A motor with a coil resistance of 12.4 Ω is attached to a voltage supply of 66.4 V.What is the current in the motor when it is running at its maximum speed with a back emf of 22.6 V?
A)zero
B)3.53 A
C)7.18 A
D)5.35 A
E)543 A

Question

Quizplus · Accepted Answer

The current in the motor can be calculated using Ohm's Law, $I = \frac{V - E_{\text{back}}}{R}$, where $V$ is the supply voltage, $E_{\text{back}}$ is the back electromotive force (emf), and $R$ is the resistance. Substituting the given values, $I = \frac{66.4V - 22.6V}{12.4\Omega} = 3.53A$.

A Motor with a Coil Resistance of 12