# Source Transformation

1. Mar 13, 2014

### dwn

1. The problem statement, all variables and given/known data

Image Attached

2. Relevant equations

Ohm's

3. The attempt at a solution

Combined the two resistors in series : 250 + 550 = 800 kΩ
Source Transformation (Current Source): V = 140,000(2*10^-6)= 0.28 V
Combine the voltage sources : 6 - 0.28 = 5.72 V

But then I recreated the circuit with these figures and it just doesn't appear correct to me.
Voltage source = 5.72 V
Resistor = 800 + 140 = 940kΩ

Ans: V = 3.35 V R = 228.19 kΩ

I don't see how they're getting these figures.

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2. Mar 13, 2014

### dwn

Can someone please explain to me what is going on here?

3. Mar 13, 2014

### Staff: Mentor

The 250 k and 550 k resistors are not in series: there's another connection at the node where they join (the 1300 k resistor) so they cannot be in series.

Your transformation of the 140 k resistor and 2 uA source to a voltage source (Thevenin equivalent) is fine, and combining it with the 6 V source to yield a net 5.720 V source is good too. What's the Thevenin resistance for that new combined source?

Your next step should be to incorporate the 550 k resistor, so another Thevenin equivalent voltage and resistance should result. You should end up with the original -11 V source, the 1300 k resistor and the new Thevenin resistance and voltage all in series.

The steps are summarized in the following figure. Working from right to left, convert and combine/incorporate components into Thevenin equivalent models as you go:

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4. Mar 13, 2014

### dwn

gneill, thank you very much for the clear and concise explanation. However, I do have a question...I thought that we were supposed to create a "black box" and create an open circuit at the end (the node where the 250 and 550 R meet) which would make them in series...apparently this is not the case and I misunderstood.
Do you alway start from the outside and work your way towards the center?

5. Mar 13, 2014

### Staff: Mentor

You work in the direction that achieves your goal I know that doesn't seem to help much... But in this case you are apparently looking to find the power dissipated by the 1300 k resistor so you know that you need to leave that one alone --- you can't transform it away into the guts of an equivalent circuit (Thevenin or Norton) because then you couldn't write any equations about it to find current or voltage for it; once you transform-away a component it becomes inaccessible to further analysis. So leaving it alone, you look towards the rest of the circuit. Most of it lies to the right of the 1300 k resistor, so you start at the far end and work back. That usually works well as a general approach.

6. Mar 13, 2014

### dwn

My apologies for the poor quality in the photo. Where have I gone wrong in this problem? Still not arriving at the correct solution.

Note: all resistors are kΩ

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7. Mar 13, 2014

### Staff: Mentor

you've opted to transform to a Norton equivalent, but didn't include the 140 k resistance that is also in series with the voltage source. That means you would end up with something like this instead:

In this problem you have the option of just sticking with Thevenin equivalents all the way through.

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8. Mar 13, 2014

### dwn

I was able to figure out the resistors and come up with the correct Rx -- what method are they using to find the voltage?

I apologize for the series of questions with this problem, but circuits are not coming easily for me.

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9. Mar 13, 2014

### Staff: Mentor

Well, your thumbnail diagram shows a classic voltage divider scenario... that would yield the Thevenin voltage for that subcircuit.

No worries, that's why we're here

10. Mar 13, 2014

### dwn

Fantastic! Love it when this works. Starting the journey is a pain in the butt though.

Thanks for the help gneill