# Why this source transformation leads to different result?

I've been stuck for this source transformation, it seems quite simple, but I just can't tell why the results are different.

As pictures below, I transformed (-3mA par 5Kohm) to (-15V series 5Kohm), but Vab just not the same. I appreciate if anyone can help me, thank you.

I am trying to think the way you simplifed the circuit but got stuck at the final step. I am fluent with solving problems by applying Nodal, Mesh and Thevenin methods(Vth, Isc, Rth), but lacking of schematic sense. So I try to solve this by cogitation instead of calculation only. Below is what I think, I know I am totally wrong, but can you give me some advise to clear the concept? Thank you very much.

• Is your first image and the blue colored diagram the problem you were given and the red ink being your work product on the basis of that problem? Further, in the second imagine, is the blue ink from the person who corrected you? Or? – jonk May 3 '17 at 17:09
• Hi, Jonk, the first image is the problem I was given. The second image (both blue and red ink) is what I try to do source transformation. – Joey Chen May 3 '17 at 17:15
• What then is the second blue diagram about? – jonk May 3 '17 at 17:16
• (The reason I asked is that I think you correctly solved the first blue diagram problem since the output voltage between A and B actually is -3.75 V, for the first diagram.) – jonk May 3 '17 at 17:18
• The top diagram is correctly solved but, it's not representative of the bottom diagram i.e. you can't do the voltage to current source transform that way @jonk – Andy aka May 3 '17 at 17:19

Joey, I'll write up another short way of thinking about things here. Follow the blue arrows for each step-wise transformation.

simulate this circuit – Schematic created using CircuitLab

1. $V_1$ and $R_1$ have no impact at all on the node controlled by $V_2$. You can just eliminate them, entirely.
2. Re-arrange the schematic a little to highlight a Thevenin opportunity.
3. Apply the Thevenin transformation.

At this point, it is pretty easy. You know you must "pull" $3\:\textrm{mA}$ through $3750\:\Omega$, leading to a voltage drop of $11\frac{1}{4}\:\textrm{V}$. If you apply that drop to $V_{TH}$, you have your answer.

Another way to do this is to simply use nodal analysis on node A:

$$\frac{V_A}{R_2}+\frac{V_A}{R_3}+3\:\textrm{mA}=\frac{30\:\textrm{V}}{R_2}$$

or,

$$V_A=\left(30\:\textrm{V}-I_1\cdot R_2\right)\cdot\frac{R_3}{R_2 + R_3}$$

That solves out as $V_A=-3.75\:\textrm{V}$ as expected.

Just another way of proceeding through a problem like this. You need to acquire lots of different tools that you are fluent with so that you can make quick work of any problem you see by pulling out the right tools for the task. So this is just another way to add to others for you.

Note: I've added some boxes to help a little on the earlier schematic. You might be able to see why I can perform the transformation this way. I'm also adding another schematic series to show you how to perform your transformation correctly:

simulate this circuit

Here, the final result is just:

\begin{align*} V_{TH_{FINAL}}=V_A&=\frac{30\:\textrm{V}\cdot 5\:\textrm{k}\Omega-15\:\textrm{V}\cdot 15\:\textrm{k}\Omega}{15\:\textrm{k}\Omega+5\:\textrm{k}\Omega}=-3.75\:\textrm{V}\\\\ R_{TH_{FINAL}}&=\frac{15\:\textrm{k}\Omega\cdot 5\:\textrm{k}\Omega}{15\:\textrm{k}\Omega+5\:\textrm{k}\Omega}=3.75\:\textrm{k}\Omega \end{align*}

Which is just either one of these two:

simulate this circuit

• Thank you, Jonk. I am trying to think the way you simplifed the circuit but got stuck at the final step. I am fluent with solving problems by applying Nodal, Mesh and Thevenin methods(Vth, Isc, Rth), but lacking of schematic sense. So I try to solve this by cogitation instead of calculation only. Below is what I think, I know I am totally wrong, but can you give me some advise to clear the concept? Thank you very much. – Joey Chen May 3 '17 at 21:21
• Dear Jonk, I can't embed a new image either here or in my question, so I embed it in your answer, hoping it won't be offended. – Joey Chen May 3 '17 at 21:54
• @JoeyChen You can simply add an answer here. – jonk May 3 '17 at 21:56
• @JoeyChen Added extras. – jonk May 3 '17 at 22:50
• Thank you, Jonk, that's crystal clear. This helps me clearing out voltage divider w/ 2 sources. – Joey Chen May 4 '17 at 8:51

You can't convert a voltage source to a current source the way you have done it because the 15 kohm in the middle limb becomes connected to the output (node a). You have to work from the left to the right. In other words you broke the rules of conversion by ignoring the 15 kohm's circuit position.

However, just stop and think about the circuit - the output connects to a 15 volt source via a 5 kohm resistor - all the components left of the 15 volt source are therefore meaningless to the analysis.

• Thank you for your kindly response, and truly the second part of your answer gives me a different point of view on the circuit : ) – Joey Chen May 3 '17 at 17:11