Kirchhoff's Voltage Law question

Thread Starter

Darrell75

Joined Apr 19, 2016
16
I'm kind of working through Grob's Basic Electricity (9th edition) and just finished Chapter 9, on Kirchhoff's Laws. Most of the material I am ok with, but in the problems for the chapter, I'm having some difficulty with numbers 10, 11, and 12. Number 10 requires using the method of branch currents to solve for the current and voltages across a number of resistors. Here's a copy of the illustration provided:

Grobcircuit.jpg
I've included my work with the answers I got, below. I'm probably not following the method to a T, but I'm not terribly worried about that. The answers aren't in the back of the book, but I modeled the circuit on a simulator and its results agreed with mine.

V1-V(R1)-V(R3)=0
V2-V(R3)-V(R2)=0

V2-V(R3)-V(R2)=0
15-15*I3-10*I2=0
15-15*(I1+I2)-10*I2=0
15-15*I1-15*I2-10*I2=0
15-15*I1-25*I2=0
15*I1+25*I2=15
I2=(15-15*I1)/25=3/5-3/5*I1

V1-V(R1)-V(R3)=0
10-10*I1-15*I3=0
10-10*I1-15*(I1+I2)=0
10-10*I1-15*I1-15*I2=0
10-25*I1-15*I2=0
25*I1+15*I2=10 -----------------------------> 25*1/16+15*I2=10
25*I1+15*(3/5-3/5*I1)=10..................... 10-25/16=15*I2
25*I1+45/5-45/5*I1)=10........................ 160/16-25/16=15*I2
25*I1+9-9*I1=10................................... 135/16=15*I2
16*I1=1................................................. I2=9/16 A
I1=1/16 A



V1-V(R1)-V(R3)=0
10-10*1/16-15*I3=0
10-10/16=15*I3
160/16-10/16=15*I3
(150/16)/15=I3
I3=10/16 A

V(R1)=10*1/16=10/16 V=0.625 V
V(R2)=
10*9/16=90/16=5.625 V
V(R3)=
15*10/16=150/16=9.375 V

Number 11 says to use KVL to prove that the sum of the voltages is zero in each of the three closed loops. For the left loop, going counter-clockwise from the negative terminal of V1, it says the answer is -10 V + 16.875 V - 15 V + 8.125 V = 0; for the right loop, counter-clockwise from the positive terminal of V2, it says the answer is 15 V - 16.875 V + 1.875 V = 0; for the outside loop, going counter-clockwise from the negative terminal of V1, it gives -10 V + 1.875 V + 8.125 V = 0.

Honestly, I have no idea what to do with the left loop, with its two voltage sources, but for the right loop, I would think 15 V - 15*10/16 - 10*9/16 = 0 (or 15 V - 9.375 V - 5.625 V = 0) should be the answer. For the outside loop, I would think - 10 V + 10*1/16 + 15*10/16 = 0 (or 10 V + 0.625 V + 9.375 V = 0) should be the answer. As I mentioned, these are not the answers the book gives. Could someone please help me understand where I am going wrong?

Number 12 involves analyzing the same circuit using the method of mesh currents, but I'll try to work through that one again before asking.

Thank you!
 

hp1729

Joined Nov 23, 2015
2,304
I'm kind of working through Grob's Basic Electricity (9th edition) and just finished Chapter 9, on Kirchhoff's Laws. Most of the material I am ok with, but in the problems for the chapter, I'm having some difficulty with numbers 10, 11, and 12. Number 10 requires using the method of branch currents to solve for the current and voltages across a number of resistors. Here's a copy of the illustration provided:

View attachment 116191
I've included my work with the answers I got, below. I'm probably not following the method to a T, but I'm not terribly worried about that. The answers aren't in the back of the book, but I modeled the circuit on a simulator and its results agreed with mine.

V1-V(R1)-V(R3)=0
V2-V(R3)-V(R2)=0

V2-V(R3)-V(R2)=0
15-15*I3-10*I2=0
15-15*(I1+I2)-10*I2=0
15-15*I1-15*I2-10*I2=0
15-15*I1-25*I2=0
15*I1+25*I2=15
I2=(15-15*I1)/25=3/5-3/5*I1

V1-V(R1)-V(R3)=0
10-10*I1-15*I3=0
10-10*I1-15*(I1+I2)=0
10-10*I1-15*I1-15*I2=0
10-25*I1-15*I2=0
25*I1+15*I2=10 -----------------------------> 25*1/16+15*I2=10
25*I1+15*(3/5-3/5*I1)=10..................... 10-25/16=15*I2
25*I1+45/5-45/5*I1)=10........................ 160/16-25/16=15*I2
25*I1+9-9*I1=10................................... 135/16=15*I2
16*I1=1................................................. I2=9/16 A
I1=1/16 A



V1-V(R1)-V(R3)=0
10-10*1/16-15*I3=0
10-10/16=15*I3
160/16-10/16=15*I3
(150/16)/15=I3
I3=10/16 A

V(R1)=10*1/16=10/16 V=0.625 V
V(R2)=
10*9/16=90/16=5.625 V
V(R3)=
15*10/16=150/16=9.375 V

Number 11 says to use KVL to prove that the sum of the voltages is zero in each of the three closed loops. For the left loop, going counter-clockwise from the negative terminal of V1, it says the answer is -10 V + 16.875 V - 15 V + 8.125 V = 0; for the right loop, counter-clockwise from the positive terminal of V2, it says the answer is 15 V - 16.875 V + 1.875 V = 0; for the outside loop, going counter-clockwise from the negative terminal of V1, it gives -10 V + 1.875 V + 8.125 V = 0.

