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KVL Exercise - Basic

by Ammar w
Tags: basic, exercise
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Ammar w
#1
Feb6-13, 09:30 AM
P: 28
1. The problem statement, all variables and given/known data
Click on the link
https://www.diigo.com/item/image/2sb3i/yjim


2. Relevant equations
KVL Law
Similar Example :
https://www.diigo.com/item/image/2sb3i/orwa

3. The attempt at a solution
(a)
In the left loop :
-8-12+VR2 = 0
VR2 = -20 v

(b)
In the big loop :
-8-12+7-9-V2-3+VR1 = 0
-8-12+7-9-V2-3+1 = 0
V2 = -24

Why is my solution wrong?

If I take the right loop for (b) using the right answer for (a) :
-4+7-9-V2-3+1 = 0
V2 = -8 which is true.

Is the wrong with the negative voltage (-8) ??

thanks.
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gneill
#2
Feb6-13, 09:48 AM
Mentor
P: 11,618
Quote Quote by Ammar w View Post
Is the wrong with the negative voltage (-8) ??
Yes, it looks like whoever created the circuit diagram created confusion when they labeled the leftmost resistor's potential drop. As it is marked, strictly speaking there should be a -8V potential drop across that resistor proceeding from bottom to top, which translates into a +8V potential rise as you've properly interpreted it. However, it would appear that they (confusingly) intended the "+ -" indicators to show the actual direction of the potential change of 8V.

Bottom line: assume that the "-8V" is really "8V" oriented according to the "+ -".
Ammar w
#3
Feb6-13, 10:03 AM
P: 28
thank you
but I see that there is no difference if it was "+8" instead of "-8" ???

gneill
#4
Feb6-13, 10:15 AM
Mentor
P: 11,618
KVL Exercise - Basic

Quote Quote by Ammar w View Post
thank you
but I see that there is no difference if it was "+8" instead of "-8" ???
Um, how can it make no difference if you change the sign of a potential change?

For part (a) your equation summing the potential drops would become: +8-12+VR2 = 0
Ammar w
#5
Feb6-13, 10:27 AM
P: 28
ok, how the solution of (a) will be if :
https://www.diigo.com/item/image/2sb3i/y0us
gneill
#6
Feb6-13, 10:35 AM
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P: 11,618
Quote Quote by Ammar w View Post
ok, how the solution of (a) will be if :
https://www.diigo.com/item/image/2sb3i/y0us
Just solve your equation that I presented in my previous post...
Ammar w
#7
Feb6-13, 10:44 AM
P: 28
I'm sorry
but can you tell me the equation of (a) if the potential change was "-8"?
I thought that the equation +8-12+VR2 = 0 is used when the potential change = -8.
gneill
#8
Feb6-13, 11:41 AM
Mentor
P: 11,618
Quote Quote by Ammar w View Post
I'm sorry
but can you tell me the equation of (a) if the potential change was "-8"?
I thought that the equation +8-12+VR2 = 0 is used when the potential change = -8.
Yes, that's right. The problem is not with your understanding, it's with the problem itself; the resistor is not labeled correctly for the given answers. The resistor should have its potential labeled as "8V", not "-8V", or alternatively, the + and - tags should be reversed.
Ammar w
#9
Feb6-13, 11:44 AM
P: 28
aha
thank you Mr. gneill


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