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

[AbhinavJ]

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?

 

[cnh1995]

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..

 

[AbhinavJ]

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

 

[cnh1995]

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

[ mentor note: content deleted ]

 

[Kinta]

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, V_AX 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?

 

[SammyS]

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?

 

[Kinta]

@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.

 

[cnh1995]

Please read the rules for offering homework help.

Do NOT give complete solutions.

My apologies..

 

[cnh1995]

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..

 

[SammyS]

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.

 

[Kinta]

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.

 

[cnh1995]

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..

 

[Kinta]

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?

 

[cnh1995]

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)..

 

[Kinta]

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.