Potential difference question - Immediate

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MathewsMD
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The emf and internal resistance are shows in the attached image. When the terminal voltage, Vba, is equal to -40V, what is the current through the battery, including its direction?

Solution:

VT = emf - iR (general)

In this case, VT = 40V, when considering VT = Vab = Vb - Va

VT = Vext - 28V - i(1.5Ω)

Now, I'm not entirely certain of where I went wrong, but don't seem to know much about the current besides that it is going from a to b in the cell.

The solution shows the answer is 8.0A but I can only get that if I do 0 = 40V - 28V - i(1.5Ω) but I don't quite see how you can apply the 40V inside (using Kirchoff's loop rule) if that's the terminal voltage instead of the applied voltage from an external source. Any clarification on what I am doing wrong or what I should do, and clarification on VT would be helpful.

This is what I quickly used as a reference: http://www.physics.ucla.edu/demoweb...d_magnetism/dc_circuits/terminal_voltage.html
 

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rude man said:
Vba : is A positive compared to B, or vice-versa?

All I'm given is what I have given you. Since Vba = -40V, Vab = 40V and I was under the impression then that b is the + side and a is -. Am I mistaken?

Also, when they say current goes from a to b, do they mean in the cell itself or outside? If it is outside, then this makes sense and I think they are trying to say then that it is a that is + and b that is -. But any explanation for how one would come to that conclusion simply given everything I have given you would be great! Thanks!
 
NascentOxygen said:
Not correct.




This doesn't look right.

EDIT
You'll find it easier if you picture b as 0v, and making every other point positive wrt end b.

The voltage between b and a is the terminal voltage.

Oh, so Vba = Va - Vb? If so, that's where I went wrong. I just don't know why I've haven't up on this convention...
 
Imagine charging a 28 V battery using a 40 V source.
If Vba = - 40 V, Vab is 40 V. 28 of those are over the battery, the remainder over the 1.5 Ω. There is only 1 current and it's going all round.