Electric circuit calculator

Draw a circuit with resistors, capacitors, batteries and current sources and get the equivalent resistance, the voltage and current of every element and the power. It solves by series–parallel reduction, delta–wye transformation, superposition or nodal analysis, showing every step.

What you can do

Animated series–parallel and Δ–Y reduction.
Superposition and nodal analysis with matrices.
Capacitor charge and discharge (τ = RC).
Ohm's and Kirchhoff's laws step by step.

Frequently asked questions

How is equivalent resistance calculated?

In series they add (R = R1 + R2); in parallel their reciprocals add (1/R = 1/R1 + 1/R2).

What is nodal analysis?

Applying Kirchhoff's current law at every node with the node voltages as unknowns.

Guide: how a DC circuit is solved

Solving a circuit means finding the current through each element and the voltage across it. Kirchhoff's two laws and Ohm's law solve any resistive circuit; for those built from series and parallel pieces there is a shortcut, reduction. This guide explains the two methods the tool uses and works a complete example.

The basic laws

Ohm's law: V = I · R. The voltage across a resistor is proportional to the current through it.
Kirchhoff's current law (KCL): at each node, the currents entering equal the currents leaving.
Kirchhoff's voltage law (KVL): around each closed loop, the voltage rises and drops add up to zero.
Power: P = V · I = I²R = V²/R. In a circuit, the power delivered by the sources equals the power absorbed by the elements.

Series and parallel

Two elements are in series if the node between them has no other branch: the same current flows through both and R_eq = R₁ + R₂. They are in parallel if they connect the same pair of nodes: they have the same voltage and 1/R_eq = 1/R₁ + 1/R₂, that is, R_eq = R₁R₂/(R₁ + R₂) for two. The parallel equivalent is always smaller than the smallest resistor.

Worked example by reduction

A 12 V battery feeds R1 = 4 Ω in series with a parallel combination of R2 = 6 Ω and the branch R3 = 1 Ω + R4 = 2 Ω.

Series: R3 and R4 form Re1 = 1 + 2 = 3 Ω.
Parallel: R2 and Re1 form Re2 = 6·3/(6 + 3) = 2 Ω.
Series: R1 and Re2 form R_eq = 4 + 2 = 6 Ω.
Total current: I = 12 V / 6 Ω = 2 A.
Back to the original circuit: V_R1 = 2 · 4 = 8 V, leaving 12 − 8 = 4 V across the parallel part. R2 carries 4/6 ≈ 0.667 A and the R3–R4 branch 4/3 ≈ 1.333 A, split as V_R3 ≈ 1.333 V and V_R4 ≈ 2.667 V.
Checks: at the node, 0.667 + 1.333 = 2 A (KCL). Powers: 16 + 2.667 + 1.778 + 3.556 = 24 W absorbed, equal to the 12 V · 2 A = 24 W the battery delivers.

The tool shows exactly these steps and values for this circuit.

When there is no series or parallel: nodal analysis

Some circuits, like a resistor bridge or those with several sources, cannot be reduced. Nodal analysis always works: choose a reference node (ground), take the voltage of every other node as an unknown and write KCL at each one, expressing each current with Ohm's law. This gives a system of linear equations, one per node, solved all at once.

How to use the calculator

Draw the circuit with the tools: resistor, battery, current source, wire, capacitor, inductor, diode… Values accept prefixes: 4.7k, 2.2M, 10u (µ), 470n.
Choose the method: reduction (series and parallel) or nodal (KCL at each node).
Read the equivalent resistance, the current and the power, the step-through reduction and the table with each element's voltage, current and power, including the power balance.

Common mistakes

Calling two resistors "in series" when a third branch reaches the node between them. They are not.
Adding parallel resistances instead of adding their reciprocals.
Using the full source voltage on an element that only gets part of it.
Forgetting signs in KVL: a voltage drop in the direction of travel subtracts.

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