Kirchhoff's Law Issue??

Discussion in 'General Electronics Chat' started by sjgallagher2, Nov 4, 2014.

Feb 6, 2013
111
7
I was checking my notebook section on Kirchhoff's law. It's been a long time since I've used it, I wanted to review. For an example, I found this circuit:

A basic circuit. Component values:
Voltage source (Vs) = 9V
R1 = 100 ohms
R2 = 2.2k
R3 = 10k

I'm solving for current I1.

With simulation software, I know that the current going through the battery should be around 4.7mA. I get the same answer when simplifying the circuit V/(R2||R3 + R1). But here comes the issue.

Using Kirchhoff's laws, I get three equations:
I1 - I2 - I3 = 0
R1*I1 - R2*I2 = Vs
R2*I2 - R3*I3 = 0

Apparently these equations are wrong! I am totally baffled as to why, I've been hitting my against a while trying to find the error in them. Because when I solve for those equations, I get I1 = 3mA, I2 = 4mA, and I3 = 1.7mA. Or more exactly, I1 = 3.08mA, I2 = 3.95mA, I3 = -1.73mA.

You can check that math by solving for a matrix:
1 -1 -1 0
100 -2200 0 9
0 2200 -10000 0

You'll get the same answers. So the issue is in the equations. But where?? I've been over this a million times. Two loops. In loop 1, Vs is polarized positive, Vr1 is negative, Vr2 is negative. In loop 2, Vr2 is positive (assuming clockwise current) and Vr3 is negative. Where am I going wrong? What am I missing? Help is needed. Thanks.

2. MikeML AAC Fanatic!

Oct 2, 2009
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What are the directions of the currents I1, I2, and I3? Mark them on the diagram before writing the equations. Then write the equations....

3. Jony130 AAC Fanatic!

Feb 17, 2009
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The current entering any junction is equal to the current leaving that junction.
I1 is entering and I2 and I3 leaving the node.

I1 = 9V/(R1 + R2||R3 ) = 9/(0.1 + 1/(1/2.2 + 1/10) = 4.7286mA

I2 = I1 * R3/(R2+R3) = 4.7286* 10/(10 + 2.2) = 3.875mA

And I3 = I1 * R2/(R2 + R3) = 0.852mA Or I3 = I1 - I2 = 4.7286mA - 3.875mA = 0.8536mA

And finally
I1 - I2 - I3 = 0
4.7286 - 3.875 - 0.8536 = 0A

Last edited: Nov 4, 2014
4. wayneh Expert

Sep 9, 2010
12,123
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Why subtract?

Feb 6, 2013
111
7
Wayneh, good catch, in my calculations though I added. I just typed it wrong.
The equations I meant to write were:
I1 - I2 - I3 = 0
R1I1 + R2I2 = 9
R2I2 - R3I3 = 0

Jony, I like what you've done. But, I'm not trying to simply solve the circuit, I want to do it using KVL and KCL with the circuit as is. Humor me?

Mike, I assumed I1 is going through the battery from negative to positive and through R1, I2 is going through R2 to the negative end of the battery, and I3 is going through R3 to the negative end of the battery. I assumed this when creating my equations though, so no help there, is there? Is there something I did wrong in the equations because of the way I assumed the currents?

6. BobTPH Active Member

Jun 5, 2013
782
114
Depends on the direction he made I1 I2 and I3. Which he has not shown.

Bob

Feb 6, 2013
111
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Bob, please refer to my previous post. The software I used to quickly draw up the schematic had no arrows

Oct 2, 2009
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9. WBahn Moderator

Mar 31, 2012
17,747
4,796
Then use Paint or some other easily available and simple to use program to draw them.

Based on your first equation, this is the way you are assuming your currents:

Took less than a minute to annotate your diagram, including adding Vs to it. Always prepare complete diagrams.

Now, with those current definitions before you, and not sitting in the back of your mind, look at your other two equations.

Your equations in Post #5 are consistent with this diagram and yield the correct answer, so what's the problem?

Feb 17, 2009
3,957
1,097