# Branch current method

Discussion in 'Homework Help' started by cassyoscar, Jan 6, 2015.

1. ### cassyoscar Thread Starter New Member

Jan 6, 2015
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0
Hello,

Was wondering if anybody could help with this question?

I think I get 2 KVL equations which are...

3I1 + 1I2 + 0I3 = 6

0I1 - 1I2 - 5I3 = -0.5

for some reason I can't complete the question, am I going wrong anywhere?
any help is appreciated thanks.

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2. ### WBahn Moderator

Mar 31, 2012
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We are not mind readers. How are we supposed to have a clue as to which branches I1, I2, and I3 refer to, let alone what direction you've assigned them?

Feb 19, 2010
3,505
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Yes, we are.

4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Must be, since I just knew you were going to say that.

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5. ### Dr.killjoy Well-Known Member

Apr 28, 2013
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My Crystal ball says the answer you seek is 4...

6. ### WBahn Moderator

Mar 31, 2012
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Huh, mine always spits out '42' for some reason.

7. ### WBahn Moderator

Mar 31, 2012
18,087
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Note to the OP: We are just spinning our wheels and having a bit of fun until you get a chance to come back and clear things up a bit -- don't take any of the bantering too seriously or as being directed at you.

8. ### cassyoscar Thread Starter New Member

Jan 6, 2015
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Sorry about that, I thought I'd uploaded my workings out aswell... I've just attached them now. Am I on the right track?

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9. ### WBahn Moderator

Mar 31, 2012
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Thank you.

You are violating the passive sign convention and this appears to be messing you up.

Let's look at the voltage and current in R1. You have I1 entering the negative terminal (based on the indicated definition of Er1). Thus Er1 = -I1·R1 because if I1 is positive, then the voltage will drop as you go from the right side of R1 to the left side. Now, I'm assuming you use conventional current. Is that correct?

You really, REALLY need to track your units properly.

Jan 6, 2015
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11. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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Ignore any mention of current direction of current flow in the Ebooks. Here are the correct voltages and currents using the Conventional Current Convention, as used by the LTSpice simulator, and the Engineering and Scientific Community.

Note that nodes C, D, and E are subsumed into the Gnd node. Note that voltages at the other nodes are reported with respect to the Gnd node, which is considered to be at 0V. Note the positive direction of the currents. Note the sign of the current through the two sources.

12. ### cassyoscar Thread Starter New Member

Jan 6, 2015
7
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Thanks a lot for that! the only thing is I need to show my working out and not quite sure where I'm going wrong?

13. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
I posted the sim to demonstrate the correct solution for Conventional Current. You should be able to solve the network algebraically, and then test your equations against the simulation. The trick is to choose the mesh currents (direction is arbitrary), but then write the equations (including signs) consistent with the convention.

14. ### WBahn Moderator

Mar 31, 2012
18,087
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The E-book here uses electron current (and, like most users of electron current does it wrong), so that makes it confusing especially for someone who is struggling with an issue related to current polarity.

The first thing you want to do is assign your voltages and currents such that they are comply with the passive sign convention. This basically means that for any device you deem to be a load, such as a resistor, the current flowing through the device is positive when it flows into what you have designated as the positive voltage terminal of the device. If you do this, then you are far less likely to make the kind of mistakes that you made in your attempt (and which I pointed out to you). You don't have to follow the passive sign convention -- things will still work out; it just makes it easier to do the work without making simple mistakes.

Take things one step at a time and be very explicit until you get comfortable writing down larger chunks by inspection. You've actually started off working at an appropriate level of detail to carry this out.

So let's start with your current and voltage definitions as you defined them and then work through your approach being very careful and methodical.

You first equation is basically:

$
V_{cb} \; + \; V_{dc} \; + \; V_{ad} \; + \; V_{ba} \; = \; 0
-6V \; + \; 0V \; + \; E_{R_2} \; + \; E_{R_1} \; = \; 0
$

So you got this far just fine.

Your next step, writing KCL, is just fine, too.

$
-I_1 \; + \; I_2 \; - \; I_3 \; = \; 0
$

It's your next step that you messed up (and that I pointed out to you previously). So be more explicit:

$
E_{R_1} \; = \; -I_1 R_1
E_{R_2} \; = \; -I_2 R_2
E_{R_3} \; = \; -I_3 R_3
$

Do you see where these are coming from? The minus signs are there because your current definitions and your voltage definitions for each of the resistors violates the passive sign convention.

