Kirchhoff's Laws

Node Law - Kirchhoff’s Current Law (KCL)

• The sum of currents at a node is zero.

$$I_1 = I_2 + I_3$$

Mesh Law - Kirchhoff’s Voltage Law (KVL)

• The sum of voltage drops around a closed mesh is zero.

$$\sum_k U_k = 0$$

Example:

$$E_1 + R_1 I_1 - R_2 I_2 - R_3 I_3 - E_2 = 0$$

Writing Equations Using Kirchhoff’s Laws

• If a circuit has n nodes and b branches:
  • There are (n-1) independent node equations.
  • There are (b - (n-1)) independent mesh equations.

Example with a circuit

First we check for nodes and branches

In the example, there is 2 nodes and 3 branches, so we have:

$$n=2\\ b=3\\ (n-1) = 2 - 1 = 1\\ (b - (n-1)) = 3 - (2-1) = 3 - 1 = 1$$

Here is the node equation:

$$I = I_1 + I_2$$

And here is the 2 meshes equation:

$$E_1 - R_1 I_1 - R I = 0$$ $$E_2 - R_3 I_2 - R I = 0$$

Some with have now 3 equations:

$$I - I_1 - I_2 = 0\\ E_1 = R_1 I_1 + R I\\ E_2 = R_2 I_2 + R I$$

Then we extract $I_1$ and $I_2$ from the equations:

$$I_1 = \frac{E_1 - R I}{R_1}\\ I_2 = \frac{E_2 - R I}{R_2}$$

Then we can replace $I_1$ and $I_2$ in the first equation:

$$I = \frac{E_1 - R I}{R_1} + \frac{E_2 - R I}{R_2}$$ $$I = \frac{E_1}{R_1} - \frac{R I}{R_1} + \frac{E_2}{R_2} - \frac{R I}{R_2}$$ $$I = \frac{E_1}{R_1} + \frac{E_2}{R_2} - \frac{R I}{R_2} - \frac{R I}{R_1}$$ $$I = \frac{E_1}{R_1} + \frac{E_2}{R_2} - I \left(\frac{R}{R_1} + \frac{R}{R_2}\right)$$

Then we can solve for $I$:

$$I \left(1 + \frac{R}{R_1} + \frac{R}{R_2}\right) = \frac{E_1}{R_1} + \frac{E_2}{R_2}$$

Finally, we can express $I$:

$$I = \frac{ \frac{E_1}{R_1} + \frac{E_2}{R_2}}{1 + \frac{R}{R_1} + \frac{R}{R_2}}$$