# Superposition Practice

**Practice question for ECE201: "Linear circuit analysis I" **

By: Chinar Dhamija

Topic: Superposition

## Question

Determine the value of $ I_x $ using superposition.

### Answer

The circuit has two sources, a voltage and a current source. In order to find $ I_x $ we need to look at two instances.

1. Deactivate the voltage source (V = 0)

2. Deactivate the current source (I = 0)

Let's find $ I_x $ first by deactivating the voltage source as seen in the picture below.

We can find $ I_x $ by using loop analysis. The left loop can be $ I_1 $ and the right loop can be $ I_2 $.

When doing loop analysis we can come up with the following equation:

$ \begin{align} I_1: 10*I_1 + 20(I_1 - I_2) = 0\\ 30*I_1 = 20*I_2\\ I_2: I_2 = 3A\\ \end{align} $

After finding $ I_1 $ and $ I_2 $ we can find $ I_x $.

$ \begin{align} I_x = I_1 - I_2\\ I_x = 2 - 3\\ I_x = -1\\ \end{align} $

That is what we get for $ I_x $ if the voltage source is deactivated.

Now we need to do the same procedure but this time we will deactivate the current source; therefore, we will get the following circuit:

We can use source transformation to turn the voltage source into the current then use current division to solve for $ I_x $.

After doing source transformation we get the following circuit:

Now use current division to find $ I_x $.

$ \begin{align} I_x = \frac{1/10}{1/10 + 1/20} * 3\\ I_x = 2\\ \end{align} $

Finally, the last step would be to add up the two $ I_x $ values we obtained by deactivating the voltage and current sources.

$ I_x = -1 + 2 $

$ I_x = 1A $

## Questions and comments

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