Honestly, I have no idea what to do with the left loop, with its two voltage sources, but for the right loop, I would think 15 V - 15*10/16 - 10*9/16 = 0 (or 15 V - 9.375 V - 5.625 V = 0) should be the answer. For the outside loop, I would think - 10 V + 10*1/16 + 15*10/16 = 0 (or 10 V + 0.625 V + 9.375 V = 0) should be the answer. As I mentioned, these are not the answers the book gives. Could someone please help me understand where I am going wrong?

Number 12 involves analyzing the same circuit using the method of mesh currents, but I'll try to work through that one again before asking.

Thank you!
Why don't they write such questions referring to real world circuits?
 

WBahn

Joined Mar 31, 2012
26,398
When you say that I1 = 1/16 A, that is ambiguous since you haven't defined what direction I1 is flowing (when it's value is positive). In fact, you are relying on the readers to assume which branch current I1 even refers to. Do force your readers to guess or assume that you know what their guesses are going to be. Engineering is NOT about guessing.

The same goes for V(R1). You have given no indication what the polarity of V(R1) is, so if it turns out to be -1.3 V, we have no idea which side of R1 is more positive than the other side.

Annotate your drawings with the definitions, including polarity, of the quantities you are going to use in your analysis.
 

WBahn

Joined Mar 31, 2012
26,398
Why don't they write such questions referring to real world circuits?
Perhaps because there aren't that many "real world circuits" consisting of batteries and resistors that allow students brand new to this stuff to start getting familiar with applying KVL, KCL, and Ohm's Law and to start developing analysis skills.

It might be similar to why, when my daughter started learning math, it was with stuff like 2 + 3 and not with "real world numbers" like sqrt(2) and ln(pi).
 

Thread Starter

Darrell75

Joined Apr 19, 2016
16
I've included a screenshot of the circuit modeled on the EveryCircuit app, below.

I know this isn't what I wrote in my first post, but for the right loop, going CCW from the positive terminal of V2, it seems to me the answer should be 15 V - 15 V (across V2) + 5.625 V + 9.375 V = 15 V. For the outside loop, going CCW from the negative terminal of V1, I would think - 10 V + 0.625 V + 9.375 V = 0 should be the answer.

As I mentioned, these answers do not agree with the book answers. For the right loop, counter-clockwise from the positive terminal of V2, it says the answer is 15 V - 16.875 V + 1.875 V = 0; for the outside loop, going counter-clockwise from the negative terminal of V1, it gives -10 V + 1.875 V + 8.125 V = 0.

Screenshot_2016-12-01-07-48-45.png
 

WBahn

Joined Mar 31, 2012
26,398
Look at your diagram and put your finger on "V2" in the right loop.

What? Can't find anything labeled "V2" in the diagram?

Well, gee, neither can we!

I'm not going to reverse engineer your notation or guess at what you mean just because you can't be bothered to define your terms on your diagram.
 

DGElder

Joined Apr 3, 2016
351
Honestly, I have no idea what to do with the left loop, with its two voltage sources, but for the right loop, I would think 15 V - 15*10/16 - 10*9/16 = 0 (or 15 V - 9.375 V - 5.625 V = 0) should be the answer. For the outside loop, I would think - 10 V + 10*1/16 + 15*10/16 = 0 (or 10 V + 0.625 V + 9.375 V = 0) should be the answer. As I mentioned, these are not the answers the book gives. Could someone please help me understand where I am going wrong?


Thank you!
The book answers are wrong or at least don't match up with your schematic. Your right loop KVL summation is correct. For the outside loop equation you lost the negative sign for -10V when writing the expression in the parenthesis. To do the left loop KVL expression the procedure is the same, so I don't understand why you can not make an attempt at it. Showing us your attempt will help identify any problems with your understanding of the principles or procedures of KVL analysis.
 
Last edited:

"Honestly, I have no idea what to do with the left loop, . . . Could someone please help me understand where I am going wrong?"


Good News, albeit much delayed!!! I've solved the Loops!!!


Recapping the Rs, Vs, and Is, we have:
V
1 = 10 Volts ; V2 = 15 Volts
R
1 = 10 Ohms ; R2 = 10 Ohms ; R3 = 15 Ohms
I
1 = .0625 Amps ; I2 = .5625 Amps ; I3 = .6250 Amps
VR1 = .625 Volts ; VR2 = 5.625 Volts ; VR3 = 9.375 Volts
Solving for the Loop Equations
Loop1 = V1 + VR1 - V2 +VR2 = 10V - .625V - 15V + 5.625V = 0 Volts
Loop2 =
-V2+ VR3 + VR2 = -15V + 5.625V + 9.375V = 0 Volts
Grand Loop = V1- VR1 - VR3 = 10V - .625V -9.375V = 0 Volts

I think this is what you were looking for. I was working the same problem yesterday. Your work showed me the way. The least I can do is to repay you by answering your question.
 
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