Equivalent Resistance

Calculate equivalent resistance for series or parallel circuits.

Inputs

R_1

R_2

R_3

U

Formula Interpretation

Series Equivalent

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

$R_{\text{eq}}$ (Ω) equals the sum of all individual resistances: $R_{\text{eq}} = R_1 + R_2 + \cdots + R_n$ . In a series circuit the same current flows through every element, and voltage is distributed in proportion to resistance — the principle behind voltage dividers and current-limiting resistors. Adding any resistor in series always increases the total resistance.

Parallel Equivalent

\dfrac{1}{R_{\text{eq}}} = \sum_{i=1}^{n} \dfrac{1}{R_i}

For parallel resistors, $1/R_{\text{eq}} = 1/R_1 + 1/R_2 + \cdots + 1/R_n$ ; for just two resistors the shortcut is $R_{\text{eq}} = R_1 R_2 / (R_1 + R_2)$ . The parallel equivalent is always less than the smallest individual branch resistance. Adding a parallel branch always decreases total resistance and increases total current demand.

Total Current

I = \dfrac{U}{R_{\text{eq}}}

With supply voltage $U$ (V) known, Ohm's Law gives total current $I$ = $U$ / $R_{\text{eq}}$ (A). This current is the starting point for verifying conductor sizing and switching-device ratings throughout the circuit.

Voltage Drop

U_i = I \times R_i

In a series circuit, each resistor's voltage drop $U_i$ (V) = total current $I$ (A) × that resistor $R_i$ (Ω). By KVL, all drops must sum exactly to the supply voltage — this check is the simplest verification that a series-circuit calculation is correct.

Branch Current

I_i = \dfrac{U}{R_i}

In a parallel circuit, every branch shares the same terminal voltage $U$ (V), so each branch current $I_i$ (A) = $U$ / $R_i$ . Smaller-resistance branches carry larger currents. By KCL, all branch currents must sum to the total current $I$ .

Knowledge Points

Series voltage divider principle and applications

In a series circuit, current is identical through all elements and voltage distributes as $U_i / U_{\text{total}} = R_i / R_{\text{eq}}$ . Two series resistors form a voltage divider — the most fundamental biasing element in analogue circuits: ADC reference voltage scaling, op-amp bias points, and signal conditioning for resistive temperature sensors (RTD, NTC thermistors) all rely on voltage dividers. A series resistor is also used as a current-limiting resistor to protect LEDs and logic-level converters from excess current.

Parallel current divider principle and applications

In parallel, voltage is identical across all branches, and total current equals the sum of branch currents (KCL): $I = I_1 + I_2 + \cdots + I_n$ . Lower-resistance branches carry higher currents ( $I_i \propto 1/R_i$ ). Two parallel resistors form a current divider — used in ammeter shunts to extend current measurement range. Multiple parallel conductors increase total current-carrying capacity. However, adding many parallel branches lowers total resistance and increases source current demand, so supply capacity must be verified.

Bounds on equivalent resistance

For series: $R_{\text{eq}} \geq \max(R_i)$ , always greater than any individual resistor. For parallel: $R_{\text{eq}} \leq \min(R_i)$ , always less than any individual resistor. For $n$ identical resistors $R_0$ : series gives $n R_0$ ; parallel gives $R_0 / n$ . These bounds provide a quick sanity check — if a calculated equivalent resistance falls outside them, an error has been made.

Delta–Wye (Δ–Y) transformation

When a resistor network contains neither pure series nor pure parallel groups (e.g., a Wheatstone bridge, three-phase load), the Δ–Y transformation converts a triangle-connected set of three resistors into an equivalent star-connected set: for equal resistors, $R_Y = R_{\Delta}/3$ . The general form is $R_1 = R_{ab} R_{ca} / (R_{ab} + R_{bc} + R_{ca})$ with the other two obtained by cyclic permutation. Applying Δ→Y or Y→Δ converts the intractable network into a reducible series–parallel form, enabling standard equivalent-resistance analysis.

