KVL Exercise: Basic Homework Statement & Solution

In summary, the homework statement is that if you take the right loop for (b) using the right answer for (a) then the equation summing the potential drops becomes: +4+7-9-V2-3+1 = 0.
  • #1
Ammar w
28
0

Homework Statement


Click on the link
https://www.diigo.com/item/image/2sb3i/yjim


Homework Equations


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

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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  • #2
Ammar w said:
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 "+ -".
 
  • #3
thank you
but I see that there is no difference if it was "+8" instead of "-8" ?
 
  • #4
Ammar w said:
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
 
  • #5
ok, how the solution of (a) will be if :
https://www.diigo.com/item/image/2sb3i/y0us
 
Last edited by a moderator:
  • #6
Ammar w said:
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...
 
Last edited by a moderator:
  • #7
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.
 
  • #8
Ammar w said:
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.
 
  • #9
aha
thank you Mr. gneill
 

1. What is KVL Exercise?

KVL Exercise stands for Kirchhoff's Voltage Law Exercise. It is a basic homework statement that involves applying Kirchhoff's Voltage Law to solve a circuit problem.

2. What is Kirchhoff's Voltage Law?

Kirchhoff's Voltage Law, also known as KVL, states that the sum of all voltage drops in a closed loop circuit is equal to the sum of all voltage sources in that same loop. This law is based on the principle of conservation of energy.

3. Why is KVL important in circuit analysis?

KVL is important in circuit analysis because it allows us to determine the values of unknown voltages in a circuit by using the known values of voltage sources and voltage drops. It is a fundamental tool in understanding and solving complex electrical circuits.

4. How do you apply KVL to solve a circuit problem?

To apply KVL, you need to first identify the closed loop in the circuit. Then, you write down the equation that states the sum of all voltage drops in that loop is equal to the sum of all voltage sources. Finally, you solve the equation for the unknown voltage.

5. Can KVL be applied to any circuit?

Yes, KVL can be applied to any circuit, as long as it is a closed loop circuit. This means that all the components and voltage sources in the circuit are connected in a loop without any breaks or open paths.

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