How do i find the difference in voltage from point A to B

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Homework Help Overview

The discussion revolves around finding the difference in voltage between two points, A and B, in a circuit involving resistors and batteries. The original poster presents their calculations and expresses confusion regarding the signs of the currents and the resulting voltage difference.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss their interpretations of current directions and voltage polarities, with some questioning the signs assigned to the currents based on their circuit diagrams. There is an exploration of Kirchhoff's laws and how they apply to the circuit setup.

Discussion Status

Some participants are actively comparing their equations and results, noting discrepancies in their calculations. The original poster acknowledges a mistake in their equations but does not elaborate further. The discussion appears to be progressing towards clarification of the circuit analysis.

Contextual Notes

There is mention of potential confusion regarding the polarity conventions of the batteries, which may affect the signs of the currents. Participants are also working under the constraints of homework rules, which may limit their ability to share complete solutions.

Dell
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i have the following
http://lh3.ggpht.com/_H4Iz7SmBrbk/Sg_Mj_v4_LI/AAAAAAAAA_c/Egdgr0F8hfY/s720/Untitled.jpg

the numbers are the resistance of the resistors in ohms

and have already found the following

I1=-18/23A
I2=-10/23
I3=8/23

and now need to find the difference in voltage between points A and B,

what i did was take the loop from point A, going clockwise to point B, and i get

2I1 + I3 =1 + v

so i get v=-36/23 +8/23 -1 = -15/23V

but the correct answer is meant to be -67/23V, and i notice tat if i had said v=-36/23 -8/23 -1 i would have reached the correct answer, but i cannot see why there should be a negatice sign on I3 since I3 is positive 8/23 and i chose I3 clockwise
 
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Dell,
To begin with I don't get the same signs as you on the currents. In fact I get all as positive (as shown in your diagram). But I may have the polarity convention on the batteries reversed in which case all the currents would be negative.

Intuitively (assuming I haven't reversed battery convention) the three volt battery outweighs the 1 volt battery which would drive I1 positive. Similarly I3 should be positive and given more resistance in the I3 segment more I1 than -I3 would contribute to I2 making it positive. (assuming I have reversed battery convention this all holds but with pos<->neg).

Remember that the voltages on resistors oppose current flow.

To set up the system to solve for currents I took the Kirchoff's law (sp?)
I2+I3=I1
and the top loop:
3v - 2I1 -1v - 1I2 = 0
and the bottom loop:
1I2 - I3 + 2v - 6I3 = 0
 
okay, getting there, i seem to have a mistake in my equations,,
i got
I3 = I1 - I2 which is like what you got
7I3 - I1 = -2v for the bottom loop, also like you got
2I1 +I2 = -2v which is opposite sign to what you got


so let's examine the top loop, i took a clockwise loop and chose both I1,I2 to be clockwise, so they will be positive, now i go into 3v's positive side and into 1v's negative so in the end i get
I2+3v+2I1-1v=0
I2+2I1=-2v

but you someohow got the opposite,
 
never mind, i got it,, thanks
 

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