How Do You Calculate Terminal Voltage and Potential Difference in a Circuit?

[Adriano25]

Homework Statement

I = 0.47 A
ε₁ = 16.0 V
ε₂ = 8.0 V
R1 = 5.0 Ω
R2 = 9.0 Ω
r₁ = 1.6 Ω
r₂ = 1.4 Ω

a) Find the terminal voltage Vab of the 16.0 V Battery
b) Find the potential difference Vac of point a with respect to point c

Homework Equations

Vab = Va - Vb = ε₁ - Ir₁
Vac = Va - Vc = IR1 + ε₂ + Ir₂

The Attempt at a Solution

Current is flowing counterclockwise since the battery 16.0 V determines the direction starting from the positive plate.
I'm just a little confused on part b) in why are we adding ε₂ + Ir₂ instead of subtracting them.
Also, I know that if try to find Vac by doing (Va-Vb) + (Vb-Vc), we would get a negative value, correct? In this case, how would the - and + sign change?

Thank you

 

[gneill]

I'm just a little confused on part b) in why are we adding ε2 + Ir2 instead of subtracting them.
Also, I know that if try to find Vac by doing (Va-Vb) + (Vb-Vc), we would get a negative value, correct? In this case, how would the - and + sign change?

Start by adding the current (blue) and the polarities of the resultant potential drops (red) to the figure:

Then you can easily write the equation for the potential difference between locations by doing a "KVL walk" from one to the other. You'll be sure to get the signs of the terms correct this way.

For example, Taking the left side path from c to a you write: $V_{ac} = +ε2 + I r_2 +I R1$

It should not matter which path you take between to locations in the circuit, you should always arrive at the same potential difference with the same sign.

 

[Adriano25]

Start by adding the current (blue) and the polarities of the resultant potential drops (red) to the figure:
View attachment 113486
Then you can easily write the equation for the potential difference between locations by doing a "KVL walk" from one to the other. You'll be sure to get the signs of the terms correct this way.

For example, Taking the left side path from c to a you write: $V_{ac} = +ε2 + I r_2 +I R1$

It should not matter which path you take between to locations in the circuit, you should always arrive at the same potential difference with the same sign.

Thank you. It's very clear now. I have two more questions if you would be kind to answer them:

1) If we take the left side path from c to a, why do we follow the current path like if we were starting from a to c? Looking at the diagram, if we start form c to a, I would be putting negative signs, thus going clockwise.

2)
If taking Vac counterclockwise:
Vac = Va-Vc
Vac = ε₂ + Ir₂ + IR1
Vac = 8.0 V + (0.47 A)(1.4 Ω) + (0.47 A)(5.0 Ω) = 11.008 V

If taking Vac clockwise:
Vac = (Va-Vb) + (Vb-Vc)
Vac = ε₁ - Ir₁ - IR2
Vac = 16.0V - (0.47 A)(1.6 Ω) - (0.47A)(9.0 Ω) = 11.018 V

Thus, I'm getting slightly different results. Am I doing something wrong?

Again, thank you for your help.

 

[gneill]

Thank you. It's very clear now. I have two more questions if you would be kind to answer them:

1) If we take the left side path from c to a, why do we follow the current path like if we were starting from a to c? Looking at the diagram, if we start form c to a, I would be putting negative signs, thus going clockwise.

You can choose to sum potential drops or potential rises along your "KVL walk" so long as you are consistent and understand what it is you are summing. You can "walk" in any direction you choose. Once you've drawn in the current, that fixes the polarities of the potential drops on the components that are due to that assumed current direction. When you do your KVL walk, you sum the potential changes dictated by those polarities (choosing drops as either positive or negative, your choice so long as you are consistent).

2)
If taking Vac counterclockwise:
Vac = Va-Vc
Vac = ε₂ + Ir₂ + IR1
Vac = 8.0 V + (0.47 A)(1.4 Ω) + (0.47 A)(5.0 Ω) = 11.008 V

If taking Vac clockwise:
Vac = (Va-Vb) + (Vb-Vc)
Vac = ε₁ - Ir₁ - IR2
Vac = 16.0V - (0.47 A)(1.6 Ω) - (0.47A)(9.0 Ω) = 11.018 V

Thus, I'm getting slightly different results. Am I doing something wrong?

No, everything's fine. But you're trying to squeeze too much accuracy out of values that you've truncated or rounded along the way to the finish line. For example, the current isn't really 0.47 A. That's a rounded value. If you want your results to have a closer match, keep more decimal places in intermediate results. So, for example, you might take $I = 0.470588$, and only round your results at the end.