Batteries and Internal resistance

AI Thread Summary
The discussion centers on calculating the internal resistance of a battery and the energy dissipated within it, given specific voltage and current values. The internal resistance was determined to be 0.556 ohms, and the external resistor was found to be 32.78 ohms. Participants are seeking help with energy dissipation over 4.4 minutes, using the formula I^2r for power, and then calculating energy. Additionally, they discuss how to apply Kirchhoff's voltage law when adding a second identical battery in series with a different external resistor value. Clarifications on circuit setup and the application of Kirchhoff’s laws are provided to assist in solving the problems.
sabak22
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A 6.0 V battery has an internal resistance r. The measured voltage (emf of the battery) is 6.0 V. When connected to a resistor R the terminal voltage is 5.9 V and the current is 0.18 A.
I managed to get a) What is the internal resistance r of the battery?
0.556 ohm

as well as b) What is the value of the external resistor R?
32.78 ohm

but i can't figure out what to do when they ask me
c) What is the energy dissipated inside the battery in 4.4 minutes?

and d)When a second identical battery is added in series and the external resistor is R = 29 Ohms what is the resulting current?

Please help me. I have been trying this for an hour!
 
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c) Power dissipated inside the battery would be

I^2r

where r is the internal resistance. once you get power, find the energy consumed inside the battery...

for part d) , set up Kirchhoff's equation again...
 
Thank you very much isaacNewton. But I am not quite sure how to solve the d part. Could you please expand a little bit? I am having trouble setting up the second identical battery in the circut.
 
have you learned how to use Kirchhoff's rule ?
 
umm all we've learned this week was that Iin=Iout. Is that supposed to help me? I don't know what to with it.
 
you should get some understanding of the voltage law to solve the problem
 
  • #10
just a quick question. will the two internal resistances be the same now that another identical battery has beenhadded? If yes then will the r= 0.556 like it was in the first case?
 
  • #11
yes, the internal resistance is the same... its the property of the battery...
but now you have two similar batteries in series...

Edit: which book your teacher is following ?
 
  • #12
so what I am getting is something like this :

6-0.556-32.78i1-6-0,556-29i2 = 0

Does this make any sense? if not can you please boost me with a little help.
 
  • #13
see, the batteries and the external resistor are in SERIES, so there is only one current throughout.
 
  • #14
draw a circuit to help you
 
  • #15
Moderator's note: thread moved to Homework & Coursework Questions.
 
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