# Kirchhoff problem

Discussion in 'General Electronics Chat' started by Dritech, Oct 19, 2012.

1. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
812
6
Hi,

I was working Kirchhoff's law problem from the link below (section: Example 1: Calculating Current: Using Kirchhoffs Rules):

http://cnx.org/content/m42359/latest/?collection=col11406/latest

Non of the results from I1, I2 and I3 did match. Can someone guide me on what am i doing wrong?

Attached is my working. Note that it took I1, I2 and I3 at different positions.

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2. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
812
6
I have a mistake in step 5, where i divided by 6 instead of subtracting. But anyway, the other the answers are still not matching.

3. ### Veracohr Well-Known Member

Jan 3, 2011
654
100
Using mesh current analysis, all loops have to be in the same direction. You have one clockwise and one counterclockwise. They need to be the same direction (either way is fine).

4. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
812
6
Thanks for the reply. But how can i the current direction? by the way, i am trying to work this using kirchhoff law not mesh current analysis.

5. ### WBahn Moderator

Mar 31, 2012
22,996
6,887
You can assign the current direction for each branch either way. The same with the component voltages. Then you just have to be careful with the signs when you apply KVL and KCL. By convention, assign currents so that they enter loads (components that absorb power) at the positive side and leave at the negative side (of the voltage assigned across that component). Power sources are, by convention, the opposite -- current enters the negative side and exits the positive side. This is assuming you are using conventional current flow and not electron current flow. It is generally easier to assign a current direction to the branch and then assign the component voltages in the branch accordingly.

Don't worry aboiut assigning the directions correct with regards to the actual voltages and currents. If the current turns out to be negative, you know it is simply flowing in the opposite direction compared to the direction you assumed.

6. ### ErnieM AAC Fanatic!

Apr 24, 2011
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As long as you are consistent in your work you can assume the currents to be in any direction you so choose.

Some problems lend themselves to defining the currents in directions other then the "all clockwise" convention. In this case my intuition tells me the currents will be positive in the directions you chose, so I would do just the same.

In fact I did do it the same way. My two loop equations I agree with yours, so it must be some silly algebraic error. I ain't gonna check *that*, that is real work. <grin>

7. ### Veracohr Well-Known Member

Jan 3, 2011
654
100
Those two loopy arrows inside the circuit - that's the standard way of displaying the mesh current analysis method. Kirchoff's rules are rules to follow for correct analysis results; mesh current analysis is the method you used.

8. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,909
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When did Kirchoff's Laws get downgraded to rules?

9. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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I never understand why people learn this crap, just use an ammeter or voltmeter!

10. ### WBahn Moderator

Mar 31, 2012
22,996
6,887
So, you're saying that when someone needs to design a voltage divider to provide a reference voltage fo an opamp or to bias a transistor, that they should find suitable resistor values through trial and error? That their is no value in being able to determine suitable values by design on paper?

11. ### blassajox New Member

Oct 21, 2012
1
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ur mistake is i1+i2=i3 this is wrong because current direction(+to _ 0r _ to +) is same for all loop whatever direction u can use. conventional direction is +to _ if u use this direction u must use for it for all loop don't change. the correct equation is i1=i2+i3

Dritech likes this.

Sep 21, 2011
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