Kirchhoff’s Voltage Law
Once around, the voltages sum to zero
One source and three resistors form a loop. The source raises the voltage; the resistors drop it. Slide the unknown voltage drop V3 until the sum of voltages once around the loop is exactly zero.
The potential returns to itself
Every point in a circuit has its own potential. Moving from a point to its neighbor along the loop, the potential goes up or down at each step. But once you go all the way around back to the start, that point’s potential must be exactly what it was when you left — a single point cannot hold two potentials at once. So all the rises and falls along the way add up to zero.
Rises are +, drops are −
Travel the loop in one direction and mark a rise in potential as +, a fall as −. Coming out of a source (EMF) at its + terminal raises the voltage; passing through a resistor in the direction of current drops it. Adding every term with its sign gives the one line ΣV = 0. Put another way, the rises supplied by sources equal the drops taken by the resistors.
One equation solves an unknown voltage
If a loop has a single unknown voltage, the one equation ΣV = 0 gives it directly. Here the source raises +12 V and two resistors drop 5 V and 4 V, so the remaining third resistor’s drop must be 12 − 5 − 4 = 3 V. To solve a whole circuit, writing a KVL equation for each loop and a KCL equation for each node and solving them together is the heart of mesh and nodal analysis.
Back to the first screen
As you slid the unknown drop, the loop’s voltage sum started on one side, crossed zero, and went over to the opposite sign, and where the sum was exactly zero, V3 was 3 V. The 12 V raised by the source is shared as 5 V and 4 V by two resistors, and the remaining 3 V is taken by the third. The simple fact that a loop returns to where it began pinned the unknown voltage at once.