Resistor Network Solver

Build and solve

Pre-loaded with 4 Ω in series with (6 Ω ∥ 12 Ω) at 24 V. Use the SER / PAR toggle on each row to change how it joins the network.

Quick experiments

• Prove a parallel pair drops below its smallest member. Set two resistors in parallel at 6 Ω and 12 Ω. The total is 4 Ω — smaller than either. Now make the second 1 kΩ: the total barely moves off 6 Ω, because a much larger branch carries almost no current.
• Watch series voltages divide in proportion. Put 2 Ω, 4 Ω and 6 Ω in series across 12 V. The current is the same 1 A everywhere, and the drops come out 2 V, 4 V and 6 V — each resistor takes exactly its share of the total resistance.
• Split a current in inverse proportion. Two branches, 4 Ω and 12 Ω, across 12 V. The 4 Ω branch takes 3 A and the 12 Ω branch takes 1 A — three times the resistance, one third of the current. The branch currents still sum to the 4 A the source delivers.
• Make two equal resistors halve. Two identical resistors in parallel always give exactly half. Set both to 10 Ω and confirm 5 Ω, then both to 100 Ω for 50 Ω. This is the one parallel case worth doing in your head.
• Turn a series chain into a series-parallel one. Start with three resistors all in series, then flip the third row to PAR. Total resistance drops, source current rises, and the drops redistribute — one toggle showing why the topology matters as much as the values.

Formula reference

Series resistance

$$R_{total} = R_1 + R_2 + \cdots + R_n$$

Same current through each; resistances add.

Parallel resistance

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

Always smaller than the smallest branch.

Two resistors in parallel

$$R_{total} = \frac{R_1 R_2}{R_1 + R_2}$$

The product-over-sum shortcut. Two only — it does not extend to three.

Voltage divider (series)

$$V_n = V_{source} \times \frac{R_n}{R_{total}}$$

Each resistor takes its share of the total resistance.

Current divider (parallel)

$$I_n = I_{total} \times \frac{R_{total}}{R_n}$$

Inverse proportion — the smallest branch takes the most.

N equal resistors in parallel

$$R_{total} = \frac{R}{N}$$

Two 10 Ω give 5 Ω; four give 2.5 Ω.

Common mistakes

• ✗Adding parallel resistors the way you add series ones.

  6 Ω and 12 Ω in parallel is 4 Ω, not 18 Ω. Add the reciprocals and invert, or use the product-over-sum shortcut for exactly two. A quick sanity check: the answer must be smaller than the smallest branch.

• ✗Using the product-over-sum shortcut on three or more resistors.

  R₁R₂/(R₁+R₂) is derived for two branches only. For three, either apply it twice — combine two, then combine that result with the third — or go back to the reciprocal sum.

• ✗Forgetting to invert at the end of the reciprocal sum.

  Summing 1/6 + 1/12 gives 0.25, which is the conductance in siemens, not the resistance. The resistance is 1/0.25 = 4 Ω. Leaving it as 0.25 Ω is off by a factor of sixteen here.

• ✗Assuming the biggest resistor carries the most current.

  In parallel it carries the least — current divides in inverse proportion. In series every resistor carries exactly the same current; there the biggest resistor takes the biggest voltage drop instead.

• ✗Reducing a series-parallel network from the outside in.

  Start with the innermost parallel group and collapse outwards, otherwise you will combine resistors that are not actually in the same relationship. Redraw the circuit if the topology is not obvious — most errors here are reading errors, not arithmetic.

Frequently asked questions