# Using Kirchhoff's curent law

Discussion in 'Homework Help' started by mic0, Sep 4, 2012.

1. ### mic0 Thread Starter New Member

Sep 4, 2012
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0
Determine the currents I1 to I3 in the circuit below.

I came up with 2 equations for the 3 unknowns, but i couldn't find a third one.

KVL: -16 + $3I_{1}$ + $4I_{2}$ = 0
KCL $I_{1}$ = $I_{2}$ + $I_{3}$

2. ### Austin Clark Active Member

Dec 28, 2011
413
47
First off, combine the voltage sources.

From there, calculating current is easy, just use ohms law.

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3. ### mic0 Thread Starter New Member

Sep 4, 2012
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I thought you could only combine voltages when they were in series?

4. ### mlog Member

Feb 11, 2012
276
36
I agree with Austin. Look closely at your schematic, and you'll see that all 3 of the voltage sources are in series.

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5. ### mic0 Thread Starter New Member

Sep 4, 2012
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I don't see it. My textbook says that devices are in series when the current only has one pathway to travel through. If I1 splits into I2 and I3, then how exactly are the voltage sources in series?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If that doesn't help your understanding, then put aside the idea of a series connection for the moment ....

From the lower conductor [reference potential / ground], consider what rise or fall in potential you see as you move through the two sources [12V & 8V]. So starting from 0V at the common, as you pass through the 12V source the potential rises to +12V, at the node where the two sources and the 2kΩ meet. As you pass on through the 8V source towards the other node, the potential will fall by 8V. So, you go from +12V down to +4V, at the left-hand node where the 3kΩ, 4kΩ and 8V source [-ve] terminal meet.

In effect, you could take the the result as the same as the algebraic addition of the two sources to obtain the result of +4V at the left-hand side node. So the result is the same as you obtain with series connection of the sources with the polarity orientations shown.

The rest is simplicity itself, with a potential difference of 12V across the 3kΩ and 4V across the 4kΩ. This will lead to the currents I1 & I2 and on to I3 from KCL.

Last edited: Sep 5, 2012
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7. ### WBahn Moderator

Mar 31, 2012
20,737
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They aren't, at least not strictly speaking. But Austin and mlog are correct as far being able to leverage them almost as though they are. The key to understanding this is that an ideal voltage source will generate whatever current is necessary in order to maintain the specified voltage across it, so it doesn't matter that they don't have the same current. As long as they are connected to each other directly, then you can directly determine the voltages at the nodes involved (at least relative to each other). What you can't do, and which makes them so that they can't really be treated as though they were in series, is assume anything about the current in one even if you know the current in one of the others.

Look at your diagram again. You have four nodes. One of them you get to call 0V (you can pick any one you want). From there, you can directly give the node of any voltage that is connected to that node via a voltage source. Once you have those nodes given, you can do the same for any and all nodes connected to them via a voltage source. In this case, all four of your nodes can be solved for just based on the voltage sources. To see this more formally, do KVL on the outer rim of the circuit and what do you end up with: one equation and one unknown (the current in the 3ohm resistor)! Do it about the rightmost loop (the 12V source and the 2ohm resistor) -- again, one equation and one unknown! Now, can you find a loop that goes only through voltage sources and the 4ohm resistor?

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8. ### mlog Member

Feb 11, 2012
276
36
Your book isn't wrong. It's just incomplete. It's not telling you the whole story. If 2 voltage sources are end to end, they're in series. Your book is correct in terms of resistances in series. If you are to combine 2 resistors in series, neither can be in parallel with another resistor. But we aren't talking about resistors. We are talking about voltage sources. The rules are different for voltage sources.

In your example, begin at the node of I1, I2, and I3, and move to the right and pass through the 8V source. Continue clockwise around the outer loop. Next you pass through the 12V source. Keep going clockwise and you pass through the 16V source. So, the 8V is touching the 12V is touching the 16V. They're all 3 in series. It doesn't matter that the 2k resistor is in parallel with the 12V source. The 12V is still in series with the 8V and the 16V.

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9. ### mic0 Thread Starter New Member

Sep 4, 2012
7
0
Okay, that makes sense. But, the values I get when I do KVL across the 3 different loops don't add up.

-16V + 3I - 8V + 12V = 0 (outer)
-12V +3I = 0
3I = 12V
I1 = 4

12V -2I = 0 (right)
I2 = 6

-16V +3I1 +4I2 =0 (left)

-16V +12+24 ≠ 0

10. ### Austin Clark Active Member

Dec 28, 2011
413
47
I'm not quite following what you're doing. To calculate the voltage around the entire loop, you don't need to worry at all about current. Just pay attention to the polarity and voltage of the individual sources, and calculate from there.

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11. ### mic0 Thread Starter New Member

Sep 4, 2012
7
0
Is the voltage around the the entire loop 12v?

12. ### WBahn Moderator

Mar 31, 2012
20,737
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Your problem is that you aren't using the labelled currents consistently. I2 is the current flowing down through the 4kohm resistor, but in your second equation you use it as the current flowing down through the 2kohm resistor. Then in the next equation you use the I2 that you solved for using the 2kohm resistor as though it really were the current in the 4kohm resistor. Can't do that!

Your first equation is fine, except for the units. I1 is not 4 (saying that a current is "4" is meaningless. That;s like saying the distance between two cities is 245. I1 is 4mA.

Your second equation is only correct if I2 is the current flowing down in the 2kohm resistor, which it isn't. There is no label provided for this current, so define one! I4 is hereby the current flowing downward through the 2kohm resistor. So you have I4 is 6mA.

Your third equation involves an unknown, namely I2. So you will use that equation to solve for I2. With I2 in hand, you can then apply KCL to the node at the top of the 4kohm resistor to find I3. With I3 and I4, you can apply KCL to the node at the top of top of the 2kohm resistor to find the current in the 12V supply (let's call it I5).

As a check, you can do KVL around the inner loop to see if it works out.

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13. ### mic0 Thread Starter New Member

Sep 4, 2012
7
0
Alright then. So,

I1=4mA

Using the equation, -16V +3I1 +4I2 =0

I2 = 1mA
I1 = I2 + I3

4mA = 1mA + 3mA
I3 = 3mA

I4 =6mA

So, if I5 is the current going into the 12V source the kcl equation for the node above the 2kΩ resistor is:

i3= i4 + i5
3mA= 6mA + i5

Does this mean that the current is negative?

14. ### WBahn Moderator

Mar 31, 2012
20,737
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We cross posted, so this response is a bit dated, but I thought it important to point out that the answer is, "No, because the voltage around ANY loop (talking conservative fields here) is 0V!" That's what KVL is based on.

Instead, ask your quesiton like. So, is the voltage drop from Point A to Point B 12V?

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15. ### WBahn Moderator

Mar 31, 2012
20,737
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Yes! So the 6mA in the 2kohm resistor is coming equally from the two supplies it is connected to.

Note that, when it's an option, we generally try to use the 'passive sign convention', which means that we pick the directions of our currents so that they flow out of the positive terminal of a voltage source. But you can pick which every you want as long as you are careful to properly track the resulting minus signs. For instance, if asked to compute the power delivered by the 12V source, it would be (-I5)(12V).

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16. ### mic0 Thread Starter New Member

Sep 4, 2012
7
0
Cool! Thanks for your help. I'll do more practice with these types of problems.