Question

The motor in a toy car operates on 6.00 V, developing a 4.50 V back emf at normal speed. If it draws 3.00 A at normal speed, what current does it draw when starting?

a) 6.00 A
b) 8.00 A
c) 9.00 A
d) 12.00 A

Answer 1

The toy motor with a 6.00 V supply voltage and a 4.50 V back emf at normal speed, draws 3.00 A at normal speed and 12.00 A when starting, as the back emf is not present at startup.

Step-by-step explanation:

The student's question involves the concept of back electromotive force (emf) in an electric motor. Back emf is the voltage that opposes the input voltage and is generated by the motor's operation due to electromagnetic induction. The motor operates on 6.00 V, develops a 4.50 V back emf at normal speed, and draws 3.00 A at this speed. When the motor is starting, the back emf is zero because the motor has not achieved normal operating speed yet. Therefore, the current draw when starting can be computed by using Ohm's Law (I = V/R), where V is the input voltage and R is the resistance of the motor.

At normal speed, the effective voltage (input voltage - back emf) is 6.00 V - 4.50 V = 1.50 V. With the given normal operating current of 3.00 A, using Ohm's Law, the resistance R of the motor can be calculated as R = V/I = 1.50 V / 3.00 A = 0.50 ohms. When the motor is starting, the full input voltage of 6.00 V is applied across this resistance. Thus, the starting current is I = V/R = 6.00 V / 0.50 ohms = 12.00 A, which corresponds to option (d).