Kirchhoff's junction rule

Current is defined as I = dq/dt, the charge crossing a chosen surface per second, so one ampere is one coulomb per second. Draw a small closed surface around the point where several wires meet and count what crosses it, taking arrival as positive. Charge is conserved, so the surplus arriving must be the rate at which charge collects inside: Σ Iᵢₙ − Σ Iₒᵤₜ = dq_node/dt. The junction rule sets that right-hand side to zero, leaving Σ Iᵢₙ = Σ Iₒᵤₜ. Five wires meeting at a node, three of them delivering 1.20 A, 0.40 A and 0.75 A and a fourth carrying 1.90 A away, therefore leave the fifth no freedom at all: it must carry 1.20 + 0.40 + 0.75 − 1.90 = 0.45 A away from the node, whatever components sit in the five branches. The rule holds instant by instant rather than only on average, so it applies to alternating currents and to the middle of a transient exactly as it does to a steady direct current, provided the instantaneous values are used.

A junction of wires is a conductor, so it has some capacitance to its surroundings, but a very small one: a few millimetres of exposed metal comes to roughly 1 pF. Suppose the currents did not balance and 1 A of surplus flowed in. The potential of the node would rise by ΔV = I Δt / C_node, reaching one volt after Δt = 1 × 10⁻¹² × 1.0 ÷ 1.0 = 1 × 10⁻¹² s. A picosecond. Long before that, the field of the accumulated charge would have slowed the incoming current and accelerated the outgoing one until the imbalance vanished, which is why the rule is self-enforcing rather than merely assumed. A capacitor plate is the deliberate exception. At 47 μF and an imbalance of 0.1 A, the same one-volt rise takes 47 × 10⁻⁶ × 1.0 ÷ 0.1 = 4.7 × 10⁻⁴ s, about 4.7 × 10⁸ times longer, which is a timescale a circuit does care about. The other place the rule strains is at radio frequencies, where stray picofarads and signal delay across the board matter.

Give every branch one symbol and one arrow. The arrow does not claim to know the direction of flow; it defines what positive means for that symbol. Then at each node write the rule in signed form, counting a current as positive when its arrow points toward the node and negative when it points away: Σₖ Iₖ = 0. Suppose two arrows point into node P carrying I₁ = 2.0 A and I₂, and one arrow points out carrying I₃ = 0.8 A. Then 2.0 + I₂ − 0.8 = 0 gives I₂ = −1.2 A. The minus sign is information, not an error: 1.2 A really flows out along that branch, opposite to the arrow drawn. Two disciplines follow. Never redraw an arrow part-way through a solution to make a number come out positive, because every other equation was written against the old convention. And never quote a magnitude while discarding the sign, because the sign is the only thing carrying the direction.

Nothing in the derivation required the region to be a point. Draw any closed boundary that does not cut through a capacitor and the currents crossing it must still sum to zero, because what it encloses stores no appreciable charge either. Two consequences are useful. First, a sealed sub-circuit crossed by only two wires must carry the same current in one and out the other, however complicated its insides are, which is exactly why elements joined in series share one current. That is a junction-rule result, not a separate law. Second, a boundary that swallows several nodes and the wires between them yields one equation that is the sum of those nodes' equations, with the internal currents cancelling out because each crosses the boundary zero times. Engineers call this a supernode and reach for it when an awkward element sits between two nodes.

Write the junction equation at every node of a network and add them all together. Every branch runs between exactly two nodes, so its current appears twice in that total: once positive, at the node it points toward, and once negative, at the node it points away from. The whole sum collapses to 0 = 0. So the last node's equation is always a consequence of the others and supplies no new information. With n nodes you may keep n − 1 of them and must find the rest of your equations elsewhere. A bridge network with four nodes and six branches makes the count concrete: six unknown branch currents, three independent junction equations, and therefore 6 − 4 + 1 = 3 independent loops required. Do still write the leftover node equation down at the end, not as an equation but as a check: a solution that fails it has an arithmetic error in it.

An ammeter measures the current through a branch, so it must be put in series: break the branch and insert it, choosing a meter whose burden resistance is small compared with the branch. Connecting an ammeter across a branch instead of into it puts a near short circuit between two points meant to be at different potentials, and blows the fuse. A clamp meter reads a single conductor without breaking it, which is often easier. Then test the rule on three branches meeting at a node: 2.48 ± 0.02 A and 1.26 ± 0.02 A in, 3.71 ± 0.03 A out. The residual is 2.48 + 1.26 − 3.71 = 0.03 A, while the combined uncertainty is √(0.02² + 0.02² + 0.03²) = 0.041 A. The residual is smaller, so the readings are consistent. A residual several times the combined uncertainty is not a failure of the rule but evidence of a branch you have not counted: an earth return, a cable shield, a leakage path.