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Electric Circuits

  • Engineering
  • Thread starter _Bd_
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  • #1
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Homework Statement


[PLAIN]http://img810.imageshack.us/img810/807/circuitxm.png [Broken]
http://img810.imageshack.us/img810/807/circuitxm.png [Broken]
set up all the equations (Kirchoffs current and voltage law)

Homework Equations



Sum of i's = 0
Sum of V's = 0

The Attempt at a Solution



there are 4 nodes, therefore I can use 3 KC laws (Current in = current out)

so at node A (top left) Im wondering if the current in is 2A (due to the current generator in the bot) and therefore there are 2 currents flowing out
or should I just label it as current 1 (i1)
also for the same thing in node C(bot left) the current comming from the 12V generator. . .is it the same current as the one running trough the 8 Ohm resistor? or is it a different one? so then the current out is 2A and the current in is . . .?

Im kinda lost. . . Are there 3 currents in the entire circuit? or are there more?

I dont have problems iwth the voltage law but the current "concept" is the one that gets me lost =/

thank you.
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi _Bd_! :smile:

(two hs in Kirchhoff! :wink:)
Are there 3 currents in the entire circuit? or are there more?

I dont have problems iwth the voltage law but the current "concept" is the one that gets me lost =/
There is a separate current for each section of wire (between two junctions) …

so there are six currents to be marked on your diagram, and for you to use in Kirchhoff's rules :smile:
 
  • #3
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A good rule of thumb to solve for all unknowns (currents and voltage drops) is to apply Kirchhoff's voltage law (loop rule) to each mesh, and to apply Kirchhoff's junction rule to all but one of the junctions of a circuit. This general method works for all circuits that are confined to a plane. Don't forget that the current source also has an unknown voltage drop that must be accounted for when writing down the corresponding loop rule.

You can see that we have 4 junctions in our circuit, yes?

Remember that a mesh is a closed part of a circuit that doesn't have anything inside it. Can you see your circuit has three meshes?

Your circuit has the added advantage that one of the branch currents is already known, thanks to the current source. Can you see why?

This means we have a total of 6 unknowns, because we have 6 branch circuits (one of these is already known, so 6-1 = 5), and 1 voltage drop in the current source; and 3 + (4-1) = 6 equations.

I am sure you will have less difficulty with all this in mind.

Good luck. :wink:
 
  • #4
CEL
656
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In reality, the problem is much simpler than it seems.
For example, the 8 ohm resistos is is parallel with a 8 V source, so you know the current through it.
The 6 ohm resistor is in parallel with a series of a -8V, a +8V and a 12V sources, so you know its currents also.
The remaining currents are very easy to find.
 
  • #5
4
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In reality, the problem is much simpler than it seems.
For example, the 8 ohm resistos is is parallel with a 8 V source, so you know the current through it.
The 6 ohm resistor is in parallel with a series of a -8V, a +8V and a 12V sources, so you know its currents also.
The remaining currents are very easy to find.
it isn't that simple dude!:D
 
  • #6
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Thanks for the replies.

In reality, the problem is much simpler than it seems.
For example, the 8 ohm resistos is is parallel with a 8 V source, so you know the current through it.
The 6 ohm resistor is in parallel with a series of a -8V, a +8V and a 12V sources, so you know its currents also.
The remaining currents are very easy to find.
well Im supposed to do it the "long-hand way"
not by simplifying resistors and sources by parallel/series

I understand all the equations, how they go and stuff but where I get lost is in labeling my currents. . .so just to be sure, can someone verify that this are the currents?
[PLAIN]http://img19.imageshack.us/img19/5037/currents.jpg [Broken]

http://img19.imageshack.us/img19/5037/currents.jpg [Broken]
 
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  • #7
tiny-tim
Science Advisor
Homework Helper
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Hi _Bd_! :smile:
I understand all the equations, how they go and stuff but where I get lost is in labeling my currents. . .so just to be sure, can someone verify that this are the currents?
Yes, but it is very important to mark an arrow on each current so that you know which way round (+ or -) to put it in each junction or loop equation. :wink:
 
  • #8
Have you ever learned to do nodal analysis?
I found it less confusing than the methods I was taught in high school or first year.

[PLAIN]http://img833.imageshack.us/img833/3799/method.png [Broken]

You pick any node and call it the reference node or the ground, i.e. 0V.

If the branch only has a V-source in it, we can make some associations:
V3 = -12V
V2 = V1+8V
(Though we can't say V2 is 8V higher than ref because there is also the 8ohm branch there)

But there are a few more important rules:
- If there is a current source in a branch, we just take that as the current (the node from V3 to V1 contains current of 2A, ignore the 6V and 4ohms for writing the KCL equations).
- If the branch (eg. the one at V1 to V2) contains only a voltage source, then the current is an unknown. (In total there would be 3 unknown currents in this question).
- For branches without a current source, you can designate the current to go either direction. Its arbitrary and doesn't matter. The red arrows I have drawn show the current directions that I have chosen, but it really doesn't matter which direction as long as you write the equations correctly in the end.

Now we can write a KCL equation for every node.

Sum up all the currents going in = all the currents going out of the node.

For example for node V1:
2A going in = I12 + (V1-V3)/6 going out // I12 is the unknown current from V1 to V2

For V2:
I12 + I2 going in = (V2-0)/8 going out // I call I2 the current going through the 8V source from ref to V2

and so forth, there should be enough equations to solve for all the unknown variables.
 
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