Resistors in Parallel Calculator

Resistors are in parallel when both ends of each connect to the same two nodes, so every branch sees the same voltage. Conductances add:

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}$

For exactly two resistors this simplifies to the product-over-sum form:

$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$

Units: every $R$ in ohms (Ω); the result is in ohms.

In series, the same current flows through each and resistances add: $R_{\text{eq}} = R_1 + R_2 + \cdots$

The assumption people forget: you must take the reciprocal at the end. $1/R_1 + 1/R_2$ is the conductance in siemens, not the resistance. A useful sanity check — the parallel result is always smaller than the smallest branch, because you have added another path for current.

In a parallel network:

• Voltage is identical across every branch: $V_1 = V_2 = V_{\text{supply}}$
• Current divides between branches, $I_n = V/R_n$, and the branch currents sum to the total (Kirchhoff's current law): $I_{\text{total}} = I_1 + I_2 + \cdots$
• The lowest-resistance branch carries the most current

In a series network the roles swap: current is common, voltage divides.

For a combined series-parallel circuit, work from the inside out — collapse each purely parallel or purely series group into a single equivalent resistor, then repeat until one value remains. Then work back outwards with $V = IR$ to recover individual voltages and currents.

The assumption people forget: this treatment assumes ideal conductors — zero wire resistance and an ideal source with no internal resistance. A real battery's internal resistance sits in series with the load and pulls the terminal voltage down as current rises.

• Forgetting the final reciprocal — the sum of reciprocals is $1/R_{\text{eq}}$, so invert it.
• Using product-over-sum for three or more resistors — $R_1R_2/(R_1+R_2)$ is a two-resistor shortcut only. Apply it pairwise, or use the general reciprocal sum.
• Getting a result larger than the smallest resistor — an immediate sign of an arithmetic slip in a parallel calculation.
• Adding parallel resistances directly — that is the series rule.
• Mixing kΩ and Ω — convert to a single unit before summing reciprocals.
• Assuming equal branch currents — they are equal only when the branch resistances are equal.
• Missing the equal-resistor shortcut — $n$ identical resistors of value $R$ in parallel give exactly $R/n$, so four $100\ \Omega$ resistors in parallel come to $25\ \Omega$.
• Miscounting what is genuinely in parallel — two components are in parallel only when both of their ends meet at the same pair of nodes. Sharing a single node is not enough.