Potential difference across two points in a circuit

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plazprestige
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Homework Statement


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Find the potential difference between points b and a in the circuit below
Circuit schematic with nodes a and b joined by three parallel branches: top branch 1.00 Ω resistor, 20.0 V battery and 6.00 Ω resistor; middle branch 4.00 Ω resistor, 1.00 Ω resistor and battery labelled E1; bottom branch 1.00 Ω resistor, battery labelled E2 and 2.00 Ω resistor. Arrows show 1.00 A and 2.00 A currents flowing down into node a.

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


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


 
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.
 
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?
 
plazprestige said:
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.
 
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)
 
plazprestige said:
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 = ? :)
 
ehild said:
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.
 
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?
 
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.
 
Could you explain how to get the sign right? I don't understand why it's negative rather than positive.
 
plazprestige said:
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