Need help on solving this circuit..

PowerEngineer

Diamond Member
Oct 22, 2001
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A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.
 

RESmonkey

Diamond Member
May 6, 2007
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Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?
 

polarmystery

Diamond Member
Aug 21, 2005
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Originally posted by: RESmonkey
Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?

No, current follows the path of least resistance. R2 is less than R3 so the current would split at that node (c). Current will go through all of the resistors there. It's not as complex as it looks I promise :)
 

RESmonkey

Diamond Member
May 6, 2007
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Originally posted by: polarmystery
Originally posted by: RESmonkey
Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?

No, current follows the path of least resistance. R2 is less than R3 so the current would split at that node (c). Current will go through all of the resistors there. It's not as complex as it looks I promise :)

Yeah, but R2 isn't alone. (R1 + R2) > R3. Am I wrong?
 

bobert

Senior member
Dec 6, 2004
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Originally posted by: RESmonkey
Originally posted by: polarmystery
Originally posted by: RESmonkey
Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?

No, current follows the path of least resistance. R2 is less than R3 so the current would split at that node (c). Current will go through all of the resistors there. It's not as complex as it looks I promise :)

Yeah, but R2 isn't alone. (R1 + R2) > R3. Am I wrong?

just because R3 is less resistance doesn't necessarily mean all the current is going to "flow out" there
 

blahblah99

Platinum Member
Oct 10, 2000
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Easy. Find the sum of the currents at node c. Next find the voltage around the loop with Vs, R1, R2, R3 . That has to equal 0.

Substitute and you'll find I1, along with the current through R3.

After that, everything else is plug & chug.
 

blueshoe

Senior member
Mar 6, 2005
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Originally posted by: RESmonkey
Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?

I don't know if you're thinking that the current source is the only source of current in the circuit. If you are, you shouldn't see it that way, there are contributions from the voltage source as well.

I think you're trying to simplify the entire circuit somehow to get your answers, but you really don't need to do this. You can do what blahblah99 has said above, but it can be done even easier than that. This problem can be solved without the loop method that he mentioned and without superposition even I believe. You are given a lot of info in this problem.

I believe you can solve the entire thing using Ohm's Law only once. V=IR.

If you are trying to find I1, you know that the same current is going through R1. The current going out of node a = the current going into node a. It has nowhere else to go. You know R and you can easily find V. V is the voltage across R1, which is Va-Vb.

Now you know V and R and can solve for I.
 

RESmonkey

Diamond Member
May 6, 2007
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I just did the problem w/ help of a friend. He said it took him 8 tries, but he got it. It didn't involve the stupid current source at all. Nor did it involve R2 and R3. It was just Va - Vb = I1 x R1. Finding I1, I just multiply and enter answers.

Thanks for all your help guys!

edit = looks like blueshoe's got it ^ :)
 

RESmonkey

Diamond Member
May 6, 2007
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Originally posted by: Goosemaster
1. REALLY study hard if you find this problem difficult.
2. make sure you understand the theory behind finding the answer.

It was confusing because of the ideal current source. WTF is that doing there, anyway?
 

Goosemaster

Lifer
Apr 10, 2001
48,775
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Originally posted by: RESmonkey
Originally posted by: Goosemaster
1. REALLY study hard if you find this problem difficult.
2. make sure you understand the theory behind finding the answer.

It was confusing because of the ideal current source. WTF is that doing there, anyway?

As with any science, the development of problem-solving techniques are used in ideal situations first.
 

Goosemaster

Lifer
Apr 10, 2001
48,775
3
81
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)
 

Itchrelief

Golden Member
Dec 20, 2005
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Originally posted by: RESmonkey
Originally posted by: PowerEngineer

A current source is analogous to a voltage source. For a voltage source, you know that the voltage stays constant regardless of the current that passes through it. For a current source, the current passing through it stays constant regardless of the voltage across it.

So does current even go thru R1 and R2? Wouldn't it just seep out thru R3?

Caveat: I'm probably much less electrically qualified than anyone else posting in this thread, so take this with many grains of salt.

Besides what other people said that there will still be current (just less, maybe much less) going through the path of greater resistance, I think you also have to consider that the voltage source is going to drive a current through R1 and R2, regardless of whether any of the current from the current source will go through them. This current driven by the voltage source has no choice but to go through R1 and R2.

Disregarding the above, you can also see just by the provided voltage numbers that there is a voltage drop across resistors R1 and R2, implying that there is indeed current flowing through them. You can use this voltage drop to calculate the current (assuming this is a DC circuit).

