KCL Node Analyzer

Quick experiments

• Check the balanced default. Three branches: 3 A in, 1 A in, 4 A out. The equation reads 3 + 1 = 4 and the node balances. This is KCL in its simplest honest form.
• Break it, then fix it. Change the outgoing branch to 5 A. The analyzer flags the imbalance immediately — 4 A arriving cannot supply 5 A leaving. Raise the first branch to 4 A and it balances again.
• Add a fourth branch. Add a branch at 2 A in. Now 3 + 1 + 2 = 6 A must leave, so set the outgoing branch to 6 A. KCL scales to any number of branches without changing form.
• Try to make every branch incoming. Flip the outgoing branch to 'in'. The widget warns that no charge can leave, because that node is physically impossible. Flip one back to 'out' and it recovers.
• Push the ratio to extremes. Set 9.8 A in, 0.2 A in and 10 A out. The node balances exactly. KCL is about the sum, not about the branches being comparable in size — a tiny branch alongside a huge one is perfectly normal.

Formula reference

Kirchhoff's Current Law

$$\sum I_{in} = \sum I_{out}$$

Everything arriving at a node equals everything leaving.

Signed form

$$\sum_{node} I = 0$$

Inbound positive, outbound negative — one equation per node.

Current divider — two branches

$$I_1 = I_{total} \times \frac{R_2}{R_1 + R_2}$$

Note the opposite resistance on top: current favours the easier path.

Branch current from node voltage

$$I_n = \frac{V_{node} - V_n}{R_n}$$

The basis of nodal analysis.

Independent node equations

$$\text{equations} = N_{nodes} - 1$$

One node is the reference, so it contributes nothing new.

Common mistakes

• ✗Treating each dot on the schematic as a separate node.

  Everything joined by plain wire is one node, however far apart it is drawn. Trace the copper before writing equations — two junctions that look distinct are often the same node, and treating them separately produces equations that cannot be solved.

• ✗Mixing up which resistance goes on top in the current divider.

  For two branches, I₁ = I_total × R₂/(R₁+R₂) — the other branch's resistance. It is the opposite arrangement to the voltage divider, because current takes the easier path. If your answer sends more current down the larger resistor, you have flipped it.

• ✗Assuming current is 'used up' as it passes through components.

  It is not. Current leaving a resistor is identical to current entering it; what the resistor consumes is energy, seen as a voltage drop. A lamp does not draw current away from parts further down the circuit.

• ✗Writing an equation for every node, including the reference.

  With N nodes only N − 1 equations are independent; the last is a linear combination of the others and adds nothing. Choose a reference node — usually ground — and write equations for the rest.

• ✗Being thrown by a negative branch current.

  It just means you guessed the direction backwards. The magnitude is correct and the node still balances. Assumed directions are bookkeeping, not physics — you do not have to guess right, only stay consistent.

Frequently asked questions