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Batteries and Internal resistance

  1. Oct 7, 2011 #1
    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 cant 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!
  2. jcsd
  3. Oct 7, 2011 #2
    c) Power dissipated inside the battery would be


    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........
  4. Oct 7, 2011 #3
    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.
  5. Oct 8, 2011 #4
    have you learned how to use Kirchhoff's rule ?
  6. Oct 8, 2011 #5
    umm all we've learned this week was that Iin=Iout. Is that supposed to help me? I dont know what to with it.
  7. Oct 8, 2011 #6
  8. Oct 8, 2011 #7
  9. Oct 8, 2011 #8
  10. Oct 8, 2011 #9
    you should get some understanding of the voltage law to solve the problem
  11. Oct 8, 2011 #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?
  12. Oct 8, 2011 #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 ?
  13. Oct 8, 2011 #12
    so what im 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.
  14. Oct 8, 2011 #13
    see, the batteries and the external resistor are in SERIES, so there is only one current throughout.
  15. Oct 8, 2011 #14
    draw a circuit to help you
  16. Oct 8, 2011 #15


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