Position: Index > Electrical Circuit >

Kirchhoff’s Laws

2014-11-01 22:53  
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.

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 .

Bipolar transistor circuit diagramBipolar transistor symbol Bipolar transistor circuit diagramBipolar transistor symbol

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