Using the voltage drop across R1, I find an approximate I1 current of 4.67 mA. R2 gives a very slightly different result, but I doubt it matters much, as long as you follow whatever sig fig rules and whatnot your instructor explained.

If what I remember from freshman physics is correct, this current driven by the voltage source flows clockwise around the circuit. Use the node at c to find the current through R3. I1 flows in, Is flows in, so I think it works out so that I3 comes out of the node, then use I3 and R3 to find V3. V3 should be = Vs. [edit: duh. by method of blueshoe, since you know zero at ground, you don't even need to use ohm's law to calculate V3 at all)]

Is is given. Thus you have all the variable values you need to solve this, assuming I didn't F* this up somehow.

 

RESmonkey

Diamond Member
May 6, 2007
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Originally posted by: Goosemaster
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)

No, it's just our school doesn't use a textbook. Instead, we have a compilation
of "lecture notes" xeroxed and binded together.
 

Goosemaster

Lifer
Apr 10, 2001
48,775
3
81
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)

No, it's just our school doesn't use a textbook. Instead, we have a compilation
of "lecture notes" xeroxed and binded together.

wow...I hate when they do that crap.

<---passed by going to the library and reading every circuits book he could find:D
 

RESmonkey

Diamond Member
May 6, 2007
4,818
2
0
Originally posted by: Goosemaster
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)

No, it's just our school doesn't use a textbook. Instead, we have a compilation
of "lecture notes" xeroxed and binded together.

wow...I hate when they do that crap.

<---passed by going to the library and reading every circuits book he could find:D

What school did you go to?

 

blahblah99

Platinum Member
Oct 10, 2000
2,689
0
0
Originally posted by: RESmonkey
I just did the problem w/ help of a friend. He said it took him 8 tries, but he got it. It didn't involve the stupid current source at all. Nor did it involve R2 and R3. It was just Va - Vb = I1 x R1. Finding I1, I just multiply and enter answers.

Thanks for all your help guys!

edit = looks like blueshoe's got it ^ :)

Not sure why you think the current source doesn't affect the circuit... but I guarantee you the right answer involves solving the circuit with the current source.
 

Goosemaster

Lifer
Apr 10, 2001
48,775
3
81
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)

No, it's just our school doesn't use a textbook. Instead, we have a compilation
of "lecture notes" xeroxed and binded together.

wow...I hate when they do that crap.

<---passed by going to the library and reading every circuits book he could find:D

What school did you go to?

long story;)

gotta go finish watching a tv epp. before I go to sleep. I have to be at work again pretty early:(
 

RESmonkey

Diamond Member
May 6, 2007
4,818
2
0
Originally posted by: Goosemaster
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
Originally posted by: Goosemaster
Originally posted by: RESmonkey
It would help if we had a textbook. :(

:Q

Please elaborate. If it is a money thing I'm sure we can scrounge up a way to help you out:)

OR at least find an online resource that would help :)

No, it's just our school doesn't use a textbook. Instead, we have a compilation
of "lecture notes" xeroxed and binded together.

wow...I hate when they do that crap.

<---passed by going to the library and reading every circuits book he could find:D

What school did you go to?

long story;)

I'm not currently sure ECE is what I want to do, even. I'm going to see after this year how I like it. If not, I can got to engineering Physics, or something fundamental like math.
I want to know 'why' before 'how.' They aren't explaining anything in scientific terms.

The class is full of international students that have already taken this in India or China, and the class just "does" the things without wanting to know why.


Worst part is the labs. Me and my partner almost failed today's. Our equipment didn't work, reading the labs was confusing with no diagrams and just assumptions that you
know what is going on.


edit = Goodnight, thanks for the help :)


 

blueshoe

Senior member
Mar 6, 2005
414
0
76
Originally posted by: blahblah99
Originally posted by: RESmonkey
I just did the problem w/ help of a friend. He said it took him 8 tries, but he got it. It didn't involve the stupid current source at all. Nor did it involve R2 and R3. It was just Va - Vb = I1 x R1. Finding I1, I just multiply and enter answers.

Thanks for all your help guys!

edit = looks like blueshoe's got it ^ :)

Not sure why you think the current source doesn't affect the circuit... but I guarantee you the right answer involves solving the circuit with the current source.

Yes it definitely affects the circuit, but it isn't needed to solve the problem because the node voltages are already given. If the node voltages weren't given he'd have to use a method like KVL or KCL.

I'm guessing the node voltages were given because it is near the beginning of an introductory circuits class and methods using KVL, KCL, and superposition haven't been taught yet.