# KCL & KVL & maybe something else, proof vs plausible

Discussion in 'Homework Help' started by groundcontrol, Apr 29, 2016.

1. ### groundcontrol Thread Starter New Member

Sep 1, 2015
24
1
Hello,

I am hoping someone might be able to take a look at the attached question (Question 2) and let me know if my validation of the values asked for are 'worthy'. Is what I've shown proof or just shown that it is plausible? The question asks to show that:

i1 = 0.714A (714mA)
i2 = 1.29A
E = 12.6V

Is there a way to have come up with the values for i1, i2 and E, by using the given info alone? If the values for i1, i2, and E were not provided in the question, could you still solve for i1, i2, and E ?

(GIVEN: V1 =15V, R1 = 7ohm, R2 = 2ohm, R3 = 5ohm, and i3 = 2A)

I have tried to use KVL and KCL (and a little bit of Ohms law to find a couple of voltage drops to support the KVL equations). So is what I have come up with proving that i1=714mA etc or is it just plausible. There are many possibilities that could add up to 2A kind of thing. Just because
0.714 + 1.29 = 2, well .. 0.5 + 1.5 = 2.. and so does 1.25+0.75 = 2 and so on.

I thought at first looking at it that maybe it was a current divider circuit, but I couldn't get the figures to work. I'm still trying to get my head around what happens/when to use the current divider equation. I think R3 is what makes the circuit in Question 2 not a current divider, or it could also possibly be that there are two power sources... eek. More to learn re current divider circuits - even though every parallel circuit is a current divider circuit so I am looking forward to understanding /utilising the logic better hopefully soon!

Thanks for having a look

2. ### WBahn Moderator

Mar 31, 2012
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4,917
Your work is valid since you show that the offered answers result in KVL being satisfied in a complete set of essential loops and KCL being satisfied at a complete set of essential nodes.

But your work is a bit hard to follow because it doesn't flow in an organized manner -- you have to look at the work at the bottom to see where the values come from that you use in the middle. Also, you ignore units and just tack them on to the end (meaning that you aren't using the units that the answer works out to, but rather the units that you hope it works out to). You need to track your units through your work and report the units that the answer actually has.

Oct 4, 2015
92
6
4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
One thing to note is that the question is actually ambiguous.

It states that the meter reads 2 A, but I does not indicate the polarity of the meter, so it is unstated whether the meter is indicating 2 A flowing left-to-right or right-to-left. We tend to assume that it is reading the symbolic current I3, which is referenced flowing left-to-right, but this is not actually a given. Notice that I1 and I2 are similarly indicated with no meters anywhere around, so there is no reason to expect the direction of I3 to have any relationship to the orientation of the meter.

But you have to assume something and this is as good an assumption as any. Just keep in mind that if the numbers hadn't worked out, that you should have then assumed the other polarity and worked the problem again from that perspective. This would be when working the problem symbolically would have really paid off because then you could have just flipped the sign on the given current and turned the crank on three equations and been done instead of having to reanalyze the entire circuit.

5. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Yes, mesh works nicely. So does node. But for this one I think ad-hoc is probably the easiest.

Given the current I3 you know the voltage on the left node relative to the right node

Vright = 0 V
Vleft = I3·R3 = (2 A)(5 Ω) = 10 V

From this you can directly find the current I1 from Ohm's Law:

I1 = (V1 - Vleft)/R1 = (15 V - 10 V)/7 Ω = 714 mA

From this KCL directly yields I2:

I2 = I3 - I1 = 2 A - 714 mA = 1.296 A

Finally, Ohm's Law gives E1.

I2 = (E - Vleft)/R2 => E = Vleft + I2·R2 = 10 V + (1.296 A)(2 Ω) = 12.59 A

It all comes down to which approach you are most comfortable with, which argues for forcing yourself to use all of the techniques you know on a regular basis so that you are reasonably comfortable with as many of them as possible, which allows you to choose techniques that are good fits for the specific problem at hand.

groundcontrol likes this.
6. ### groundcontrol Thread Starter New Member

Sep 1, 2015
24
1
Thank you for taking a look at this and for your input.

I am reading chapter 10 as advised by thumb2 - thanks for the direction

I am going through your responses WBahn line at a time, while working on the problem. It will take me some time to properly go through what you have said and really learn it. My apologies that my workings are not clear, I will try and ensure that they are presented more methodically next time! Thanks for the heads up.

Thank you for still looking into the problem and helping me with it. I learn so much from this forum every time. I really really appreciate it.

Thank you !!!

7. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Good luck to you. It takes time to learn to present your work clearly -- it's a learned art that you will never stop learning. It helps to take a step back and pretend that you are someone else trying to understand the work based solely on what you have presented. You'll seldom spot all the weaknesses (because you can't help but see what you've written in the "tone" of how you wrote it), but you can learn to spot most of them.