# Source Transformation to find i_x

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1. Oct 14, 2014

### 4Mike

1. The problem statement, all variables and given/known data
I'm to use source transformation to find the current through the 24 Ohm resistor

2. The attempt at a solution
I used source transformation on the left 12V source and got a .5A current upwards. The 24 and 30 ohm resistor are in parallel so I found an equivalent resistance of 13.33.
From here i used source transformation again and ended up with a voltage source of 6.67V on the left in series with the 13.33 and 60 ohm resistors.
Here's where I'm a little lost, with the dependent current source on the right, is it safe to assume that it just becomes Vx with the 10 Ohm resistor in series?
If so, then one of the equations I was able to find using KVL was -6.67 + 83.33ix + Vx = 0

Any idea's on how I can find the next equation that includes Vx?

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2. Oct 14, 2014

### Staff: Mentor

Hi 4Mike, Welcome to Physics Forums.

Since the dependent source depends upon the current ix, it's not a good idea to transform away ix so that you can't get at it when writing your equations. If you transform the 12 V source and its 24 Ohm series resistor, ix disappears from the circuit.

Instead, why don't you start at the other end of the circuit, transforming the dependent current source and "swallowing up" the components in between? Yes, you'll have to carry the ix along as part of the expression for any new sources' value.

Last edited: Oct 14, 2014
3. Oct 14, 2014

### 4Mike

Thanks for the quick response!

Is it right that when I transform the 0.7ix current source it turns into Vx? What happens to the 0.7?

4. Oct 14, 2014

### Staff: Mentor

Vx would be a new variable. How would it relate to the circuit? You would have to define it it terms of the given circuit parameters. If you just declare the new voltage source as Vx it won't contain any information about where it came from.

What you want to do is use the 0.7ix expression as the "value" of the current source and use it in the transformation of the current source into a voltage source when you construct the Thevenin equivalent.

5. Nov 3, 2014

### 4Mike

I ended up getting the correct answer. I didn't realize that when I used source transformation, the ix variable remained the same, only the coefficient changed. Thanks for the help!