i cant understand those kirchhoff's equations..

hgmjr

Joined Jan 28, 2005
9,029
http://img504.imageshack.us/my.php?image=img8856io0.jpg

when i go round the I1 current when i go threw the diamond shape
battery from - to + i get in the equation +Vr
but when i go from + to - (opposite) on the 100v battery i get +100 in the second equation(the value is taken as positive again)

why???
The image posted is woefully illegible. Is this an accurate reading of the equations that you have provided?

KVL for LOOP \(I_{1}:\;4*(I_{2}-I_{1})\;+\;V_{R}\;-\;4*I_{1}\;+\;4*(I_3-I_1)\;=\;0 \)

KVL for LOOP \(I_{2}:\;100\;-\;4*(I_{3}-I_{1})\;+\;4*(I_2-I_3)\;=\;0 \)

hgmjr
 

recca02

Joined Apr 2, 2007
1,214
but when i go from + to - (opposite) on the 100v battery i get +100 in the second equation(the value is taken as positive again)
You went from -ve to +ve. We consider the direction of traverse through the element.
 

recca02

Joined Apr 2, 2007
1,214
Not only in voltage source, anywhere when you travel a circuit and go '*through'(*enter and leave) the element from -ve to +ve side(if a polarity is already asigned, like in a voltage source), you will have to put a +ve sign for voltage across it(again depends, you can also assign a -ve to to it but then you will have to assign a -ve to any voltage that you traverse from -ve to +ve--meaning you will have to be consistent with your sign convention).

In short you are correct but it is true for any element not just voltage sources.

Actually this is why it becomes confusing if you follow some convention and then in the midst think over it.
 

Thread Starter

transgalactic

Joined Apr 29, 2008
74
regarding the I1 equation:
if we go clockwise(as I1 flows) from the minus of the voltage sorse
first we go from minus to plus which gives us +Vr then we go threw 4ohm resistor which gives us +4I1 then we go threw a resistor where
we are going against I3 so its 4(I1-I3) and the same thing with thenext
resistor we get 4(I1-I2)

which gives us:+Vr+4I1+4(I1-I3)+4(I1-I2)=0

but if we look at the I1 equation here:
http://img504.imageshack.us/my.php?image=img8856io0.jpg

we got the opposite equation as if we multiplied by -1(which is fine)
except they kept +Vr as it is (i expected it will become -Vr)

why they kept it as +Vr??????????????
 

recca02

Joined Apr 2, 2007
1,214
which gives us:+Vr+4I1+4(I1-I3)+4(I1-I2)=0
How can voltage rise and voltage drop add up?

Check the polarity markings for the resistors. You should take entering end for resistors as + and leaving as -ve as they are at lower potential than the entering ends.
So if you consider traveling -ve to +ve as rise and assign + sign to voltage, should you not be assigning -ve to voltage across elements where you travel from +ve to -ve?
 

hgmjr

Joined Jan 28, 2005
9,029
I think you would benefit greatly by studying several Kirchoff's equation that have already been solved. Since the problem you are attempting to solve is proving to be difficult, it may be time to pause and take a look at a simplier problem. There are a number of simpler examples in the AAC e-book. If you study these examples closely until the technique for setting up and solving the equations is clear to you then you will be ready to tackle the tougher problem you have been assigned.

hgmjr
 

Thread Starter

transgalactic

Joined Apr 29, 2008
74
i solved lots of simple KCL and KVL
where is my mistake in my understanding
i specifically explained each step of my way in solving this
where is my mistake???
 

recca02

Joined Apr 2, 2007
1,214
where is my mistake???
Your mistake lies in how you added voltage drops.
See the link that was pointed out by Mr. HGMjr. Notice the voltage polarities for voltage drops( the '+' and '-' signs n the resistors). In the image that you provide, you haven't considered any voltage polarity for drops. Usually, one would assume a direction of current and then for the 'entering point" assign a +ve polarity and -ve to the leaving side. Now traverse the loop in any direction of your choice and assign +ve sign for either +ve to -ve(drop in voltage) Or -ve to +ve(rise in voltage) . Once you chose this sign convention be consistent and assign signs appropriately. One for drops and other for rise.
 
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