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 (I1 +I3) while the sum that leaves it is (I2+I4+I5). Applying Kirchhoff’s Current Law, we can write
I1 +I3 = I2+I4+I5
Currents, Impedance And Associated Voltages
Consider an impedance Z, carrying a current I, connected between two terminals marked 1 and 2.
A voltage E12 having a magnitude IZ, will appear across the impedance. The question of polarity now arises. Is E12 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 E12 = +IZ. Conversely, when moving across an impedance against the direction of current flow, the voltage IZ is preceded by negative sign. Thus, E21 = -IZ. The current can be AC or DC and the impedance can be resistive (R) inductive (jXL) or capacitive (jXC).
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 E13 and E24 are connected to four known impedance’s Z1,Z2,Z3 and Z4. 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:
+I4Z4 + E31 – I1Z1 = 0
Voltage I4Z4 is preceded by a (+) sign, because in going around the loop we are moving in the direction of I4. On the other hand, voltage I1Z1 is preceded by a negative sign because we are moving against the direction of I1.
Loop 3423, starting with node 3 and moving counterclockwise:
+I3Z3 – I2Z2 + I4Z4 = 0
Voltages I3Z3 and I4Z4 are preceded by a (+) sign, because in going around the loop we are moving in the direction of the respective currents. Voltage I2Z2 is preceded by a negative sign because we are moving against current I2.
Loop 242, starting with node 2 and moving clockwise:
E24 – I2Z2 = 0
Kirchhoff’s Current Law at node 2
I5 = I1+I2+I4
Kirchhoff’s Current Law at node 3
I4+I1 = I3