What would be the current delivered to the motor if the batteries were connected in series?
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(.03 in parallel with .03) = effective internal resistance of 0.015 ohm.
Total circuit resistance = (1.5 + 0.015) = 1.515 ohms.
Current = (12/1.515) = 7.921A.
(.03 + .03) = .06 ohm.
Total circuit resistance = (1.5 + 0.06) = 1.56 ohms.
Current = (24/1.56) = 15.385A.
Total circuit resistance = (1.5 + 0.015) = 1.515 ohms.
Current = (12/1.515) = 7.921A.
(.03 + .03) = .06 ohm.
Total circuit resistance = (1.5 + 0.06) = 1.56 ohms.
Current = (24/1.56) = 15.385A.