Source Transformation to find i_x

[4Mike]

Homework Statement

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 V_x with the 10 Ohm resistor in series?
If so, then one of the equations I was able to find using KVL was -6.67 + 83.33i_x + V_x = 0Any ideas on how I can find the next equation that includes V_x?

 

[gneill]

Hi 4Mike, Welcome to Physics Forums.

Since the dependent source depends upon the current i_x, it's not a good idea to transform away i_x 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, i_x 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 i^x along as part of the expression for any new sources' value.

 

[4Mike]

Thanks for the quick response!

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

 

[gneill]

Thanks for the quick response!

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

V_x 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 V_x it won't contain any information about where it came from.

What you want to do is use the 0.7i_x 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.

 

[4Mike]

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