Charging current - jumpstarting a car

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To calculate the charging current when connecting a 12-V battery to a 10.5-V battery in parallel, the potential difference is 1.5 V, derived from the voltage difference between the two batteries. The total resistance in the circuit is the sum of the internal resistances, which is 0.17 Ω. Using Ohm's Law (I=V/R), the current can be calculated as I=1.5/0.17, resulting in approximately 8.8 amps. There is confusion among students regarding the correct method, with one suggesting an average voltage approach that yields a different current. Proper application of the voltage difference and total resistance is essential for accurate calculations in this scenario.
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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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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
 
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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