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Kirchhoff’s Laws

Using Kirchhoff’s Laws we can analyze one electric circuit and determinate the values of some physics units current , voltages ,resistance .
There are two Kirchhoff’s Laws : for one circuit loop and for one circuit node .
The first Kirchhoff’s law is for one circuit node and says : The algebraic sum of all currents entering and exiting a node must equal zero .

first kirchhoff law

I1+I2-I3=0

Example of the first Law .

We  have I1 = 2A I2 = 3A and I3 = 6A and we need to find the value of I4

We choose the way for I4  current ( arbitrary ) like in the figure bellow  and we have :

example of kirchhof's law

I1 + I2 - I3 + I4 = 0
=> I4 = I3 - I2 - I1 = 6 - 3 - 2 = 1A =>
The way for I4 is correct . If the value obtained for I4 is with minus the way chosed for I4 is reverse .

 

A circuit node is one point from the electric circuit where is connected more than two sides of the circuit ( minim three ) .
Kirchhoff’s second law says : The algebraic sum of all voltages in a loop must equal zero .
A circuit loop is a part from the circuit with at least two sides which makes a closed polygonal line.
Example for the second Law .
kirchhoff's second law

U1 + E1 = R1I1
U2 + E2 = R2I2
U3 + E3 = R3I3
U4 + E4 = R4I4
U1 - U2 + U3 + U4 = 0
E1 - E2 - E3 - E4 = R1I1 - R2I2 + R3I3 + R4I4

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