Determine the voltage at each point with respect to ground

[Lay1]

Homework Statement

Determine the voltage at each point with respect to ground in the figure provided.

Relevant Equations

V=I/R, V(drop)=Rx/Rt * Vs

This is the figure stated for problem 57.

 

[BvU]

Is there a question here ?

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[Lay1]

Is there a question here ?

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Yes, I have stated in the homework statement. It is also shown again in the picture shown in the attempt section. Thank you.

 

[phinds]

Yes, I have stated in the homework statement. It is also shown again in the picture shown in the attempt section. Thank you.

Yes, you have stated what the question is asking YOU. What is needed is, what are you asking US?

 

[nasu]

The figure in your drawing is not the same as in the book image. It is not even clear what problem are you asking about.

 

[Lay1]

The figure in your drawing is not the same as in the book image. It is not even clear what problem are you asking about.

The first photo is just my attempt showing that I have tried. The probem number 57 is what I am asking about. Sorry for your confusion.

 

[Lay1]

Yes, you have stated what the question is asking YOU. What is needed is, what are you asking US?

As you can see I got the wrong answer. I dont know where I did wrong. That is what I am asking about.

 

[gneill]

@Lay1 combined the two sources into one, so the image in the picture is okay, it just didn't mention the additional step.

One way to do this problem is to first work out the total current and then use KVL around the loop, adding in the successive resistors as you go. For example, finding the voltage at point C you could start at the negative terminal of the 15 V battery and work your way around the loop clockwise until reaching point C.

$V_C = 15 V - (56 kΩ + 560 kΩ + 100 kΩ) ~ I$

 

[nasu]

As you can see I got the wrong answer. I dont know where I did wrong. That is what I am asking about.

Where can I see that you got the wrong answer?

 

[Lay1]

@Lay1 combined the two sources into one, so the image in the picture is okay, it just didn't mention the additional step.

One way to do this problem is to first work out the total current and then use KVL around the loop, adding in the successive resistors as you go. For example, finding the voltage at point C you could start at the negative terminal of the 15 V battery and work your way around the loop clockwise until reaching point C.
View attachment 324750

$V_C = 15 V - (56 kΩ + 560 kΩ + 100 kΩ) ~ I$

Thank you for your very thorough reply. I will work out using this way. However may I know where did I do exactly wrong because I did not get the result that is described in the textbook. Thank you for your help so far.

 

[Lay1]

Where can I see that you got the wrong answer?

It is in the first photo with handwritten figure and characters. Thank you.

 

[bob012345]

Thank you for your very thorough reply. I will work out using this way. However may I know where did I do exactly wrong because I did not get the result that is described in the textbook. Thank you for your help so far.

What did the textbook give?

 

[Lay1]

What did the textbook give?

V(A)=14.82V
V(B)=12.97V
V(C)=12.64V
V(D)=9.34V
The answers are these.

 

[bob012345]

V(A)=14.82V
V(B)=12.97V
V(C)=12.64V
V(D)=9.34V
The answers are these.

Then can you figure what is $V_{AC}$ and $V_{AC}$ from these values?

 

[Lay1]

Yes, it is 2.18V.

 

[bob012345]

Yes, it is 2.18V.

Isn't that what you had above which you claimed was wrong?

 

[Lay1]

No. It is problem 58. The problem I am asking about is problem 57.

 

[bob012345]

No. It is problem 58. The problem I am asking about is problem 57.

Do you understand it now? Are there still issues?

 

[DaveE]

When you redrew the schematic, you combined the two batteries. In that step you redefined what the ground potential was. So imagine you were asked what is the voltage from the leftmost battery (connection to R1) to ground. In the original schematic it's 15V. In your schematic it's 6V.

I haven't checked for other mistakes. But I'd imagine that all of your answers are off by 9V.

 

[gneill]

Thank you for your very thorough reply. I will work out using this way. However may I know where did I do exactly wrong because I did not get the result that is described in the textbook. Thank you for your help so far.

Sorry, but I can't quite make out your figures in the image provided. I believe that you're trying to form the partial resistance over the total resistance for each of the points A,B,C,D? For example:

If the total voltage is $6 V$, and $R_T$ is the total resistance, then the voltage at $C$ with respect to ground will be:
$$V_{CG} = \frac{R_4 + R_5}{R_T} 6 V$$
If that's so, then we'll need to see a better image of your work to find the error.

 

[Lay1]

Do you understand it now? Are there still issues?

