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

  • Thread starter Dell
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  • #1
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i have the following
http://lh3.ggpht.com/_H4Iz7SmBrbk/Sg_Mj_v4_LI/AAAAAAAAA_c/Egdgr0F8hfY/s720/Untitled.jpg [Broken]

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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Answers and Replies

  • #2
jambaugh
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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
 
  • #3
590
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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 lets 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,
 
  • #4
590
0
never mind, i got it,, thanks
 
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