# Kirchhoff’s Current Law (KCL) Kirchhoff’s Circuit Laws

**“Kirchhoff’s current law states that the algebraic sum of the currents that arrive at a point is equal to zero.”**

It is the sum of the currents that flow into a terminal is equal to the sum of the currents that leave it.

Let’s consider five currents arriving at a common terminal or node. The sum of the currents flowing into the nodes is (I_{1} +I_{3}) while the sum that leaves it is (I_{2}+I_{4}+I5). Applying Kirchhoff’s Current Law, we can write

**I _{1} +I_{3} = I_{2}+I_{4}+I5**

## Currents, Impedance And Associated Voltages

Consider an impedance Z, carrying a current I, connected between two terminals marked 1 and 2.

A voltage E_{12} having a magnitude IZ, will appear across the impedance. The question of polarity now arises. Is E_{12} equal to **+IZ** or **-IZ**? The question is resolved by the following rule.

When moving across an impedance Z in the same direction as the current flow I, the associated voltage IZ is preceded by a positive sign. Thus, we write E_{12} = +IZ. Conversely, when moving across an impedance against the direction of current flow, the voltage IZ is preceded by negative sign. Thus, E_{21} = -IZ. The current can be AC or DC and the impedance can be resistive (R) inductive (jX_{L}) or capacitive (jX_{C}).

In most circuit it is impossible to predict the actual direction of current flow in various circuit elements. Consider the circuit in which two known voltage sources E_{13} and E_{24} are connected to four known impedance’s Z_{1},Z_{2},Z_{3} and Z_{4}. Because the actual directions of current flows are presently unknown, we simply assume arbitrary directions. It is a remarkable fact that no matter what directions are assumed, the final outcome after solving the equations is always correct.

Loop** 2312**, starting with node 2 and moving clockwise:

**+I _{4}Z_{4} + E_{31} – I_{1}Z_{1} = 0**

Voltage I_{4}Z_{4} is preceded by a (+) sign, because in going around the loop we are moving in the direction of I_{4}. On the other hand, voltage I_{1}Z_{1} is preceded by a negative sign because we are moving against the direction of I_{1}.

Loop** 3423**, starting with node 3 and moving counterclockwise:

**+I _{3}Z_{3} – I_{2}Z_{2} + I_{4}Z_{4} = 0**

Voltages I_{3}Z_{3} and I_{4}Z_{4} are preceded by a (+) sign, because in going around the loop we are moving in the direction of the respective currents. Voltage I_{2}Z_{2} is preceded by a negative sign because we are moving against current I_{2}.

Loop **242**, starting with node 2 and moving clockwise:

**E _{24} – I_{2}Z_{2} = 0**

Kirchhoff’s Current Law at node 2

**I _{5} = I_{1}+I_{2}+I_{4}**

Kirchhoff’s Current Law at node 3

**I _{4}+I_{1} = I_{3}**