Actually, I haven't sorted out the problem yet. I will try again using KVL as suggested by a mentor here. Thank you for your concern:)

 

[Lay1]

When you redrew the schematic, you combined the two batteries. In that step you redefined what the ground potential was. So imagine you were asked what is the voltage from the leftmost battery (connection to R1) to ground. In the original schematic it's 15V. In your schematic it's 6V.

I haven't checked for other mistakes. But I'd imagine that all of your answers are off by 9V.

I will check it out again. Thank you.

 

[Lay1]

Sorry, but I can't quite make out your figures in the image provided. I believe that you're trying to form the partial resistance over the total resistance for each of the points A,B,C,D? For example:
View attachment 324753
If the total voltage is $6 V$, and $R_T$ is the total resistance, then the voltage at $C$ with respect to ground will be:
$$V_{CG} = \frac{R_4 + R_5}{R_T} 6 V$$
If that's so, then we'll need to see a better image of your work to find the error.

V(A)=(R2+R3+R4+R5)/Rt *6V
V(B)=(R3+R4+R5)/Rt *6V
V(C)=(R4+R5)/Rt *6V
V(D)=R5/Rt * 6V
It is how I work out for the problem.

 

[bob012345]

You can reduce the book schematic to a very simple form by using a well known property of resistors in series.

 

[Lay1]

You mean Rt=R1+R2+....
If it is so, I will also consider this fact

 

[DaveE]

V(A)=(R2+R3+R4+R5)/Rt *6V
V(B)=(R3+R4+R5)/Rt *6V
V(C)=(R4+R5)/Rt *6V
V(D)=R5/Rt * 6V
It is how I work out for the problem.

First, let's name the other nodes in the original schematic (the one with 2 batteries). So "E" is between R5 and the 9V battery. "F" is between R1 and the 15V battery.

V(A)=(R2+R3+R4+R5)/Rt *6V is what I would call V_AE, the voltage across R2, R3, R4, and R5; or the voltage from node A to node E. They want the voltage at node A with respect to ground, not with respect to node E. Voltmeters always have two probes. For convenience in communication we assign (sometimes a bit arbitrarily) one node in the circuit to be "ground" and all voltages that are expressed as the voltage at node "X" are assumed to have the other (-) voltmeter probe at the predefined ground location.

Your problem is to find the voltage at node A, B, ...etc. with respect to ground; meaning that the - meter lead is at ground. What if there was no ground symbol. What if we just called that node "G". Could you find the voltage from node A wrt node G. This is essentially the same as problem 58. Find the voltages between two nodes, one of which is between the two batteries in this case.

BTW: anyone that is using your simplified schematic with only one battery isn't helping. It's problem #57 people! They probably haven't really read the problem in question. Best to ignore them and focus on my previous comment about redefining "ground".

 

[Lay1]

First, let's name the other nodes in the original schematic (the one with 2 batteries). So "E" is between R5 and the 9V battery. "F" is between R1 and the 15V battery.

V(A)=(R2+R3+R4+R5)/Rt *6V is what I would call V_AE, the voltage across R2, R3, R4, and R5; or the voltage from node A to node E. They want the voltage at node A with respect to ground, not with respect to node E. Voltmeters always have two probes. For convenience in communication we assign (sometimes a bit arbitrarily) one node in the circuit to be "ground" and all voltages that are expressed as the voltage at node "X" are assumed to have the other (-) voltmeter probe at the predefined ground location.

Your problem is to find the voltage at node A, B, ...etc. with respect to ground; meaning that the - meter lead is at ground. What if there was no ground symbol. What if we just called that node "G". Could you find the voltage from node A wrt node G. This is essentially the same as problem 58. Find the voltages between two nodes, one of which is between the two batteries in this case.

BTW: anyone that is using your simplified schematic with only one battery isn't helping. It's problem #57 people! They probably haven't really read the problem in question. Best to ignore them and focus on my previous comment about redefining "ground".

Yes. Thank you so much for your detailed explanation. I will do as you say.

 

[TonyStewart]

I hope you understand, Ground= 0V always unless it is not 0V. i.e. "a bad ground"

You have R5 "grounded" which is incorrect for absolute KVL Kirchoff Voltage Loop questions like this, but OK for KCL current loop calculations as you tried (Ʃ=equiv. sum)
I (loop) = Ʃ V / Ʃ R

 

[Lay1]

I hope you understand, Ground= 0V always unless it is not 0V. i.e. "a bad ground"

You have R5 "grounded" which is incorrect for absolute KVL Kirchoff Voltage Loop questions like this, but OK for KCL current loop calculations as you tried (Ʃ=equiv. sum)
I (loop) = Ʃ V / Ʃ R

Thank you for your guidance. I very much appreciate it.