Discussion in 'General Electronics Chat' started by odm4286, Jun 17, 2016.

Sep 20, 2009
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2. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
Could you be a bit more specific? There are six circuit diagrams in that problem.

What would really help would be for you to take you best shot at what you think they should be. Then we can look them over and see where you are going wrong and address that specific issue.

3. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
Sorry about that. Referencing this schematic

My node is the collector of the NPN transistor. Letting i be the current drawn from the 6v source and 5 - i the current drawn from the 7.2 source I've set my KCL equation as follows.
-i + 5 -i + 5 = 0

I know this is wrong but I'm not too sure how to setup the equation to get 1.9mA (the correct answer for i).

I hope that was a bit clearer! Thanks for the reply!

4. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
Your problem appears to be that you want to keep track of the currents in your head and, as a result, didn't properly account for which currents were flowing into the node and which were flowing out.

I1 = Current drawn from 6 V source
I2 = Current drawn from the 7.2 V source
I3 = Current in transistor collector.

From the annotated diagram:

I1 + I2 - I3 = 0

I1 = I
I2 = 5 mA - I
I3 = 5 mA

(I) + (5 mA - I) - (5 mA) = 0

5. ### DGElder Member

Apr 3, 2016
351
87
Question 2:

KCL: at center node
(6-V)/1k -5mA + (7.2-V)/1K = 0
I = (6-V)/1K

KVL: around outer loop
-6 + I*1K - (5mA-I)1K + 7.2V = 0

Last edited: Jun 17, 2016
6. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
I believe I have the same equation as you, only difference is the -i. Is this incorrect? I'm under the assumption current entering the node is -, electrons are negative, and leaving the node are positive. Also, I still don't see how this would equate to i = 1.9mA, based off your equation it would end up being 0 = 0.

Sorry, this is new to me. Trying to wrap my head around how they reached the final answer of i = 1.9mA.

7. ### DGElder Member

Apr 3, 2016
351
87
"I'm under the assumption current entering the node is -, electrons are negative, and leaving the node are positive."

When setting up KCL equations you can call current entering a node positive or negative, it doesn't matter as long as you are consistent. Since the diagram shows I entering the node it is convenient in this case to call currents entering the node positive.

But you should understand, that by convention, current flows from a positive potential to negative potential - which unfortunately is the opposite direction of electron flow. If you are not comfortable with this convention you should get used to it right away.
See this explanation: https://en.wikipedia.org/wiki/Passive_sign_convention

Last edited: Jun 17, 2016
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8. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
Right, I understand so far but how in the world is i = 1.9mA? I don't see how you could get that from either of the equations, KCL or KVL

9. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
Then it isn't the same equation, is it?

Being off by a minus sign is about the biggest mistake one can make. Jump start a car battery and make a minus sign error and see what happens.

It has absolutely nothing to do with the sign of the charge on electrons. Consider people carrying bags of concrete into and out of a warehouse. We only know how much each person is carrying and we want to determine whether the number of bags in the warehouse is staying the same, increasing, or decreasing. So we say that we will keep track of bags entering the warehouse (so someone entering the warehouse is carrying a positive number of bags) while someone leaving the warehouse is deemed to be the same as someone entering the warehouse with a negative number of bags. We then add up all of the bags entering the warehouse (some of which are positive and some of which are negative) and the total is the net number of bags that entered the warehouse. If the number is positive then there are more bags in there than there were while if the number is negative there are fewer. If the number is zero, then there are the same number as there were to start with. For KCL, we set the net current to zero because a node in (almost any) circuit can't collect any significant charge, so whatever current goes in must come back out.

Bottom line, we simply choose to call flowing into a node as positive and if the actual current is flowing out of the node it will have a negative value to indicate that (but we are free to define it the other way around).

All that equation is doing is showing that KCL is satisfied. Without realizing it, you applied KCL when you said that the current coming out of the 7.2 V battery was (5 mA - I). How did you come up with that? You looked at the node and intuitively reasoned that it had to be that if I was flowing in from the left and 5 mA was flowing out down the center. THAT is KCL! All the equation did was confirm that you intuitively reasoned it out correctly. KCL is satisfied for ANY value of I. But only ONE value of I will satisfy the other constraints imposed on the circuit -- by Ohms Law, in this case.

We know that the center node will have some voltage -- we don't know what it is, so let's call it Vx.

Ohm's Law says that the current in the left resistor is the voltage at the left side of it (the tail of the current arrow) minus the voltage at the right side of it (the head of the current arrow) divided by the value of the resistance. So

I = (6 V - Vx)/(1kΩ)

Ohm's Law also says that the current in the right resistor is the voltage at the right side of it (the tail of the current arrow) minus the voltage at the left side of it (the head of the current arrow) divided by the value of the resistance. So

(5 mA - I) = (7.2 V - Vx)/(1kΩ)

You have two unknowns (I and Vx) and two equations. Solve them and you have it.

10. ### DGElder Member

Apr 3, 2016
351
87
odm, If you are talking to me, I don't see the problem. You just solve the equations I wrote out with algebra. For the KCL you solve for V (the node voltage) and substitute that (4.1 volts) in the second equation to get I. Or with the KVL equation just solve for I - it is one equation with one unknown. ???

Or don't you agree with my equations?

Last edited: Jun 17, 2016
11. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
Thanks for the help, I still don't quite understand but I'll continue to play around with it until I get it! Sometimes I just need to see it done in order to grasp it. I'm one of those "work backward" types to understand things.

12. ### DGElder Member

Apr 3, 2016
351
87
What is it you don't understand? How to setup the equation or how to solve it? If you would do it the way you think it should be done and show your work we could correct any specific misconceptions.

13. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
Sorry for the late reply, I've been busy between work and life. Here is what I did to solve to Vx in hopes to solve for i. Doesn't take much to know I screwed up, Vx cannot be -2493.8 haha.

The question is, where did I screw up? Thanks again

14. ### DGElder Member

Apr 3, 2016
351
87
I don't understand your first equation which quickly reduces to the trivial case 0=0.

Anyway, your KCL equation variable setup is correct, but pay attention to the units for the current source!

Last edited: Jun 21, 2016
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15. ### odm4286 Thread Starter Senior Member

Sep 20, 2009
232
12
Ah now I see! Thank you for the help!