Solving for Vemf and Internal Resistance

[pleasehelpme6]

When a battery is connected to a 110.-ohm resistor, the current is 3.98 A. When the same battery is connected to a 400.-ohm resistor, the current is 1.11 A.

Find the emf supplied by the battery and the internal resistance of the battery.





Equations:

Vemf = i*R

I'm not sure how to approach this since you get two voltages (438.7 and 444) from the different resistors.

Should I treat these resistors like they are in the same circuit? In series or in parallel?

Please help!

 

[collinsmark]

I'm not sure how to approach this since you get two voltages (438.7 and 444) from the different resistors.

Yes, you'll get two different voltages, by using different resistors, because there is another resistance involved that is always there: the internal resistance of the battery.

(By the way, check your math regarding the voltages you calculated.)

Should I treat these resistors like they are in the same circuit?

Not the 110 and 400 Ω resistors, no.

But the battery's internal resistance is always there, so yes, the internal resistance is within the circuit in both cases.

In series or in parallel?

What do you think? (Hint: look in your textbook for "internal resistance" of a battery, and the answer should be clear.)

 

[pleasehelpme6]

Thanks! that explanation makes a lot of sense. I'm sure I can figure the rest out from there!