# Theorems of superposition, Thévenin and Norton

In this post are shown theorems of superposition, Thévenin and Norton. Only DC circuits with resistors are used as examples.

### Superposition theorem

Before to know Thévenin and Norton, is necessary to learn this theorem. It is useful to analyze circuits with many voltage and current sources. In this method, voltage source can be replaced by short-circuits and a current source can be replaced by an open circuit.

#### Example

How to use this theorem? Let’s use this circuit as example. We have to calculate the voltage in current source I1.

Step 1: Calculate the desired value with only 1 source. Let’s short-circuit voltage sources and analyze only with current sources. If you wish, you can start with one of voltage sources and put current in open circuit.

Step 2: Find the desired voltage value in step 1 circuit. It is 16 V.

Step 3: Analyze the circuit with another source and calculate result.

Step 4: The same of step 3, but with other source.

Step 5: Algebraic sum of results from steps 2, 3 and 4.

$16-3,33-2=10,67 V$

### Thévenin theorem

This theorem is very useful to simplify complex circuits to facilitate analysis and apply other theorems. Any circuit can be replaced by equivalents voltage source and resistance.

Procedure to apply this theorem.

Step 1: Define which part of circuit must get Thévenin equivalent. This part must have two terminals.

Step 2: Remove any component between these terminals which are not part of circuit to obtain Thévenin equivalent.

Step 3: Find Thévenin’s equivalent resistance, transform all voltage sources in short-circuit and current sources in open circuit.

Step 4: Restore sources in original positions. Calculate voltages between terminals of equivalent circuit and make step 2 again. This is Thévenin voltage.

#### Example

This circuit will be the example.

Steps 1 and 2: We must obtain Thévenin equivalent in this part of circuit.

Step 3: Calculating Thévenin’s equivalent resistance $R_{Th}$.

$\frac{5600\cdot 2200}{5600+2200}=1579,48 \Omega$

$1579,48+3300=4879,48\Omega$

$\frac{4879,48\cdot 6800}{4879,48+6800}=2840,91\Omega$

$R_{Th}=2840,91+1200=4040,91\Omega$

Step 4: Calculate Thévenin voltage $V_{Th}$. In this case is equal to 9.73 V.

This is the Thévenin equivalent circuit.

### Norton Theorem

In this theorem, obtain equivalent resistance is the same method of Thévenin. An option is convert Thévenin equivalent circuit in Norton simply using source conversion method shown in “Circuit analysis (Part 1)”.

$I_{N}=\frac{V_{Th}}{R}$

Where $I_{N}$ is Norton current and $R$ is equivalent resistance equal to both Thévenin and Norton.

Other option is repeat steps 1, 2 and 3, which are the same of previous method. But in step 4, must calculate Norton current In, which is a short-circuit current in terminals of equivalent circuit.

#### Example

This is example circuit, we have to calculate Norton current which passes through RL.

To find Norton current, RL must be replaced by short-circuit and calculate In current.

Norton current is 5 A.

The equivalent circuit of example above.