CR · DC analysis

Kirchhoff’s Current Law

At a point in a circuit, currents split apart and merge. The simple fact that no charge piles up at that point becomes a powerful rule tying the incoming and outgoing currents together, and this unit shows how to use that rule to find an unknown current.

Bring the node current sum to 0

Four branches meet at the node N in the middle, and the currents in three of them are already known. Counting current flowing in as + and current flowing out as −, change the current I4 in the fourth branch until the current flowing into the node equals the current flowing out. The sum of the currents is then exactly 0.

Unknown current I4 (in +)I4 = −3 A

Algebraic sum of node currents

ΣI = −5 A

(+5) + (−3) + (−4) + (−3)

Unbalanced

Charge does not pile up at a node

A node is just a point where wires meet; it has no room to hold charge. If more current flowed in than out, charge would pile up continuously by the difference and the potential would shoot up without limit, yet that never happens in practice. So the current flowing in and the current flowing out must be exactly equal at every instant; this is conservation of charge as it appears at a node.

Signs tie in and out together

Take incoming current as + and outgoing current as −, and the statement that in equals out tidies into a single line: the sum of all currents, signs included, is 0. ΣI = 0. However many branches there are, and however the current directions were assumed at the start, the equation works out by itself as long as the signs are kept consistent. If a current actually flows the other way, its value simply comes out negative.

One equation solves an unknown current

If a node has only one unknown current, the single equation ΣI = 0 gives its value at once. Add up the known currents with their signs, and the unknown current is whatever makes that sum 0. In this figure, +5, −3, and −4 add to −2, so the fourth current must be +2. When solving a whole circuit, writing a KCL equation like this at every node is the starting point of nodal analysis.

ObserveΣI = 0

The algebraic sum of the node currents is 0.

ChooseI_in ? I_out

The current in equals the current out.

Fill in5 − 3 − 4 = ?

The known currents add to −2.

On your ownI4 = ?

The unknown is the value that makes the sum 0.

Back to the first screen

As you changed the unknown current, the algebraic sum at the node started with one sign, crossed 0, and passed to the opposite sign, and at the point where the sum was exactly 0, I4 was +2A. That is the value that fills in −2, the sum of the known +5, −3, and −4. A single simple fact, that charge does not pile up at a node, pinpointed the unknown current in one step.

Kirchhoff's current law (KCL) states that the current flowing into a node equals the current flowing out. Taking in as + and out as −, ΣI = 0: the sum of the currents, signs included, is 0. Because a node cannot store charge (conservation of charge), the law always holds, and writing this equation at every node is the foundation of nodal analysis.