Kirchhoff’s Voltage Law:
Kirchhoff’s voltage law states that the sum of the electrical potential differences is equal to 0. Simply put, in a closed circuit the incoming and outgoing voltages add up to 0. The following explanation refers to the drawing below with 2 resistors:
In a closed circuit, the total current through the circuit can be calculated using the voltage and resistance values.
-12 + 5 x I + 10 x I = 0
(-12 volts + 5 ohms x the current + 10 ohms x the current = equal to 0)
Because this is a series circuit, the resistances can be added together; 5 + 10 = 15Ω. The current I can then be calculated:
I= U / R
I = 12 / 15
I = 0.8A
The total current through the circuit is 0.8A. The current is the same from the battery as it is at R1 and R2. Now the voltage across the resistors must be calculated. The current and resistance values are known, so with Ohm’s law the voltage can now be calculated:
UR1 and UR2 are the voltages (U) across the resistors R1 and R2:
UR1 = I x R
UR1 = 0.8 x 5
UR1 = 4v
UR2 = I x R
UR2 = 0.8 x 10
UR2 = 8v
Now Kirchhoff’s law can be applied;
-Ubattery + UR1 + UR2 = 0
-12v + 4 + 8 = 0
U = 12v
R1 = 5 Ω
R2 = 10 Ω
I = unknown
This proves that the Kirchhoff voltage equation is correct, because if you start at the bottom left of the diagram at the battery, you begin at the negative terminal of the battery. That is why you start with -12. If you then continue reading the diagram (clockwise), you first arrive at the + of R1 and then at the + of R2. Hence, the minus of the battery voltage (entry at the voltage source) is equal to (plus) the sum of all (outgoing) loads. In this case the resistors. This can be a checking tool in complex diagrams where, for example, equivalent resistances have been calculated. By applying Kirchhoff’s Voltage Law, it can be checked whether the calculated data are correct.
Kirchhoff’s Current Law:
Kirchhoff’s current law states that all currents at a node add up to 0. All currents that enter the node must also leave it.
I1 + I2 + I3 + I4 = I5 (all currents leave the node via I5)
With multiple currents leaving the node, the diagram and formula become as follows:
I1 + I2 + I3 = I4 + I5 (the currents I1, I2 and I3 are distributed over I4 and I5).
