Electric Potential across charging battery.

[Amadeo]

Homework Statement

[/B]
A car battery with a 12v emf and an internal resistance of .040 ohms is being charged with a current of 50A. A.) What is the potential difference across the terminals? B.) The rate of energy dissipation in the battery Pr. C.) the rate of energy conversion to chemical form? D.) When the battery is used to supply a 50A current tot he starter motor, what is V and Pr?

Homework Equations

[/B]
V=IR
P=iε
Pd=i^2R

The Attempt at a Solution

[/B]
for A

I thought that since the emf of the battery is 12 and the potential drop in potential due to the resistor is (50A)(.04ohms) = 2v the total potential across the battery would be 12-2 = 10. However, the book says that it is 14 (which is 12+2). I am not able to figure out why it is plus instead of minus. (Or, if I am correct in thinking that it should be minus in the first place.)

for C:

I thought that the power supplied to the recharge of the battery's chemical energy store would be the total power of the device minus the power dissipated by heat. So, since the total power would be 50(12) = 600w and the power of heat dissipation would be (50)^2(.04) = 100, the power of the recharge would be 600-100 = 500. However, the book says that it is 600.

for D:

I thought that the V would be what it was in part a, namely 10, and for the same reasons. I thought that Pr would be the same as in B, or 100w. These are the answers the book gives.

The only thought I have about the difference between situations in A/B and D would be the direction of the charge flow. But I don't see why a resistance would create an increase in potential just because the current flowed though it form a different direction.

Thank you for your assistance.

 

[haruspex]

why it is plus instead of minus

Draw the battery as a resistance in series with an ideal 12V battery. If 50A flow through that, what is the end-to-end p,d.?

I thought that the power supplied to the recharge of the battery's chemical energy store would be the total power of the device minus the power dissipated by heat.

Yes, but the total power supplied was 14Vx50A.

 

[Amadeo]

Draw the battery as a resistance in series with an ideal 12V battery. If 50A flow through that, what is the end-to-end p,d.?
.

In my textbook (Halliday) there is this diagram. This is what I was thinking about in trying to solve the problem. As in the diagram, it seems that Vb would be ε-ir, or, for this problem, 12-2=10.

 

[haruspex]

As in the diagram,

That is the discharge diagram.

 

[Amadeo]

So the V of a real battery is different when the current is flowing one way through it versus the other. From the side of higher potential to lower potential, the charges drop potential after the resistor, and then drop potential again moving over the battery from positive to negative. Alternatively, if the charges are moving from the lower potential end of the battery to the higher (left to right in the diagram), they will gain potential after the battery, and then some after the resistance (in the battery.) So one way is two drops, the other way is one gain and one drop.

 

[haruspex]

So the V of a real battery is different when the current is flowing one way through it versus the other.

Yes. When discharging the battery has to overcome its own resistance, so the voltage delivered is below its nominal voltage, but when being charged the applied voltage has to overcome the battery's nominal voltage and its resistance.