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**Home ⇒ Overview Courses ⇒ Electrical engineering ⇒ Superposition Theorem**

## Superposition Theorem to simplify electric networks

The superposition Theorem is used to **analyze networks** with **multiple energy sources** (current or voltage sources).

In principle, it involves considering the effect in terms of current or voltage of the individual current or voltage sources on their own and combining them at the end.

### Example - Simplifying according to the superposition theorem

The following circuit is given (figure a) with a voltage source and a current source. Your task: Calculate the individual currents of this circuit

Example - Simplifying according to the superposition theorem

The following circuit is given (figure a) with a voltage source and a current source. Your task: Calculate the individual currents of this circuit.

**Step 1: Viewing the voltage source on the left - current source on the right is "removed"**

Figure b shows the following:

I_{11} = I_{31} = V_{S} / R_{2} + R_{2} + R_{3} = 12 V / 60 Ω = 0,2 A

**Step 2: Viewing the current source on the right - voltage source on the left is removed**

Figure C shows the following:

I_{32} = I_{q} * R_{123} / R_{3} = 5A * 15 Ω / 30 Ω = 2,5A Current divider rule

I_{q} + I_{12} - I_{32} = 0 Junction rule

⇒ I_{12} = I_{32} - I_{q} = 2,5A - 5A = - 2,5A

Additional calculation: R_{12} = R_{1} + R_{2} = 30 Ω; R_{123} = R_{12} II R_{3} = 15Ω

**Step 3: Add up the individual effects of the sources**

I_{1} = I_{11} + I_{12} = 0,2A - 2,5A = -2,3A

I_{3} = I_{31} + I_{32} = 0,2A + 2,5A = 2,7A

Note: Because the current arrow of I_{1} points to the node, the value of I_{1} is negative.