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Homework Help: Charging of a battery

  1. Apr 25, 2016 #1
    • Poster has been reminded to post schoolwork in the Homework Help forums
    A battery has an Emf 6 Volts. It is completely discharged. It is charged by maintaining a potential difference of 9 Volts across it. If the internal resistance of the discharged battery is 10 ohms, find the current through the battery, just after the connections are made.
    My textbook says that the net potential difference across the battery is 3 volts, but if the battery is discharged why would we subtract the emf of the battery while calculations? Wouldn't it act like a conductor when it is discharged completely?
  2. jcsd
  3. Apr 25, 2016 #2


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    There are 3 answers that I can come up with:

    A. You've left out details from the text, which doesn't allow us to answer the question properly. (somewhat likely)
    B. You've misinterpreted what the text was saying. (most likely)
    C. The textbook is wrong. (most unlikely)
  4. Apr 25, 2016 #3


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    I agree that the interpretation is uncertain.
    But the significant point here is that batteries do not have a well defined point when they should be called discharged. I can read this as saying, "in this case, the battery is regarded as completely discharged at emf 6V."

    I would hope that, as per OmCheeto's point A, there is more info to follow: otherwise I can't see any point in giving both the emf and the "completely discharged".
    If "completely discharged" were intended to mean the battery was at 0V EMF, then to what does the 6V EMF refer? Surely not the nominal voltage of the battery, which would not be much use in the calculation?

    I suspect, as OmCheeto, that all would be much clearer with the full question.
  5. Apr 25, 2016 #4


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    It's possible that people aren't aware of the fact that fully discharged doesn't mean that a battery will measure zero volts.
    (Good grief! That's like a quadruple negative in one sentence: are not aware... no charge... does not mean... no volts):redface:

    Perhaps an example would be clearer:
    lead acid battery under no load conditions [ref]:
    100% charge = 12.65 volts
    0% charge = 11.89 volts​
  6. Apr 25, 2016 #5
    Think of the charging process as supplying energy to the battery at a certain rate ( due to the charger) which is equal to the rate of increase in which the energy is stored as chemical energy in the battery PLUS the rate at which energy is dissipated as heat due to the internal resistance of the battery.

    VchargeIcharge = VemfIcharge + RintI2charge


    Vcharge = Vterminal = Vemf + RintIcharge
  7. Apr 25, 2016 #6


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    You have to drive a chemical reaction to charge a battery. Essentially you have to break apart some molecules. It takes a specific voltage to tear apart a given molecule. The required voltage is similiar to the voltage of a charged battery. So you won't get much current until you exceed the 'fully charged voltage' of the battery. That's why you must supply more than the nominal battery voltage to start charging.
  8. Apr 26, 2016 #7


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    This had me somewhat confused, as I'm not good with "word" answers. But I cut and pasted your description onto my interpretation of the OP's circuit diagram, and it made perfect sense. :partytime:


    ps. I'm still not sure I understand that "internal resistor" thing, so I googled it yesterday, and it appears to be different for different battery chemistries.

    per wiki;
    The internal resistance of a battery may be calculated from its open circuit voltage VNL, load voltage VFL, and the load resistance RL:

    Many equivalent series resistance (ESR) meters, essentially AC milliohm-meters normally used to measure the ESR of capacitors, can be used to estimate battery internal resistance, particularly to check the state of discharge of a battery rather than obtain an accurate DC value. Some chargers for rechargeable batteries indicate the ESR.

    In use, the voltage across the terminals of a disposable battery driving a load decreases until it drops too low to be useful; this is largely due to an increase in internal resistance rather than a drop in the voltage of the equivalent source.

    In rechargeable lithium polymer batteries, the internal resistance is largely independent of the state of charge but increases as the battery ages; thus, it is a good indicator of expected life.

    Life was so much easier, in the olden days, when the only rechargeable battery you had to worry about, was "lead acid".

    pps. Anyone seen the OP recently? Kirchhoff’s voltage law is a pretty important concept, to lock into your brain, IMHO. :angel:
    Last edited: Apr 26, 2016
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