Newbie Needs Help! Kirchoff's Voltage Law problem

Thread Starter

doggettdoggett

Joined Jan 17, 2017
6
Can anybody help point me in the right direction to solve this problem? At first i thought I would use the superposition theorem but without Resistor values I have no idea how to approach this problem. I can see that R2 and R3 are in parallel with Vs2, but i'm not sure how this would effect the total voltage of the circuit. Any help or hints would be greatly appreciated.
IMG_2586.JPG
 

WBahn

Joined Mar 31, 2012
26,398
Pick any closed loop around some portion of the circuit in which you know all but one of the voltages. Use KVL to find the unknown voltage. That voltage is now a known voltage for use in finding the value other unknown voltages.

Using that hint, show your best attempt to find Vr3 if you are still having problems.
 

WBahn

Joined Mar 31, 2012
26,398
So Vr3 would be Vs2-Vr2 = 9V - 4.3 V= 4.7V?
Voltages and currents always have polarity, so there is no way to tell if your answer is correct unless you define the polarity of Vr3 (not that the problem not only did not do this for you, but explicitly stated that you were expected to do so. Once you've done that I can review your answer, but it looks like you did it correctly.

Although in looking at your diagram, it looks like perhaps you already tried to indicate the polarity but then erased it. In either case, make it very clear.
 

Thread Starter

doggettdoggett

Joined Jan 17, 2017
6
I'm not sure how to show polarity through text. All i know is the voltage drop across the resistor is 4.7 V dc if i'm correct, with current flowing from left to right according to the diagram layout. Your response is much appreciated.
 

Thread Starter

doggettdoggett

Joined Jan 17, 2017
6
And now i'm curious how Vs1 effects this circuits voltage drops. The two sources are series aiding I believe, so would that equal a 15 Vdc source. I guess the parallel arrangement with resistors in between is what is given me the hardest time to visualize the voltage drops.
 

WBahn

Joined Mar 31, 2012
26,398
I'm not sure how to show polarity through text. All i know is the voltage drop across the resistor is 4.7 V dc if i'm correct, with current flowing from left to right according to the diagram layout. Your response is much appreciated.
There are a number of ways to describe it in text. One way is to simply state that Vr3 is defined as the voltage of the right side of R3 with respect to the left side. That means that if Vr3 is a positive value, that the right side of R3 is at a higher potential than the left side.

Note that you can flip a coin in making this assignment. So just for illustration sake, let's define Vr3 such that it is the left side with respect to the right side (in which case we know we should get an answer of - 4.7 V). Now let's go clockwise around the upper right loop starting at point B and total up all of the voltage drops as we go and set them equal to zero. We first encounter Vr3 and, as we have defined it, as we go from right to left (clockwise around the loop) we are going from the negative side of Vr3 to the positive side so it is a voltage gain and, hence, a negative voltage drop. As we go up through R2 we drop 4.3 V. Notice how it is only the presence of the polarity marks that we know whether to add or subtract the 4.3 V -- just knowing that Vr2 is 4.3 V is not sufficient to determine this very important thing. Finally we have another voltage gain (or negative drop) as we come down through Vs2. This makes our KVL equation

- (Vr3) + (4.3 V) - (9 V) = 0

Vr3 = - (9 V) + (4.3 V)

Vr3 = - 4.7 V

Regardless of how we define Vr3, the voltage ends up being such that the right side of R3 is more positive than the left side, so current is flowing from right to left, not left to right (unless you are one of the unfortunate souls that is being force to use "electron current").

Notice that in putting together the KVL loop equation I never once made any reference whatsoever to any of the components that were not along the loop path. I completely ignored them. So just define your path and then only pay attention to the components that are along that path and that path only -- because KVL applies to a closed path.
 

WBahn

Joined Mar 31, 2012
26,398
And now i'm curious how Vs1 effects this circuits voltage drops. The two sources are series aiding I believe, so would that equal a 15 Vdc source. I guess the parallel arrangement with resistors in between is what is given me the hardest time to visualize the voltage drops.
Vs1 and Vs2 are most definitely NOT in series. In order to be in series, whatever current flows in one device MUST also flow in the other device. But current that flows in Vs1 can split with some of it going through Vs2 and some of it going through R2/R3.
 

Thread Starter

doggettdoggett

Joined Jan 17, 2017
6
There are a number of ways to describe it in text. One way is to simply state that Vr3 is defined as the voltage of the right side of R3 with respect to the left side. That means that if Vr3 is a positive value, that the right side of R3 is at a higher potential than the left side.

Note that you can flip a coin in making this assignment. So just for illustration sake, let's define Vr3 such that it is the left side with respect to the right side (in which case we know we should get an answer of - 4.7 V). Now let's go clockwise around the upper right loop starting at point B and total up all of the voltage drops as we go and set them equal to zero. We first encounter Vr3 and, as we have defined it, as we go from right to left (clockwise around the loop) we are going from the negative side of Vr3 to the positive side so it is a voltage gain and, hence, a negative voltage drop. As we go up through R2 we drop 4.3 V. Notice how it is only the presence of the polarity marks that we know whether to add or subtract the 4.3 V -- just knowing that Vr2 is 4.3 V is not sufficient to determine this very important thing. Finally we have another voltage gain (or negative drop) as we come down through Vs2. This makes our KVL equation

- (Vr3) + (4.3 V) - (9 V) = 0

Vr3 = - (9 V) + (4.3 V)

Vr3 = - 4.7 V

Regardless of how we define Vr3, the voltage ends up being such that the right side of R3 is more positive than the left side, so current is flowing from right to left, not left to right (unless you are one of the unfortunate souls that is being force to use "electron current").

Notice that in putting together the KVL loop equation I never once made any reference whatsoever to any of the components that were not along the loop path. I completely ignored them. So just define your path and then only pay attention to the components that are along that path and that path only -- because KVL applies to a closed path.


Thank you for this detailed explanation . Let's check to see if I understand it correctly by trying to solve for Vr1. Assume that the left side is more positive than the right side and following the closed loop around the circuit, I've come up with :
VR1 -( 9 V) + ( 1.7 V ) - (6 V) = 0
Vr1 = +9V - 1.7V + 6 V
Vr1 = 13.3 V with the left side of R1 at a higher potential than the left side .
 

WBahn

Joined Mar 31, 2012
26,398
Thank you for this detailed explanation . Let's check to see if I understand it correctly by trying to solve for Vr1. Assume that the left side is more positive than the right side and following the closed loop around the circuit, I've come up with :
VR1 -( 9 V) + ( 1.7 V ) - (6 V) = 0
Vr1 = +9V - 1.7V + 6 V
Vr1 = 13.3 V with the left side of R1 at a higher potential than the left side .
That is correct. I think you are getting the hang of it.

Now that you are getting the idea, one thing that usually makes things go more easily and also reduces the chances of silly mistakes due to too many minus signs running around is to tweak KVL from "the sum of the voltage drops around the circuit is zero" to "the sum of the voltage drops is equal to the sum of the voltage gains". Then you simply put the drops as you go around the circuit on the left side of the equation and the gains on the other. So that allows you to immediately write your KVL equation as

Vr1 + (1.7 V) = (9 V) + (6 V)
 
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