Potential difference between two points for a circuit with two batteries

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Homework Statement



The problems tells me to determine the potential difference between between two points. I alrady know that the potential of one point is zero. The problem I am having is that the cirucit has two batteries. I do not have a question so i will describe it, it hsould be easy to visulaize. The first battery is 9V and current I1=3.0A and moves across a resistor R1=2.0 ohms. The second battery is 12.0V and current I2=2.0A and moves across a resitor R2=1.5 ohms. The current then combines at a junction where I1+I2=I3. My question is how to I find the at the junction. I3 flows out of the junction and is5.0A.


Homework Equations



V=IR
series and parallel resistor equations
Kirchoff's rule



The Attempt at a Solution


I caluclated the voltage drop at R1 to be 6.0V, therefore there is 3.0V heading to the junction from I1. I calcualted the voltage drop at R2 to be 3.0V, therefore there is 9.0V heading into the junction from I2. Are the voltages additive? Im not sure bc it would make sense?
are the voltages additive at the point.
 

Answers and Replies

  • #2
tiny-tim
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Hi medguy1234! :smile:
The problems tells me to determine the potential difference between between two points. I alrady know that the potential of one point is zero. The problem I am having is that the cirucit has two batteries. I do not have a question so i will describe it, it hsould be easy to visulaize. The first battery is 9V and current I1=3.0A and moves across a resistor R1=2.0 ohms. The second battery is 12.0V and current I2=2.0A and moves across a resitor R2=1.5 ohms. The current then combines at a junction where I1+I2=I3. My question is how to I find the at the junction. I3 flows out of the junction and is5.0A.

I caluclated the voltage drop at R1 to be 6.0V, therefore there is 3.0V heading to the junction from I1. I calcualted the voltage drop at R2 to be 3.0V, therefore there is 9.0V heading into the junction from I2. Are the voltages additive? Im not sure bc it would make sense?
are the voltages additive at the point.

Nicely described! :smile:

mmm … I always got very confused by electric circuits. :confused:

Common-sense seems to say that the potential (that is, the battery voltage minus the potential drop) should be the same from both paths.

Your voltage drop calculations look right. That gives you 9 - 6 = 3 for one path, and 12 - 3 = 9 for the other path.

Are you sure you have the batteries the right way round? :smile:
 
  • #3
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here is a picture, this should help hopefully

im tryin to find v at A by v at b
 

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  • #4
tiny-tim
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Hi medguy1234! :smile:

Thanks for the PM.
here is a picture, this should help hopefully

im tryin to find v at A by v at b

ah … that picture is not what I was expecting.

You didn't say how you got your currents … I think they must be wrong …

When you apply Kirchhoff's rules to the left and right circuits, what two equations do you get? :smile:
 
  • #5
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When i applied kirchnoffs rules for the diagram i get,


Loop 1 being- the left side
loop 2 being- being the whole thing
loop 3 being the the rights side

loop 1= E1-I1R1-I3R3=0
I solved this one for I1 and got I1=(E1-I3R3)/R1 and got I1=3.0A

for loop 3 i got E2-I2R2-I3R3-I2R4=0
i solved this one for I3 and got I3=(E2-I2R2-I2R4)/R3 and got I3=5.0A

thats how i got them
 
  • #6
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loop 2 is E1-I1R1 +I2R2-E2 +I2R4=0, but i ddin't use it
 
  • #7
tiny-tim
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loop 1= E1-I1R1-I3R3=0
I solved this one for I1 and got I1=(E1-I3R3)/R1 and got I1=3.0A

for loop 3 i got E2-I2R2-I3R3-I2R4=0
i solved this one for I3 and got I3=(E2-I2R2-I2R4)/R3 and got I3=5.0A

thats how i got them

ok … that looks right! :smile:

The question only asks you now for the potential difference between points A and B …

and they're both on the right-hand circuit …

and you know all the currents …

so just use Kirchhoff again! :smile:
 
  • #8
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wait, how would i do that, would i just make a half of a loop then or what, the best i could think of is dividing the loop in two part and solving for1 part,

i said that E2-R2I2=I3R3 + I2R4

Solving for either side gives 9 V, is that the correct answer, but most importantly the correct reasoning behind it
 
  • #9
tiny-tim
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wait, how would i do that, would i just make a half of a loop then or what, the best i could think of is dividing the loop in two part and solving for1 part,

i said that E2-R2I2=I3R3 + I2R4

Solving for either side gives 9 V, is that the correct answer, but most importantly the correct reasoning behind it

Hi medguy1234! :smile:

ah … I think I see what's worrying you …

you're thinking that you can't apply Kirchhoff to half a loop!

Think of it as applying Kirchhoff to both halves in the same direction

one half adds to 9, and the other to minus 9 … total zero … you will always get total zero for a loop, no matter where you split it in two.

The correct reasoning is simply to say "potential difference = sum of potential drops." :smile:
 
  • #10
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ok the potential difference between the two points a and b, is the sum of the potential drops between them, therefore, there is a 3 volt drop and a 6 volt drop, which adds to 9, therfore the ptential difference is 9 V

is that correct?
 
  • #11
tiny-tim
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Yes! :smile:
 
  • #12
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tim i love u, ur great, a genius i should say, i appreciate the help
 

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