Simultaneous Equation with KVL problem

Thread Starter

KrisK

Joined Apr 20, 2018
3
Hello, thank you for taking the time to look at my question and help. I am trying to self study in preparation for a course out of a book with mathematical electricity problems. I am stuck on a question involving Kirchhoff's Voltage Law. In this question you are given a circuit and are told to solve for the currents through the resistors using the KVL and simultaneous equations. The way it is explained in the book is like this "emf is considered positive if the negative terminal is met first" and "IR drop is considered positive if the current in resistor R is in the same direction as path being followed"

IMG_20180420_123132.jpg

The yellow arrows are the directions of current assumed and the red ones are the directions of the "path". The resistances by the emfs are internal to the emf and have to be taken in to account, and are considered as in series with the resistors in the branch.
Following these instructions I get two equations like this:

-4v = +(11R)I1 + (8R)I2
-8v = +(1R)I3 + (8R)I2

Since I3 = I1 + I2

-8v = (+1R)I1 + (9R)I2

Multiplying second loop equation by 11 to cancel I1 and adding the two equations.

-4v = -(11R)I1 + (8R)I2
-88v = +(11R)I1 + (99R)I2

-92 = (107)I2
-92/107= I2 = -0.859 < Not correct. Neither magnitude or polarity.

The correct I2 = 0.785 to the left, I1 = 0.935 to the left.

I noticed that if the polarity of the 88 volts were positive then in the end we would get

84/107 = I2 = 0.785 and we can derive the rest of the currents substituting I2 in the equation.

I feel as though I followed all the instructions so I don't know why I cant get 0.785 as I2.

Can anyone help? Much thanks!
 

WBahn

Joined Mar 31, 2012
26,145
You don't label your current arrows and so people have to guess which current is I1, I2 and I3. Even if they reverse engineer it out from your equations, that doesn't mean that you didn't misinterpret things when you set up your equations and thus have part of them consistent with one assignment and other parts consistent with a different assignment.

For instance, your first two equations are consistent with the yellow currents being labeled from I1 on the top to I3 on the bottom with I2 being in the middle. But your next equation is consistent with I3 being the middle current.

Don't make people from whom you are asking for free help have to guess.
 

WBahn

Joined Mar 31, 2012
26,145
Where are you getting that the correct value for I2 (whichever current that is) is 785 mA?

Have you used that current and determined whether or not it is actually a solution to the circuit?
 

Thread Starter

KrisK

Joined Apr 20, 2018
3
You don't label your current arrows and so people have to guess which current is I1, I2 and I3. Even if they reverse engineer it out from your equations, that doesn't mean that you didn't misinterpret things when you set up your equations and thus have part of them consistent with one assignment and other parts consistent with a different assignment.

For instance, your first two equations are consistent with the yellow currents being labeled from I1 on the top to I3 on the bottom with I2 being in the middle. But your next equation is consistent with I3 being the middle current.

Don't make people from whom you are asking for free help have to guess.
Thanks for responding. The currents are I1 top, I2 middle and I3 bottom.
In the third equation I replaced the current I3 with I1 + I2.

+(1RI3)=+(1R)I1 +(1R)I2

-8v = +(1R)I1 + (1R)I2 + (8R)I2

With like terms together
-8v = +(1RI1) + (9RI2)

Where are you getting that the correct value for I2 (whichever current that is) is 785 mA?

Have you used that current and determined whether or not it is actually a solution to the circuit?
The answers I gave are from the answer section of the book. It's just questions and answers at the end of the chapter, but they are not worked out problems. They don't ask for the third current going through the internal resistor of emf2 ( the bottom one ) they only ask for currents in the external resistances.

Answers from the book are: I1 = 0.935 to the left, I2 = 0.785 to the left, I3 = not given in the answers.
 

Thread Starter

KrisK

Joined Apr 20, 2018
3
I think there are a couple errors in my question regarding polarity writing down the IR rise and drop in the loop equations. Seeing as that's what I am confused about I'll forgo fixing the errors, it's what I need help with. The description of the question is correct, could someone help me based on the picture with the directions I assumed. I just need to find I2 with the correct loop equations.

I will be grateful.
 

mlv

Joined Nov 6, 2017
17
This textbook exercise is a more fundamental lesson in textbooks, I think - not in circuits. Try analyzing the circuit with the 8V battery terminals reversed.

I think you'll find that you've run across an error in the book.

One glaring contradiction in the circuit given and the answers from the book regarding current is that both batteries are positive on the left side of the circuit while the supplied answer says that current in the middle branch flows to the left.

Cheers,
M.
 

WBahn

Joined Mar 31, 2012
26,145

Thanks for responding. The currents are I1 top, I2 middle and I3 bottom.
In the third equation I replaced the current I3 with I1 + I2.


So why do you believe that I3 is equal to I1 + I2?

Think of the currents as people entering and leaving a gate (and in which no one can just stay in the gate, anyone that enters must immediately leave). The rightmost node is the gate, so the sum of the currents entering the gate must equal the sum of the currents leaving the gate.

Write the sum of the currents entering the gate on the left side of the equals sign and the sum of the currents leaving the gate on the right. What do you get?
 
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