What is the role of a grounded point in a circuit?

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Homework Help Overview

The discussion revolves around the role of a grounded point in a circuit, specifically in relation to voltage potentials and reference points within the circuit. Participants explore how grounding affects the potential difference between various points in the circuit.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants question the implications of grounding on voltage measurements, particularly how different reference points affect the perceived voltage at various nodes in the circuit. There are inquiries about the effects of removing components, such as capacitors, on the overall circuit behavior.

Discussion Status

The discussion is active, with participants offering insights and asking clarifying questions. Some have provided explanations regarding the relationship between grounding and voltage distribution, while others express their own uncertainties and seek further understanding.

Contextual Notes

There are indications of confusion regarding the concept of grounding and its implications in circuit analysis. Participants are navigating through assumptions about reference points and the effects of circuit modifications on voltage readings.

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AbhinavJ said:

Homework Statement



https://docs.google.com/file/d/0B0FIIKckKOcKM1FmXzJkYWJiZ2c/edit?usp=docslist_api

Homework Equations

The Attempt at a Solution


Shouldn't the potential be 10 as it is directly connected to the battery?
Potential is assumed w.r.t some reference called 'ground' which is assumed to be at 0V. In your circuit, point x is at 0V and not the -ve terminal of the battery. So, if you say A is at 10V, that means -ve terminal of the battery is earthed (which is not). So, taking x as reference, A is at 7.5 V and the potential distribution from A to B is not 10V to 0V ..its actually 7.5V to -2.5V..
 
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cnh1995 said:
Potential is assumed w.r.t some reference called 'ground' which is assumed to be at 0V. In your circuit, point x is at 0V and not the -ve terminal of the battery. So, if you say A is at 10V, that means -ve terminal of the battery is earthed (which is not). So, taking x as reference, A is at 7.5 V and the potential distribution from A to B is not 10V to 0V ..its actually 7.5V to -2.5V..

Can you please tell me how to calculate that? Even a hint would do. Thanks
 
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AbhinavJ said:
Can you please tell me how to calculate that? Even a hint would do. Thanks

[ mentor note: content deleted ]
 
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cnh1995 said:
PD between A and B is 10V and voltage across each capacitor will be 2.5V (equal). So, between X and A, there are 3 capacitors. So, VAX is 7.5V with x at 0V potential. This means, A is at 7.5V.

Why is placing the ground between the third and fourth capacitor any different than simply removing the fourth capacitor and unearthing the point X?
 
Kinta said:
Why is placing the ground between the third and fourth capacitor any different than simply removing the fourth capacitor and unearthing the point X?
@Kinta :

Is this a question to help the OP, or are you asking this because you don't know the answer?

IMG_20150604_140746.jpg
 
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SammyS said:
@Kinta :

Is this a question to help the OP, or are you asking this because you don't know the answer?

I'm asking because I don't know. I should know, but I don't. I figured if I had difficulty understanding this concept, someone else (maybe even OP) might have similar troubles.
 
SammyS said:
Please read the rules for offering homework help.

Do NOT give complete solutions.
My apologies..
 
Kinta said:
Why is placing the ground between the third and fourth capacitor any different than simply removing the fourth capacitor and unearthing the point X?
I think it will depend on where you assume the ground (0V) after X is unearthed. But I'm not getting your idea of removing the fourth capacitor..
 
  • #10
Kinta said:
Why is placing the ground between the third and fourth capacitor any different than simply removing the fourth capacitor and unearthing the point X?
If you simply remove the fourth capacitor, then the 10 Volt potential difference is split across only three capacitors instead of four.
 
  • #11
SammyS said:
If you simply remove the fourth capacitor, then the 10 Volt potential difference is split across only three capacitors instead of four.

I understand that. What I don't understand is if this is any different than placing the ground at the point X in OP's picture.
 
  • #12
Kinta said:
I understand that. What I don't understand is if this is any different than placing the ground at the point X in OP's picture.
If you remove the 4th capacitor, automatically point x (ground) will be connected to the -ve terminal of the battery. So, A will now be at 10V instead of 7.5V..
 
  • #13
Once the ground is placed at point X, why does it matter that there is a capacitor between it and the -ve terminal? Aren't they both grounded?
 
  • #14
Kinta said:
Once the ground is placed at point X, why does it matter that there is a capacitor between it and the -ve terminal? Aren't they both grounded?
No they aren't..Ground is just a reference point..It can be anywhere in the circuit.. Battery neither knows nor cares where you place the ground. It will just establish a 10V pd between A and B. How to "see" the 10V path is decided by the ground. If ground is at B, then it will be
10V(point A)---7.5V---5V---2.5V(point X)---0V(point B). In the problem, X is assumed to be 0V. So, the 10V path is
7.5V(A)-5V-2.5V-0V(X)- -2.5V(B)..
 
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  • #15
cnh1995 said:
No they aren't..Ground is just a reference point..It can be anywhere in the circuit.. Battery neither knows nor cares where you place the ground. It will just establish a 10V pd between A and B. How to "see" the 10V path is decided by the ground. If ground is at B, then it will be
10V(point A)---7.5V---5V---2.5V(point X)---0V(point B). In the problem, X is assumed to be 0V. So, the 10V path is
7.5V(A)-5V-2.5V-0V(X)- -2.5V(B)..

Ok. I think I've got it now. My confusion was stemming from a misunderstanding of what it means to have a grounded point in a circuit. Thanks. :biggrin:
 
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