Source Transformation to find i_x

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Discussion Overview

The discussion revolves around using source transformation techniques to find the current through a 24 Ohm resistor in a circuit involving dependent sources. Participants explore the implications of transforming sources and the relationships between variables in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant describes their initial approach using source transformation on a 12V source, resulting in a calculated current and equivalent resistance.
  • Another participant advises against transforming away the dependent current source, suggesting that it is better to start from the dependent source and retain the variable ix in the equations.
  • A participant questions how the dependent current source transforms into a voltage source and what happens to the coefficient associated with it.
  • There is a clarification that the variable ix remains unchanged during the transformation, although its coefficient may vary.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to handle the dependent source and the variable ix. There is no consensus on a single method, and the discussion reflects multiple perspectives on the transformations involved.

Contextual Notes

Participants highlight the importance of maintaining the relationship between variables during transformations and the need to define new variables in terms of existing circuit parameters. There are unresolved aspects regarding the transformation process and its implications for the dependent source.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with circuit analysis, particularly those learning about source transformations and dependent sources in electrical engineering.

4Mike
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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 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 = 0Any ideas on how I can find the next equation that includes Vx?
 

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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:
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?
 
4Mike said:
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?
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.
 
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!
 

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