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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?
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?