Norton and Thevenin theorems

Norton theorem

Any electric circuit that contains only resistors, current generators or voltage generators can be replaced by a practical current generator at any two chosen points, thus simplifying the network for further calculations.

Thevenin theorem

Any electric circuit that contains only resistors, current generators or voltage generators can be replaced by a practical voltage generator at any two chosen points, thus simplifying the network for further calculations.

Why substitute?

The equivalent circuit connected the same load resistor has the same values for voltage, current and power. With a single practical generator instead of a complex circuit, further calculations are much simpler.

Substitution steps

1. Replacement of generators and resultant resistance

All voltage generators are replaced by a short circuit and the current generators by an open circuit, and then the resultant resistance is calculated from the two points. This is the Norton or Thevenin resistance.

2. Determining the voltage between the points

The output voltage can be calculated, for example, using the voltage divider formula, or by any other method. It needs to determined what voltage a voltmeter would show if placed between the two points in the circuit.

3. Calculation of the Norton current

Knowing the voltage and resistance, the current is easy to calculate.

$I_{n} = \frac{V_{t h}}{R_{n}}$

Example task 1

$V_{1} = 10 [V]$

$R_{3} = 16 [Ω]$

$R_{1} = 12 [Ω]$

$R_{4} = 18 [Ω]$

$R_{2} = 14 [Ω]$

Solution:

Following the steps above, first calculate the resultant resistance between the two points without the voltage source, then calculate the output voltage in the original circuit, and finally calculate the Norton current.

$R_{A B} = (R_{1} + R_{2}) \times (R_{3} + R_{4})$

$R_{A B} = R_{t h} = R_{n} = 14.73 [Ω]$

$V_{o u t} = V_{1} \cdot \frac{R_{3} + R_{4}}{R_{1} + R_{2} + R_{3} + R_{4}}$

$V_{o u t} = V_{t h} = 5.667 [V]$

$I_{n} = \frac{V_{t h}}{R_{n}} = 0.385 [A]$

Results:

$V_{t h} = 5.667 [V]$

$R_{t h} = 14.73 [Ω]$

$I_{n} = 0.385 [A]$

$R_{n} = 14.73 [Ω]$

Example task 2

$V_{1} = 10 [V]$

$R_{3} = 16 [Ω]$

$R_{1} = 12 [Ω]$

$R_{4} = 18 [Ω]$

$R_{2} = 14 [Ω]$

$R_{5} = 20 [Ω]$

Solution:

The steps are the same, except that the resistance on the output wire of point A is not included in the voltage divider formula. This is because a voltmeter has infinite resistance, the output voltage is not affected by R5.

$R_{A B} = (R_{1} + R_{2}) \times (R_{3} + R_{4}) + R_{5}$

$R_{A B} = R_{t h} = R_{n} = 34.73 [Ω]$

$V_{o u t} = V_{1} \cdot \frac{R_{3} + R_{4}}{R_{1} + R_{2} + R_{3} + R_{4}}$

$V_{o u t} = V_{t h} = 5.667 [V]$

$I_{n} = \frac{V_{t h}}{R_{n}} = 0.163 [A]$

Results:

$V_{t h} = 5.667 [V]$

$R_{t h} = 34.73 [Ω]$

$I_{n} = 0.163 [A]$

$R_{n} = 34.73 [Ω]$