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How much work a battery does connected to a resistor

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A battery has an EMF of 12.0 volts and internal resistance of 3 ohms
    A 21.0 ohm resistance is connected to the battery.
    0.5 Amps flow through the battery, hence 0.5 Amps flows through the 21.0 ohm resistor
    The potential difference across the 21 ohm resistor is 10.5 Volts, hence the terminal voltage of the battery is 10.5 Volts.

    The question is How much work does the battery connected to the 21.0 ohm resisitor perform in one minute?

    2. Relevant equations

    Voltage = Joule/coloumb
    1 amp = coloumb/ second = (6.258 *10^18 electrons) second
    1 electron has a charge -1.60*10^-19 coloumbs

    3. The attempt at a solution

    current = 0.5 Amps = 3.125*10^18 electron/ second = 30 coloumbs pass wire per minute

    usisng the definition of EMF (voltage): 30 coloumbs/ minute * 10.5 volts = 315 Joules

    however it keeps on telling me that answer is wrong i know its simple i just can't see it right now thanx
     
  2. jcsd
  3. Oct 21, 2008 #2

    dlgoff

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    Science Advisor
    Gold Member

    1 amp x 1 volt = 1 coul/sec x 1 joule/coul = 1 joule/sec = 1 watt
     
  4. Mar 17, 2009 #3
    Using the EMF definition, shouldn't you use EMF and not Terminal voltage?
     
  5. Mar 17, 2009 #4
    My thought on your problem is as follows.
    You have a terminal Voltage of 10.5 V, EMF of 12V, Flow of 1/2 A..so pretty much everything you need.

    Use the equation for Work on a Resistance : [itex]W=I^2 R t [/itex] where R=r the resistance of the battery. That gives you 45 Joules produced in 60s = 1minute.

    I don't take it for granted that I am correct!
     
  6. Mar 17, 2009 #5
    Lets try not to mislead him. He had it mostly correct. Since Volt = Joules/Coulomb he already had the current in C/min...

    30 C/min * 12 (J/C) = ___J/min
     
  7. Mar 17, 2009 #6

    berkeman

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    Staff: Mentor

    Actually I think the only error is due to the question being a little tricky in wording. You calculated the work done by the battery on the external resistance. The battery also heats up due to the current flowing through its internal resistance....
     
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