Thévenin and Norton equivalent circuits

Any two-terminal network of linear resistors and independent sources can be replaced by a Thévenin equivalent — a single ideal voltage source $U_{\text{th}}$ (open-circuit voltage) in series with a Thévenin resistance $R_{\text{th}}$ , or by a Norton equivalent (short-circuit current $I_N = U_{\text{th}}/R_{\text{th}}$ in parallel with $R_{\text{th}}$ ). Computing $R_{\text{th}}$ is exactly the task of this calculator: zero all independent sources (short voltage sources, open current sources) and find the equivalent resistance looking into the terminals. This simplification is the cornerstone of load-line analysis and maximum power transfer design.

Examples

Three resistors in series: $R_1 = 10\,\Omega$ , $R_2 = 20\,\Omega$ , $R_3 = 30\,\Omega$ . The source voltage is $U = 12\,\text{V}$ .

Step 1 — Sum equivalent resistance

R_{\text{eq}} = 10 + 20 + 30 = 60\,\Omega

Step 2 — Compute total current

I = \dfrac{12}{60} = 0.2\,\text{A}

Step 3 — Voltage drops

U_1 = 2\,\text{V},\; U_2 = 4\,\text{V},\; U_3 = 6\,\text{V}

Two resistors in parallel: $R_1 = 100\,\Omega$ and $R_2 = 200\,\Omega$ . The source voltage is $U = 10\,\text{V}$ .

Step 1 — Reciprocal sum

\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{100} + \dfrac{1}{200} = 0.015\,\Omega^{-1}

Step 2 — Equivalent resistance

R_{\text{eq}} = \dfrac{1}{0.015} \approx 66.7\,\Omega

Step 3 — Branch currents

I_1 = 0.1\,\text{A},\; I_2 = 0.05\,\text{A}

Extended Knowledge

•Step-by-step simplification of series–parallel networks：To reduce a complex series–parallel network: ① identify the innermost pure series or pure parallel group; ② replace it with its equivalent; ③ repeat until a single equivalent remains. For multi-level networks (cascaded dividers, ladder networks), work from the load end toward the source, labelling intermediate results clearly to avoid confusion. The critical judgement is correct identification: resistors sharing the same two nodes are in parallel; resistors carrying the same current with no branching nodes between them are in series.
•Wheatstone bridge and balance condition：A Wheatstone bridge consists of four resistors in a diamond (Δ–Y) configuration. When the balance condition $R_1/R_2 = R_3/R_4$ is met, no current flows through the bridge arm (galvanometer), and the differential voltage across it is zero. With three known resistors, an unknown fourth can be measured to 0.01% accuracy. Modern applications embed strain gauges, thermistors, and gas sensors in Wheatstone bridge circuits, converting tiny resistance changes into differential voltages that are easily amplified and digitised.
•Resistor power rating verification：Each resistor''s power dissipation must be checked individually: in a series circuit $P_i = I^2 R_i$ (same current everywhere); in a parallel circuit $P_i = U^2/R_i$ (same voltage everywhere). Choose a resistor whose rated power exceeds the calculated value, then apply a 50–70% derating for reliability. In a voltage divider, the high-resistance leg (low current) and the low-resistance leg (high current) can have very different power demands — computing them separately and selecting appropriate ratings for each is essential.
•PCB trace and via resistance in power distribution：PCB copper traces and vias carry measurable resistance that must be included in the equivalent-resistance analysis of high-current paths. Trace resistance $R_{\text{trace}} = \rho L / (W \cdot t)$ (see the Resistance Formula calculator); a standard 0.3 mm diameter via with 35 μm copper plating has approximately 5–20 mΩ. In power-integrity (PI) analysis, DC voltage drop (IR drop) across the power-distribution network is modelled as an equivalent resistor network to verify that supply voltage at every device pin stays within specification — a critical step for modern high-current digital ICs.