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Homework Help: Easy circuit question but i get the wrong answer for PD!

  1. Oct 23, 2005 #1
    Hello everyone!! I am stuck on this simple problem...
    A car battery with a 12 V emf and an internal resistance of 0.050 is being charged with a current of 60 A.

    (a) What is the potential difference V across its terminals?
    wrong check mark V
    (b) What is the rate Pr of energy dissipation inside the battery?
    (c) At what rate is electrical energy being converted to chemical energy?
    (d) When the battery is used to supply 60 A to the starter motor, what is V?
    (e) What is Pr in this case?

    I found the V = 9V;
    I drew the simple circuit then Used
    Vb-Va = E-Ir;
    V = 12v -60*.050 = 9V, but it was wrong, any ideas? thanks
  2. jcsd
  3. Oct 23, 2005 #2
    pd = 12V

  4. Oct 23, 2005 #3
    pd = 12V

    correct me if i am wrong
  5. Oct 23, 2005 #4
    nope, its wrong it said
  6. Oct 24, 2005 #5


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    Homework Helper

    The car battery can be imagined as an ideal voltage source of emf 12 V and the internal resitance r=0.050 ohm connected in series. See the picture. A and B are the terminals of the battery, and X is the virtual junction of the virtual ideal source with the internal resistance.
    The carging current flows into the battery to supply positive charges onto the positive electrode. It causes a potential drop of 0.05 * 60 = 3 V across the internal resistance. The potential drops along the direction of the current. So B is more positive than X. X is more positive than A. So the net potential difference across the terminals A, B of the real battery is 12+3=15 V during the process of charging.
    Your formula is true when the battery supplies current to the starter motor. The direction of this current is reverse with respect to the charging current and the potential drops 3 V from X to B inside the battery. The terminal voltage is 9 V during discharge.
    Last edited: Jun 29, 2010
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