Question

Model a motor as a resistor in series with an ideal motor where:

vemf = Kv x w and Tau = Ka Tau x i

Answer 1

A motor can be modeled as a resistor in series with an ideal motor. The voltage across the motor is the difference between the driving emf and the back emf. The current through the motor is related to the voltage by the equation I = V/R.

Step-by-step explanation:

A motor can be modeled as a resistor in series with an ideal motor. The voltage across the motor is the difference between the driving emf and the back emf, given by V = Vemf - Vback emf. The current through the motor is related to the voltage by the equation I = V/R, where R is the equivalent resistance of the motor coils.

For example, if the motor is driven by a 48.0 V emf and has an equivalent resistance of 0.400 Ω, the current drawn by the motor is I = 48.0 V / 0.400 Ω = 120 A. The power dissipated by the motor coils is P = I²R = (120 A)² × 0.400 Ω = 5.76 kW.

Under normal operating conditions, if the back emf is 40.0 V, the total voltage across the coils is 8.0 V and the current drawn is determined using Ohm's law, I = V/R. The resistance of the motor can be calculated using the equation R = (Vemf - Vback emf) / I.