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Charging voltage versus battery voltage

  1. Jul 27, 2012 #1
    Why during charging a battery the terminal Potential Difference becomes greater than the EMF?
     
  2. jcsd
  3. Jul 27, 2012 #2

    Drakkith

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    Staff: Mentor

    Re: Charging

    I believe it has to in order to reverse the chemical reaction that produces the EMF in the first place.
     
  4. Jul 27, 2012 #3
    Re: Charging

    Well to be more specified.I wanted to ask HOW..
     
  5. Jul 27, 2012 #4

    Drakkith

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    Staff: Mentor

    Re: Charging

    How what? Please be specific when posting questions, or you will confuse people, and more importantly you will confuse me! :biggrin:
     
  6. Jul 27, 2012 #5
    Re: Charging

    Sorry for inconvenience...I wanted to ask:
    How the potential difference becomes greater than EMF?
     
  7. Jul 27, 2012 #6

    Drakkith

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    Staff: Mentor

    Re: Charging

    I think the charger is simply set up to apply a greater voltage than the battery produces.
     
  8. Jul 27, 2012 #7

    Bobbywhy

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    Gold Member

  9. Jul 28, 2012 #8
    Hi moatasim:)...IN the start you can't roughly say t.p.d will be greater than emf during charging.when u begin charging, the terminal p.d simply increases coz of the incoming positive charges from other source.charge accumulation leads to greater increase in potential difference(as Q is directly related to V). when charge accumulation goes on increasing then incoming positive charge from external source towards negative plate of cell(under charge) will be attracted towards it more strongly and when it reaches negative plate then it is pushed towards positive plate due to increasingly potential difference.as emf is the push of positive charge from negative to positive terminal,then with da increase in time this push becomes more and more ease and less emf though will be due to a p.d cloud.NOW,as you know emf exists even theres no current in circuit and p.d dont exist if theres no current.In start when charging begins,p.d goes on increasing but still be less than emf already present.SO in begining and till to a particular time t.p.d will be less than emf. MATHEMATICALLY FOR DISCHARGING
    t.p.d = E - Ir
    V= E - Ir ( E is the electromotive force)
    or E=V + Ir ( V is t.p.d)
    FOR CHARGING
    v = E + Ir (as for charging respective electrodes are alike)
    E= V-Ir SO you can clearly see emf decreses by a function of time Q/t.( Ir=Q/t*r) with the passage of time E goes on decreasing and consequently V goes on increasing..I hope you will get it:)
     
  10. Jul 28, 2012 #9
    Hi moatasim:)...IN the start you can't roughly say t.p.d will be greater than emf during charging.when u begin charging, the terminal p.d simply increases coz of the incoming positive charges from other source.charge accumulation leads to greater increase in potential difference(as Q is directly related to V). when charge accumulation goes on increasing then incoming positive charge from external source towards negative plate of cell(under charge) will be attracted towards it more strongly and when it reaches negative plate then it is pushed towards positive plate due to increasingly potential difference.as emf is the push of positive charge from negative to positive terminal,then with da increase in time this push becomes more and more ease and less emf though will be due to a p.d cloud.NOW,as you know emf exists even theres no current in circuit and p.d dont exist if theres no current.In start when charging begins,p.d goes on increasing but still be less than emf already present.SO in begining and till to a particular time t.p.d will be less than emf. MATHEMATICALLY FOR DISCHARGING
    t.p.d = E - Ir
    V= E - Ir ( E is the electromotive force)
    or E=V + Ir ( V is t.p.d)
    FOR CHARGING
    v = E + Ir (as for charging respective electrodes are alike)
    E= V-Ir SO you can clearly see emf decreses by a function of time Q/t.( Ir=Q/t*r) with the passage of time E goes on decreasing and consequently V goes on increasing..I hope you will get it:)
     
  11. Jul 28, 2012 #10

    NascentOxygen

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    Staff: Mentor

    Re: Charging

    The model for most batteries is of an "ideal" voltage source in series with a resistor. It's that resistor that you need to consider.

    * ideal means whatever you need it to, here. :smile:
     
  12. Jul 28, 2012 #11
    Batteries produce power via chemical reactions...charging reverses those...the higher the charge voltage the higher the current flow and the faster the reverse chemical reaction.

    BUT higher currents [via higher charge voltages] mean more heat and that can ruin batteries...so charge voltages must be limited....

    An AGM battery can be charged at it full rated amp hour capacity because it has very low internal resistance, so not much heat is generated during charging, not so for a traditional wet cell lead acid battery which is limited to about 25%. These mean for a battery rated at, say, 100 amp hours, you can safely charge an AGM at 100 amps but a conventional battery at only about 25 amps. In a conventional wet cell lead acid battery about a quarter of the charge power is lost thru heating, only 1 or 2% is lost to heat in an AGM type battery. .
     
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