Potential difference across two points in a circuit

  • #1

Homework Statement


[/B]
Find the potential difference between points b and a in the circuit below
Screen Shot 2015-01-31 at 1.09.33 AM.png

I have already solved for the voltages of the two batteries (1 and 2) in the circuit (18 V and 7 V respectively)

2. Homework Equations


Kirchhoff's Rules
1) Potential difference across any closed loop is 0
2) Current in = current out

The Attempt at a Solution


[/B]
I have read that Kirchhoff's current rule is how to approach these kinds of problems, but I am not sure how.


 

Answers and Replies

  • #2
ehild
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You have found the voltages of the batteries correctly. How did you do it? For that, you had to use the potential difference between points a and b, that you know from the uppermost branch of the circuit.
 
  • #3
I used Kirchhoffs voltage law to find the battery voltages. Could you elaborate on how I unknowingly found the potential difference between b and a please?
 
  • #4
ehild
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I used Kirchhoffs voltage law to find the battery voltages. Could you elaborate on how I unknowingly found the potential difference between b and a please?
Right the equation, please.
 
  • #5
For battery 2:

-2(1) - E2 - 2(2) - 1(6) + 20 -1(1) = 0, solve for E2 (Kirchhoff's voltage law around the large loop)

For battery 1:

E1 - 1(1) - 1(4) - 2(1) - E2 - 2(2) = 0, solve for E1 (Kirchhoff's voltage law around the bottom half loop)
 
  • #6
ehild
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For battery 2:

-2(1) - E2 - 2(2) - 1(6) + 20 -1(1) = 0, solve for E2 (Kirchhoff's voltage law around the large loop)
You started from point a and and followed the change of potential along the current. The potential change from a to b is Ub-Ua = -2(1) - E2 - 2(2) .
From b to a, to potential changes by - 1(6) + 20 -1(1)= Ua-Ub. The potential difference between b and a is Ub-Ua = ??? :)
 
  • #7
You started from point a and and followed the change of potential along the current. The potential change from a to b is Ub-Ua = -2(1) - E2 - 2(2) .
From b to a, to potential changes by - 1(6) + 20 -1(1)= Ua-Ub. The potential difference between b and a is Ub-Ua = ??? :)
Oh, I think I understand. So if I start at point b and end at point a:

-1(6) + 20 - 1(1) = Vb - Va = 13 V

Is this correct? I only I have one try left, so I don't want to squander it.
 
  • #8
ehild
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It must be correct.
 
  • #9
Ok, I got the magnitude right (13) but the sign wrong (it was - 13 V). I don't understand why. If it wanted the potential difference from b to a, why would you not start at b and work your way towards a?
 
  • #10
ehild
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It depends how the potential difference between points a and b is defined. Usually it is Ua-Ub. But the question was the potential between b and a, which is Ub-Ua, and it is negative.
 
  • #11
Could you explain how to get the sign right? I don't understand why it's negative rather than positive.
 
  • #12
ehild
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Could you explain how to get the sign right? I don't understand why it's negative rather than positive.
In Post #6, the potential changed from Vb to Va by 13 V. So Va =13+ Vb, Va-Vb =13 . When it is said "potential difference between X and Y" your book means that you have to take the change of potential from Y to X: VX-VY.
The question was
Find the potential difference between points b and a in the circuit below
You have to find the potential at b with respect to a, that is, Vb-Va. Vb-Va = - (Va-Vb) = -13 V
 

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