A 36 V 300 mAhr battery provides 15 A of current to a small

A 3.6 V, 300 mA-hr battery provides 1.5 A of current to a small electric motor for a period of 50 s. How much power is the motor taking? How much energy has been provided by the battery?

Solution

a) Given, Voltage of the battery = 3.6 V

Storage capacity of the battery = 300 mA

Curent drawn by the motor = 300 mA-hr*50/3600hr = 150/36 mA

Hence,power drawn by the motor = V*I = 3.6*150/36 mW =15 mW

b) Energy provide by the battery = VIt ,where I = constant current delivered by the battery, V=voltage provided by the battery, t= time for which the current is delivered

E= 3.6*1.5*50/3600 Wh = 0.075 Watt-hr = 0.075*10^-3 kW-hr =7.5*10^-5 kW-hr