# Thevenin Norton Problems- check work

Discussion in 'Homework Help' started by RoKr93, Jun 23, 2013.

1. ### RoKr93 Thread Starter New Member

Jun 17, 2013
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I'm reviewing for an exam by working some problems and just have a couple of problems that I'd like to verify my work for: I'd appreciate if someone could check my work for these two problems:

Problem 1:

Work (using source transformations):

Problem 2:

Work (using mesh analysis- my drawing is kind of hard to read. The top left loop is i1, the right loop is iA, and the lower left is is2.):

Thank you!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I haven't gone through all the problems. In the first [question 7] your initial reduction by source transformation already has an error. The transformed left hand branch shows Vs1 in series with the transformed resistances. This is where the error lies. The transformed source will not simply be equal to Vs1 [48V] - rather you will need to find the voltage drop across CD with the source Vs1 orientated as shown. This will require a the use of a couple of voltage divider exercises to find (say) the individual volt drops across R1 & R3 or R2 & R4 to find VCD.

3. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
One of the nice things about circuit analysis (and, in fact, many engineering problems) is that the correctness of an answer can be determined from the answer itself.

So take your answer and use it to compute quantities in the original circuit and see if they agree with the given values.

So, for instance, in your first circult you came up with a Thevenin equivalent circuit of 89.6V and 10kΩ. So what will the current and voltage at the terminals be if you put a 10kΩ load across the terminals?

Using these values, work your way back and see what you get for the voltage an Node C? Using this, then see if the currents you get through the bridge containing Vs1 and then see if KCL is satisfied on Node C and consistent with Is2.

4. ### WBahn Moderator

Mar 31, 2012
18,079
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In order to claim that two components are in parallel, they must have the same voltage across them. Does R1 have the same identical voltage across it as R2? If not, then they are not in parallel and can't be combined that way.

5. ### RoKr93 Thread Starter New Member

Jun 17, 2013
24
0
I'm not wrapping my head around what's going on between C and D correctly.

I know that if you remove the voltage source, you have R1||R2 and R3||R4 in series. That's how I know Rth is equal to 10 k. I realize now that this is NOT the case if the voltage source is present. What I don't understand is how to use voltage division in that situation. It doesn't seem like any of these resistors are in series since C is attached between R3 and R4 and D is attached between R3 and R2.

I tried redrawing the diagram and it seems that perhaps R1||R3 and R2||R4 are in series, which would make voltage division possible...but that seems to conflict with the placement of C and D and I'm not sure how it gets me closer to finding the total voltage drop.

6. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Not true. If you "remove" the voltage source, then you have (R1+R3) in parallel with (R2+R4). If you set the voltage source to 0V (which is very different from removing it), THEN you have (R1||R2)+(R2||R4).

So perhaps you can't use simple voltage division. By the time you are dealing with Thevenin and Norton equivalent circuits, you should have a pretty full bag of circuit analysis tools at your displosal, including nodal, mesh, and superposition.