Why Does Battery Terminal Voltage Calculation Depend on Current Direction?

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Homework Help Overview

The discussion revolves around calculating the terminal voltage of batteries in a circuit, specifically addressing how the direction of current affects these calculations. Participants are examining two questions related to potential differences and terminal voltages in a circuit involving multiple batteries and resistances.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between current direction and terminal voltage calculations, questioning why different signs are used in equations based on current flow. They discuss specific examples from their homework, comparing the calculations for two different questions.

Discussion Status

Some participants express confusion about the application of current direction in their calculations, particularly regarding the signs used in the equations for terminal voltage. There is an ongoing exploration of how the definitions of current direction impact the results, with some guidance provided on the arbitrary nature of current direction.

Contextual Notes

Participants are working under the constraints of homework assignments that require them to determine terminal voltages and potential differences, leading to discussions about the implications of their calculations and assumptions regarding current direction.

sunflowerzz
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Homework Statement



I've attached the diagrams for both questions

Question 24
Determine the terminal voltage of each battery

Question 25
a) What is the potential difference between points a and d
b) What is the terminal voltage of each battery?

Homework Equations



V = E - Ir

The Attempt at a Solution



I was able to solve Q24 and Q25 but only after seeing the answer and changing the signs of my equations so I'm just wondering why this is the case:

Q24:
I drew the current is in a clockwise direction
-I(1.0) + 18V - I(6.6) - 12 - I(2.0) = 0
I = 0.625A

For the 18V battery: V = 18 - (0.625 * 1.0) = 17.4V
For the 12 V battery: V = 12 + (0.625 * 2.0) = 13.3V

--> The explanation for why the 12V battery is E + Ir is because there is a voltage gain going across the internal resistance from left to right...which I understand

But for Q25:
From an earlier example, I1 = -0.858A; I2 = 2.58A; I3 = 1.73A

a) potential difference Vad = Vd - Va = -I1(30) = -25.7V --> how come you don't use the negative value of current 1, since that was what I calculated from the previous example?

b) 80V battery: the terminal voltage is the potential difference from point g to e
45 V battery: the terminal voltage is the potential difference from point d to b

V(80) = E1 - (I2)r = 80 - (2.58 * 1.0) = 77.4V

V(45) = E2 - (I3)r = 45 - (1.73 * 1.0) = 43.3V

So in this case, how come V(45) is not also E2 + (I3)r? Is it because it is going in the same direction as the current (I3) and in the previous question, it is going against the current?
 

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sunflowerzz said:
a) potential difference Vad = Vd - Va = -I1(30) = -25.7V --> how come you don't use the negative value of current 1, since that was what I calculated from the previous example?
I agree.
d has certainly a higher potential than a.

b) 80V battery: the terminal voltage is the potential difference from point g to e
45 V battery: the terminal voltage is the potential difference from point d to b

V(80) = E1 - (I2)r = 80 - (2.58 * 1.0) = 77.4V

V(45) = E2 - (I3)r = 45 - (1.73 * 1.0) = 43.3V

So in this case, how come V(45) is not also E2 + (I3)r? Is it because it is going in the same direction as the current (I3) and in the previous question, it is going against the current?
Right, the current direction is different.
 
mfb said:
I agree.
d has certainly a higher potential than a.

Right, the current direction is different.

But can you tell me why that is?
 
The current direction (as drawn in the sketch) is an arbitrary definition, you can choose it as you like. If you choose the opposite direction, the current value and all signs for it change. The voltages stay the same.
 
mfb said:
The current direction (as drawn in the sketch) is an arbitrary definition, you can choose it as you like. If you choose the opposite direction, the current value and all signs for it change. The voltages stay the same.

So for Q25 part a, I just take the value of I1 as 0.858 and not the negative sign?

And about the directions:
for Q24, the 12 V battery is V = E + Ir because it is in the opposite direction of the current BUT
for Q25, The 45 V battery is V = E - Ir because it is in the same direction as the current?
 

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