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Internal resistance of the battery and power dissipation

  1. Dec 10, 2013 #1
    I got this question ( i won't include a lot of details because this is not the hw forum) where i need to calculate the power dissipation due to internal resistance. What i did is , i calculated total V and total R and I , then i treated the battery as a normal resistance ( eg. a 1kohm resistor , of course it has way less resistance ) , was that the right thing to do ?
    Should the batteries be treated differently than resistors when it come to power dissipation?
     
  2. jcsd
  3. Dec 10, 2013 #2

    sophiecentaur

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    There are many different ways of thinking about and calculating these things. But you end up learning shortcuts and some ways are a bit shorter; whatever suits you is good. When answering exam questions, it's always good to put all your workings in and annotate them briefly so you can get some marks even if your arithmetic is wrong. That internal resistor can be treated just the same as any resistor. (Most batteries are treated as if they have a pretty well fixed equivalent ohmic resistor inside them.)
    If you want to do it that way, that's fine. Input V and total R gives current. Then Isquared times the appropriate R gives the power dissipated in the resistor.
    Sometimes it's good to calculate things in more than one way and feel smart when you get the same answer each time.
     
  4. Dec 11, 2013 #3
    i have this problem where there are 3 batteries in parallel
    and one resistor , to calculate the power dissipation i used this formula from this site ( http://farside.ph.utexas.edu/teaching/302l/lectures/node62.html )(V^2*r/(r+R)^2)
    will that work ? can i simply replace the r+R in his formula by the total resistance of the circuit?
     
  5. Dec 11, 2013 #4

    sophiecentaur

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    You can approach this in two ways. The easiest way is to assume that the batteries are identical and in the same state of charge. This means you assume the emfs are identical so they share the Load current equally. You can do the same as you have done already but, mentally, split the circuit into three batteries feeding three different parallel loads, each with resistance 3R (using ideas of symmetry). Each battery will dissipate 1/3 of the total (new value of) lost power. The voltage drop would be about 1/3 of the single battery case (when the internal resistances are low - as you'd normally expect - and the current will not change much, so the lost power will be about 1/3. Alternatively, you could also just assume a single emf and an internal resistance or r/3 (again assuming the batteries are totally identical.
    The rigorous way to approach it would be to do the whole thing with different emfs and internal resistances r1, r2 and r3. This more complicated circuit can be tackled using Kirchoff's Laws 1 and 2. Personally, I reckon that's a bit over the top but it would be interesting if you wanted to know what happens when one battery goes soggy first (imagine a standby system, after years of use)
     
  6. Dec 11, 2013 #5
    I will begin my calculations and post them, gonna post a new thread in homework section, thank you
     
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