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Charging current - jumpstarting a car

  • Thread starter tony873004
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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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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
 
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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