# Homework Help: Potential at point in circuit

1. Jun 4, 2015

### AbhinavJ

• Warned: URL/Image only problem statement
2. Jun 4, 2015

### cnh1995

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..

Last edited: Jun 4, 2015
3. Jun 4, 2015

### AbhinavJ

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

4. Jun 4, 2015

### cnh1995

• member has been warned that providing complete or almost complete solutions is forbidden at Physics Forums
[ mentor note: content deleted ]

Last edited by a moderator: Jun 4, 2015
5. Jun 4, 2015

### Kinta

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

6. Jun 4, 2015

### SammyS

Staff Emeritus
@Kinta :

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

Last edited: Jun 4, 2015
7. Jun 4, 2015

### Kinta

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.

8. Jun 4, 2015

### cnh1995

My apologies..

9. Jun 4, 2015

### cnh1995

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. Jun 4, 2015

### SammyS

Staff Emeritus
If you simply remove the fourth capacitor, then the 10 Volt potential difference is split across only three capacitors instead of four.

11. Jun 4, 2015

### Kinta

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. Jun 4, 2015

### cnh1995

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. Jun 4, 2015

### 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?

14. Jun 4, 2015

### cnh1995

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)..

15. Jun 4, 2015

### Kinta

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.

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