Question

A student is asked to design a circuit to supply an electric motor with 1.0 mA of current at 3.0 V potential difference.

a. Determine the power to be supplied to the motor.

b. Determine the electrical energy to be supplied to the motor in 60 s.

c. Operating as designed above, the motor can lift a 0.012 kg mass a distance of 1.0 m in 60 s at constant velocity. Determine the efficiency of the motor.

To operate the motor, the student has available only a 9.0 V battery to use as the power source and the following five resistors.

d. In the space below, complete a schematic diagram of a circuit that shows how one or more of these resistors can be connected to the battery and motor so that 1.0 mA of current and 3.0 V of potential difference are supplied to the motor. Be sure to label each resistor in the circuit with the correct value of its resistance.

Answer 1

a. P = IV = 3 mW = 3 × 10^-3 W

b. E = Pt = 0.180 J

c. e = “what you get”/”what you are paying for” = (power lifting the mass) ÷ (power provided by the motor) P_lifting = Fv = mgv = mgd/t = 1.96 mW so the efficiency is 1.96/3 = 0.653 or 65.3 %

d. To reduce the battery voltage of 9 V to the motor’s required voltage of 3 V, we need 6 V across the resistors.

The required resistance is then V/I = (6 V)/(1 mA) = 6000 Ω. This is done with a 1000 Ω and a 5000 Ω resistor in series.