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Charging current - jumpstarting a car

  1. Mar 12, 2008 #1

    tony873004

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    A 12-V battery with internal resistance of 0.02 Ω is used to charge a battery with an emf of 10.5 V and an internal resistance of 0.15 Ω. What is the charging current?

    We don't have any examples in the book or class notes about connecting 2 batteries of differing voltages in parallel.

    My guess would be to average the batteries' voltages to 11 V, add the resistors, then use I=V/R = 11 / 0.17 = 1.6 amps. Another student got 11.53 amps, but I'm not sure of his method. Any thoughts?
     
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  3. Mar 12, 2008 #2

    mgb_phys

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    You connect the batteries in parallel so the 12V is trying to push current into the 10.5V - this is a potential difference of 1.5V
     
  4. Mar 12, 2008 #3

    tony873004

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    Thanks. In that case, I get I=1.5/0.17 = 8.8 Amps. Is that the way to do it? The only examples in the book have junctions with 3 branches, needing Kirchhoff's Rules.
     
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