. A $2\,\Omega$ equivalent resistor across $12\,\mathrm{V}$ also draws $12/2=6\,\mathrm{A}$.

A useful check and special case

For three finite, positive resistances in parallel, the equivalent resistance must be less than the smallest individual resistance. In the example, the smallest resistance is $4\,\Omega$, and the equivalent is $2\,\Omega$, so the result passes this check. Adding a parallel path gives current another route, lowering the resistance seen by the source.

If all three resistors have the same resistance $R$, the calculation simplifies:

$$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}, \qquad R_{\mathrm{eq}}=\frac{R}{3}.$$

For example, three $9\,\Omega$ resistors in parallel have an equivalent resistance of $3\,\Omega$.

Common mistakes

• Adding the resistance values: $R_1+R_2+R_3$ is the series formula, not the parallel formula.
• Forgetting to invert: The reciprocal sum gives $1/R_{\mathrm{eq}}$, not $R_{\mathrm{eq}}$ itself. Invert the final result.
• Mixing units: Convert values to the same unit before adding. For instance, express every resistance in ohms or every resistance in kilohms.
• Assuming a drawing guarantees parallel wiring: Check that each resistor connects to the same two nodes. Components that merely appear side by side may not be in parallel.
• Accepting a result larger than the smallest resistor: That usually signals an arithmetic or connection-identification error for a network of ordinary finite, positive resistors in parallel.