Engineering Calc Tools

Circuits

Circuit Solver (Nodal Analysis, DC and AC)

Node voltages, branch currents, power and Thévenin equivalents for any DC or AC circuit, with dependent sources and op-amps.

Tool no.
ECT-K09
Rev
1.0
Updated
Oct 2026
Method
Modified nodal analysis
Examples

Given

Name, the two nodes, then the value; 0 is ground. R1 a b 4.7k · C1 b 0 10u · L1 b c 2m · Z1 a b 3 4 (3 + j4 Ω) · V1 a 0 12 (+ at a) · I1 0 b 2 (2 A from 0 into b) · AC sources take a phase: V1 a 0 10 30.
Dependent: E1 a b c d 2 (Vab = 2Vcd) · G1 a b c d 0.5 (0.5Vcd from a to b) · F1 a b R2 3 (3 × current in R2) · H1 a b R2 5 (Vab = 5 Ω × current in R2) · op-amp O1 out plus minus.

Answers

Node voltages = —
Runs in your browser. Nothing you enter is uploaded.Modified nodal analysis · DC and AC
How to use this solver

Label every node with a name or number, with 0 for ground, and type each element on its own line: its name (the first letter sets the type), the two nodes it connects, and its value. Values take the usual prefixes: 4.7k, 10u, 2m, 1meg. For mesh-analysis problems the answer is the same: the branch currents are in the element table.

Method

KCL at each nodeΣ currents leaving = 0, with I = (Va − Vb)/Z
ImpedancesZR = R,  ZL = jωL,  ZC = 1/(jωC)
ThéveninVth = open-circuit voltage,  Zth = Vtest/Itest with independent sources off
Maximum powerZL = Zth*,  Pmax = |Vth|²/(4Rth) (rms)

Worked example

12 V source, 4 Ω and 6 Ω divider, 2 Ω to node c fed by a 3 A source (the default)

  1. At node c all 3 A flows through the 2 Ω resistor to b, so Vc = Vb + 6.
  2. KCL at b: (12 − Vb)/4 + 3 = Vb/6, so Vb = 14.4 V and Vc = 20.4 V.
  3. The current in the 4 Ω resistor is (12 − 14.4)/4 = −0.6 A: it flows back into the 12 V source, which absorbs 7.2 W.

Frequently asked questions

How do I find a current for a dependent source?

Name the element that carries it: F1 a b R2 3 uses the current through R2, from its first node to its second. To control by a current in a plain wire, add a 0 V source there as an ammeter.

Peak or rms?

Phasor answers use the same basis as the sources you enter. Power uses ½V·I* for peak values and V·I* for rms values.