# Source transformation problem

Thread Starter

#### kaiosama

Joined Dec 6, 2010
29 The first way I do this, is transform the 12V and 16V voltage sources into current sources with parallel resistor, which gives me (after combining parallel resistors and current sources) R_equivalent = 5.714 ohm and current source of 4 amps, which is equivalent to voltage source of 22.85 volts with the same R_equivalent
So, answer is: 22.85 volts with 5.714 ohm resistance in series.

However, I have tried something else more complicated just to see if it would work, but It's giving me another result. I begin by transforming the 3 amps current source into a 30V voltage source in series with the 10 ohm resistor. Then I transform the 12V and 16V voltage sources into current sources with parallel resistance. All these 4 elements are thefore in parallel. 0.6 Amps current source, 0.4 amps current source, 20 ohms resistor, 40 ohm resistor.

Combining the two resistors I get one single resistor of 13.33 ohms, and the current sources are added to give 1 Amp. The current source and the resistor are in parallel. I transform this to voltage source of 13.33 Volts with resistance of 13.33 ohm

We now have 30V voltage source, 10 ohms resistor, 13.33 volts voltage source, 13.33 ohms resistor, all in series. Simplifying, this gives 43.33 volts voltage source, 23.33 ohms resistor. Which is not the same as the first answer.

What did I do wrong?

Thank you very much.

#### hgmjr

Joined Jan 28, 2005
9,029
Maybe you can sketch your alternate solution so that we can see what you are describing.

hgmjr

Thread Starter

#### kaiosama

Joined Dec 6, 2010
29
Hello, here is the sketch: The first circuit is after transforming the initial 3 amp current source with 10 ohms resistor in parallel into a 30 volts voltage source with 10 ohms resistor in series and transforming the voltages sources into current sources.. you get the point.

Thank you very much.

#### JoeJester

Joined Apr 26, 2005
4,390
I would recommend you draw an extended line from the top and bottom of the 40 ohm resistor and label them A and B. That is your reference for voltage and current.

Then do your drawings, one by one, and you will find the voltage and current at point A and B will be equal as you configure the circuit in any manner of sources.

Thread Starter

#### kaiosama

Joined Dec 6, 2010
29
I would recommend you draw an extended line from the top and bottom of the 40 ohm resistor and label them A and B. That is your reference for voltage and current.

Then do your drawings, one by one, and you will find the voltage and current at point A and B will be equal as you configure the circuit in any manner of sources.
I'll try this right now, but I just can't see what is wrong with the second solution... to me all the steps seem perfectly valid. Why exactly is it wrong?

#### steveb

Joined Jul 3, 2008
2,436
I'll try this right now, but I just can't see what is wrong with the second solution... to me all the steps seem perfectly valid. Why exactly is it wrong?
The last step of saying that everything is in series is the error. The entire idea of making equivalent power sources is to make them equivalent in the sense that an external load resistor would not "know" the difference. A load placed on the output makes it clear that your two voltage sources are in parallel in the final step. I suspect that when you come back, you will already know this answer.

With that said, I fault the statement of the problem as much as I fault you for not thinking about what the ultimate goal is. They did not tell you the ultimate goal because they did not identify the output terminals of the system you are supposed to equate. If you take the output directly across the 16 V source, you would get a different answer. And, if you take the output across the 40 Ohm resistor, you get an even different answer, still.

Thread Starter

#### kaiosama

Joined Dec 6, 2010
29
Ohhhh, I see. Thank you very much.

#### rsashwinkumar

Joined Feb 15, 2011
23
First let us assign the top node in ur original circuit as 'A' and bottom node as 'B'. Now your first solution has given the thevinin's equivalent circuit as viewed across 'AB'.

In your second solution, just before last step, you have got 13.33 V and 13.33 ohm across 'AB'. Now u cant just add the resistances and voltages, coz you are finding equivalent circuit across those two nodes. Hence to find thevinin's eq., resistance, short the voltage sources and you will find across 'AB' 13.33 ohm and 10ohm will be in parallel which gives 5.17 ohm, similarly, find the voltage across 'AB'

V(A-B) = I*13.33(ohm) + 13.33V

I=(30-13.33)/(10+13.33) = 0.714 A

Now V(A-B) = 0.714*13.33 + 13.33 = 22.85V
which matches your first solution.

REMEMBER :

When you transform sources just have in your mind that you are finding an equivalent circuit when viewed across some node pair, and be cautious about it. So mark the nodes
and proceed. You wont make these kind of mistakes.

To learn more on reduction of Thevenin's eq., ckt., , refer ' CIRCUIT ANALYSIS BY WILLIAM HAYT' (an excellent book on circuits).

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