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

  • Thread starter Dell
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  • #1
i have the following [Broken]

the numbers are the resistance of the resistors in ohms

and have already found the following


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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  • #2
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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?)
and the top loop:
3v - 2I1 -1v - 1I2 = 0
and the bottom loop:
1I2 - I3 + 2v - 6I3 = 0
  • #3
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

but you someohow got the opposite,
  • #4
never mind, i got it,, thanks