Asked by Anonymous
A battery has an internal resistance of 0.016 and an emf of 9.00 V. What is the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V?
Answers
Answered by
Henry
Ri = 0.016 Ohms.
E = 9.00 Volts.
Imax = V/Ri = (9.00-8.75) / 0.016 = 15.625 Amps.
E = 9.00 Volts.
Imax = V/Ri = (9.00-8.75) / 0.016 = 15.625 Amps.
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