Use source transformation, find Vo

  • #1

Homework Statement



http://imageshack.us/a/img26/8403/homeworkprobsg28.jpg [Broken]


a. Use source transformations to find the voltage V0 in the circuit (green).

b. Find the power developed by the 250V voltage source

c. Find the power developed by the 8A current source


Homework Equations



V = IR

KVL, KCL,

voltage division: v = (R / total series r)(v in)

current division: i = (other resistance / math. sum of parallel r)(i in)

P = IV
P = (I^2)(R)



source transformation:

current to voltage, resistor moves up but value stays the same, and voltage replacing current source = (coefficient of resistor)*(coefficient of current source)

switch between

a current source in parallel with a resistor or

a voltage source in series with a resistor

The Attempt at a Solution



I think I sort of get the basic idea behind source transformation, but it seems more complicated for this problem. The voltage source is not in series with a resistor and vice versa for the current source.

I think I can / am supposed to do source transformation with the 8A and 125Ω to get this:

http://imageshack.us/a/img843/3033/homeworkprobsg28edit.jpg [Broken]

not sure if the 125Ω is supposed to be on the bottom though. Any tips?
 
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Answers and Replies

  • #2
gneill
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The current supply is not in parallel with the 125 Ω resistor; the 25 Ω resistor is in the way, and there's a junction between them (where the voltage source connects). So you can't transform that pair.

There are a couple of things going on in this circuit that are intended to make you think a bit. One is the 125 Ω resistor in parallel with the voltage source, and the other is the 10 Ω resistor in series with the current source.

In the case of the resistor and voltage source, what would be the potential across them if the resistor was changed to another value? Does the value of that resistance affect the operation of the rest of the circuit?

How about the resistor in series with the current source? Will changing its value affect the current through and delivered by that branch to the nodes where it connects?
 
  • #3
The 8A current down in that branch would still be 8A no matter what the resistance is.

Are you saying I can take out the 10Ω resistor?


And are you saying I can take out the 125Ω too?
 
  • #4
gneill
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The 8A current down in that branch would still be 8A no matter what the resistance is.

Are you saying I can take out the 10Ω resistor?


And are you saying I can take out the 125Ω too?

Yes and Yes, for answering part (a) you can get rid of those resistances.
 
  • #5
so 125Ω is too high then right? Which is why you could take it out? (Just so I could conceptually understand better)
 
  • #6
gneill
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so 125Ω is too high then right? Which is why you could take it out? (Just so I could conceptually understand better)

It can be any value at all (except zero (short circuit) of course). No matter what resistance it is the voltage supply will maintain the same voltage across it. So the rest of the network sees the same voltage no matter what.
 
  • #7
Ok I think I see what you were saying about the potential across.

You can still treat the wire that had the 125 ohm as an open circuit then, right?


Then I did source transformation with the 250V and 25 ohm and simplified some resistors on the right so I got the circuit to look like this now:

http://img822.imageshack.us/img822/651/homeworkprobsg28edit2.jpg [Broken]

but am stuck again due to the 8A current source. I can just find the current across the 50/3 ohm and it would be the same as V knot in the original right?
 
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  • #8
gneill
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You should preserve, through any transformations, the nodes across which you want to find the potential. In this case that can best be accomplished by not including the 100 Ω resistor in the transformations.

Note also that you can combine parallel current sources.

Edit: Let me amend this a bit. Since the 25Ω resistor has been transformed away to a parallel resistor, the nodes for which you want the potential difference have become the top and bottom rails, as you've shown. So that's fine, and the need to "preserve" the 100Ω resistor's separate existence is no longer required :smile: To answer part (a) then, you could even reduce your circuit further to a single current source in parallel with a single resistor :wink:
 
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  • #9
So r eq of that is 10 ohms and total I is 2A

therefore V tot = 20V

so V knot of the original = 20V (and it's the right answer)

Thanks again gneill.


Forgot how I would find the power from each source now though, would it be P = IV?
 
