Kirchhoff’s Current Law
Bring the node’s current sum to zero
Three currents at a node are known. Count flowing in as +, flowing out as −. Slide the fourth branch current I4 until what flows into the node equals what flows out, so the sum is exactly zero.
Charge does not pile up at a node
A node is just a point where wires meet — it has no capacity to store charge. If more current flowed in than out, charge would pile up by that difference every moment, and the potential would shoot up without bound. Since that never happens, the current in and the current out must be exactly equal at every instant. This is charge conservation showing itself at a node.
Signs tie in and out together
Take currents flowing in as + and out as −, and the statement “in equals out” tidies into one line: the signed sum of all currents is zero, ΣI = 0. However many branches there are, and however you first guessed the current directions, as long as you keep the signs consistent the equation works out by itself. If a current really flows the other way, it simply comes out negative.
One equation solves an unknown current
If a node has a single unknown current, the one equation ΣI = 0 gives its value directly. Add the known currents with their signs, and the value that makes the sum zero is the unknown. Here +5, −3 and −4 add to −2, so the fourth current must be +2. When solving a whole circuit, writing a KCL equation at each node like this is the starting point of nodal analysis.
Back to the first screen
As you slid the unknown current, the node’s signed sum started on one side, crossed zero, and went over to the opposite sign, and at the point where the sum was exactly zero, I4 was +2 A. That is the value filling the −2 left by the known +5, −3 and −4. The single simple fact that no charge piles up at a node pinned the unknown current at once.