Now when you substitute these into your earlier equation you get

$
-6V \; + \; 0V \; - \; I_2 R_2 \; - \; I_1 R_1 \; = \; 0
-6V \; - \; I_2 (1 \Omega ) \; - \; I_1 (3 \Omega ) \; = \; 0
$

You should be similarly explicit with your other loop equation:

$
V_{da} \; + \; V_{ed} \; + \; V_{fe} \; + \; V_{af} \; = \; 0
- \; E_{R_2} \; + \; 0V \; + \; 0.5V \; - \; E_{R_3} \; = \; 0
- \; $$-I_2 R_2$$ \; + \; 0.5V \; - \; $$-I_3 R_3$$ \; = \; 0
I_2 R_2 \; + \; 0.5V \; + \; I_3 R_3 \; = \; 0
I_2 (1 \Omega ) \; + \; 0.5V \; + \; I_3 (5 \Omega ) \; = \; 0
$

You then have as your three equations:

$
- \; I_1 (3 \Omega ) \; - \; I_2 (1 \Omega ) \; + \; I_3 (0 \Omega) \; = \; 6V
\; I_1 (0 \Omega) \; + \; I_2 (1 \Omega ) + \; I_3 (5 \Omega ) \; = \; -0.5V
- \; I_1 \; + \; I_2 \; - \; I_3 \; = \; 0
$

To get all three equations into consistent units, simply divide each side of the first two equations by 1Ω:

$
- \; 3I_1 \; - \; 1I_2 \; + \; 0I_3 \; = \; +6.0A
+ \; 0I_1 \; + \; 1I_2 + \; 5I_3 \; = \; -0.5A
- \; 1I_1 \; + \; 1I_2 \; - \; 1I_3 \; = \; 0
$

Solve them and you are done (provided I didn't make a mistake, which is why you should always then check your answers against the original problem).

What I recommend doing is to break the problem into two phases:

(Step 1) Set the problem up applying the EE principles:

$
V_{cb} \; + \; V_{dc} \; + \; V_{ad} \; + \; V_{ba} \; = \; 0 \; \; \; \text{(KVL Left Loop)}
V_{da} \; + \; V_{ed} \; + \; V_{fe} \; + \; V_{af} \; = \; 0 \; \; \; \text{(KVL Right Loop)}
-I_1 \; + \; I_2 \; - \; I_3 \; = \; 0 \; \; \; \text{(KCL)}
$

Right there you have described the problem.

Next establish the relationships between the variables as you have defined them using Ohm's Law:

$
E_{R_1} \; = \; -I_1 R_1
E_{R_2} \; = \; -I_2 R_2
E_{R_3} \; = \; -I_3 R_3
$

Next establish the relationships between the component voltages and the voltages in your set-up equations:

$
V_{cb} \; = \; 6V
V_{dc} \; = \; 0V
V_{ba} \; = \; E_{R_2}
V_{da} \; = \; -E_{R_1}
V_{ed} \; = \; 0V
V_{fe} \; = \; 0.5V
V_{af} \; = \; -E_{R_3}
$

At this point you really have applied all of the EE principles and now everything after this is just math.

You then substitute these back into original equations to get the set of simultaneous equations that need to be solved.

This is ultra explicit and, with just a bit of practice, you will be able to write down the final set of equations by inspection. In fact, once you learn Mesh Current Analysis you will be able to write down two equations that are sufficient to solve the problem by inspection and once you learn Node Voltage Analysis you will be able to write down a single such equation by inspection, making this a very simple problem to solve.

(Step 2) Work the math

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15. ### cassyoscar Thread Starter New Member

Jan 6, 2015
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Thanks very much for the in-depth reply. I now see where I was messing up, definitely need to keep practicing these!

16. ### WBahn Moderator

Mar 31, 2012
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4,917
It's all about practice and experience. And as you gain experience, never let your ego prevent you from taking a step back to basics. I've seen numerous examples of students and practicing engineers make fundamental mistakes because they think that since they are at a point where they can write a lot of stuff down by inspection that it is somehow an unacceptable admission of weakness if they don't do so every time; it's actually the opposite in that refusing to step back when appropriate is an admission of weakness -- and is unacceptable. In particularly complicated circuits I will still sometimes start out with a KVL equation of the form Vab+Vbc+Vcd+Vde+Vea=0, relying on subscript arithmetic to ensure that I have walked around the particular path I am interested in without missing something and then expressing every voltage term explicitly before substituting them in (although more often than not I am comfortable with doing both of these at the same time with the liberal use of parens). This is particularly the case in the handful of times that I need to deal with magnetically-coupled circuits since I don't work with them very often.

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17. ### cassyoscar Thread Starter New Member

Jan 6, 2015
7
0
As you see I haven't had much practice or experience yet. I'm just getting in to electronics, trying to self-teach myself then maybe take a few years out to study. It's something that really interests me.

Thanks for your great help it's really appreciated!