  • #10
gneill
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So r eq of that is 10 ohms and total I is 2A

therefore V tot = 20V

so V knot of the original = 20V (and it's the right answer)
Excellent! :smile:
Thanks again gneill.


Forgot how I would find the power from each source now though, would it be P = IV?

You're welcome. Yes, P = IV is correct. Watch out for the signs of the V's. Also note that you'll have to re-analyze the circuit with the ignored resistors restored, since they'll dissipate power. You'll have the benefit of knowing what Vo is to help you find the currents and potentials you'll need.
 
  • #11
If it was just one source I can just find the power dissipated in each resistor (current through each resistor squared times resistance) and then use Tellegan's theorem (add them) to find out but there are two sources and I'm not sure how to separately find them.

If I just say P = IV ---> for say (8A)(20V) the real answer is way higher.

Besides that, how would I also account for the resistors ignored in part a?


I can do the same thing here but how would I tell how much comes from each source? Is there way I can sum them up like I would do for power dissipated in each resistor?
 
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  • #12
gneill
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If it was just one source I can just find the power dissipated in each resistor (current through each resistor squared times resistance) and then use Tellegan's theorem (add them) to find out but there are two sources and I'm not sure how to separately find them.

If I just say P = IV ---> for say (8A)(20V) the real answer is way higher.
Yes, because the 'V' here is way higher. While Vo is 20V, it is the sum of the potential drops across the 10Ω resistor and the 8A source (which will generate its own potential difference to drive the 8A of current) which sums to that 20V. You want to find the potential drop across the 8A source.
Besides that, how would I also account for the resistors ignored in part a?

I can do the same thing here but how would I tell how much comes from each source? Is there way I can sum them up like I would do for power dissipated in each resistor?
You'll want to find the potential across and the current through each source, then use P = VI.
 
  • #13
Sorry for replying late again but I kind of have trouble finding all the currents.


Since the total voltage on the 8A source branch is 20V (and V knot is 20V) though and the current through the 10 ohm on it is 8A it's

(10 ohm)(8A) = 80V

80V -20V = 60V through the 8A current source

Power supplied by the 8A source = (60V)*(8A) = 480W (and it's correct).


but not sure how to find the current through the 250V supply now. I get this:

http://imageshack.us/a/img839/5030/homeworkprobsg28edit3.jpg [Broken]
 
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  • #14
gneill
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Can you find the potential across the 25Ω resistor? You should know the potentials of the nodes it's connected to...
 
  • #15
I still need work with that too. But it's 20V to the right so, would it be

250V - 20V?
 
  • #16
gneill
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Yup. So what's the current through the 25Ω resistor?
 
  • #17
Thanks again...

So that was basically a little bit of nodal analysis then too (or just seeing the branched potentials add up)?


Anyways, it's then 9.2A.

The potential down through the 125Ω resistor would be the same so then the current is 1.84A.

That would make the total current out of the voltage source be 11.04A

P from the 250V source = I*V = 11.04A * 250V = 2760W, right? (the correct answer says 2800W)
 
  • #18
gneill
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Thanks again...

So that was basically a little bit of nodal analysis then too (or just seeing the branched potentials add up)?

Anyways, it's then 9.2A.
Yes, it's using a bit of nodal analysis technique; some KVL and Ohm's law. 9.2A looks good.
The potential down through the 125Ω resistor would be the same so then the current is 1.84A.

That would make the total current out of the voltage source be 11.04A
Why would you say that the potential across the 125Ω resistor is the same as that across the 25Ω resistor? What's in parallel with the 125Ω resistor? On your schematic, label the potentials at the resistor's connection points.
 
  • #19
Thanks again; my mistake, it was supposed to just be the battery supply voltage.


250V then not 230V. So 250V / 125 ohm = 2A

2A added to 9.2 A would be 11.2A through the battery

P = I*V would then be 11.2A*250V = 2800W.
 
  • #20
gneill
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There ya go! :smile:
 

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