# Superposition Practice

Practice question for ECE201: "Linear circuit analysis I"

By: Chinar Dhamija

Topic: Superposition

## Question

Determine the value of $I_x$ using superposition.

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.

$I_x = -1 + 2$
$I_x = 1A$