# KCL and Groud point

Discussion in 'Homework Help' started by drao, Mar 12, 2013.

1. ### drao Thread Starter New Member

Mar 11, 2012
16
0
Untitled.png
For this circuit, in the ground point i can apply KCL??? I3, I2 and I1???
Why do i ask you this .
I solved the second circuit, and if i consider Ie of transistor = Ie of resistor and i apply KVL for Ie*R+Veb=0 (i forgot to mark the loop) i dont obtain the same current as if i apply KVL for the marked loop . WHy???
Untitled.png
BTW, im really sorry for not posting in the right section :|.

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2. ### StayatHomeElectronics AAC Fanatic!

Sep 25, 2008
1,021
71
You have not applied KCL correctly. The currents on the left hand side are not Ie. There is an entirely different portion of the circuit to consider and that is the portion containing the supply VCC. Draw in the supply with its ground connection and try again with KCL.

Define KCL. This will help us understand your understanding of the concept.

Make sure you identify the nodes you are calculation KCL on.

3. ### drao Thread Starter New Member

Mar 11, 2012
16
0
Why???
Dont we consider Ib=0 and Ig=0???
And arent the picture circuits equivalent???
I dont know how to draw the supply. I tried, but i dont know if its what i had to do.
Untitled2.png
Supply.png

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4. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Now compare your Supply.png drawing to your untitled.png (and try to give your images some kind of meaningful title, it doesn't have to be much). In your untitled one, you concluded that any current that went down to ground in the right hand branch had to come out of ground and go up through the left hand branch. But now look at the Supply image. Do you see that this is not the case? Some of the current that current that goes into ground from the right branch can come up out of the ground but then take a left turn and go to the supply and not go up through the middle (used to be left hand) branch. In fact, ALL of the current that comes down the right hand branch can take the left turn and there can be a completely independent current that comes down the middle branch and takes the turn toward the supply before entering the ground symbol.

5. ### drao Thread Starter New Member

Mar 11, 2012
16
0
Dam, i cant belive how i didnt think at this :|.
Thx very much for clarification. I gave myself a simple exemple and im feeling bad about my circuit analysis skills, i need improvements . But i have 1 more question, if i can edit the circuit like in the untiled2 pic, what happens with the 2A current difference its going thourgh the supply branch???
pic.png

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6. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
In your pic.png you have 6A going into the ground in the left hand branch and 4A going into the ground in the right hand branch. That gives a total of 10A going into the ground. That 10A is then coming out of the ground and into the negative terminal of the 12V supply. Which is a good thing because, looking at the top of the pic, you've got a total of 10A coming out of the 12V supply.

And, yes, your circuit analysis skills do seem to be pretty weak. But that's okay -- we were all there at some point. It takes practice and being willing to make up and work your own problems shows a level of effort that relatively few are willing to put forth. So I think you will make the improvements you need and do just fine.

7. ### drao Thread Starter New Member

Mar 11, 2012
16
0
Again i made a big mistake applying KCL because of not drawing the supply. Anyway, thanks a lot of explanation. All the best!!!

8. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Most people need to start off drawing all the connections. Then, at some point, you will just naturally become comfortable leaving certain components and connections, such as supplies, implied.