1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

When do I have a dead battery?

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    When a certain 9.0V battery is connected to a voltmeter, the voltmeter reads 8.00V (assume that the voltmeter has an internal resistance of 10MΩ. We may conclude that
    (A) only 1/9 of the battery's fuel has been used, and so the battery has plenty of life left in it.
    (B) the battery is nearing the end of its life, but is still usable.
    (C) the battery is unable to provide more than a few μA of current to any circuit (even a short circuit), and so is therefore (for all practical purposes) completely dead.

    2. Relevant equations
    V = IR

    3. The attempt at a solution
    No idea honestly. It seems that the battery has plenty of life in it, but that's the wrong answer.
     
  2. jcsd
  3. Apr 11, 2016 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Usually, a voltmeter has high resistance,more than 1000 ohms, or even much higher. If it reads 8 V across the terminal of a 9 V battery, what can be the internal resistance of the battery in terms of the voltmeter resistance? If you load the battery with a light bulb of about 1 ohm resistance, what will be the voltage across it?
     
  4. Apr 11, 2016 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    What current flowed when connected to the voltmeter? (You can work that out.).
     
  5. Apr 11, 2016 #4

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    One way to model a battery is to think of it as a constant ideal voltage source of 9V and an internal series resistor. When the battery is fully charged the internal resistance is small. As it runs down the internal resistance gets larger.

    Try drawing a battery like this with another resistor connected to represent the voltmeter. How large must the internal resistor be to give 8v on the terminals?
     
  6. Apr 11, 2016 #5
    8/10e6 = 8e-7. But what does that say? Of course there will be like no current through the voltmeter.
     
    Last edited: Apr 11, 2016
  7. Apr 11, 2016 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Of course the current DOES flow through the voltmeter.

    Model the battery as being an ideal voltage source of 9.0 V in series with an internal resistor. Therefore, 1.0 V is dropped across that internal resistor. What is the resistance value of that resistor?
     
  8. Apr 11, 2016 #7
    V = IR
    1.0 = 8e-7*R
    R = 1250000 ohms ?
     
  9. Apr 11, 2016 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Correct.

    Now consider using that battery to power a 1Watt light bulb which is designed to be used at 9 V .
     
  10. Apr 11, 2016 #9
    SammyS,
    P = IV
    1 = 9*I
    I = 0.11 A

    We need 0.11 A
    We have 8e-7 A
    Is that right?
     
  11. Apr 11, 2016 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That's essentially correct.

    At operating conditions, the bulb has a resistance of about 80 Ω . ( P = V2/R )

    That corresponds to a voltage drop of 8 or 9 micro-volts and a power output on the order of 1 micro-watt.
     
  12. Apr 11, 2016 #11
    So because the battery cannot even supply a μA of current to even a short circuit, the battery is completely dead. The answer is (C).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: When do I have a dead battery?
Loading...