Question

A 120 v, series-wound motor has a field resistance of 84 Ω and an armature resistance of 18 Ω for a total resistance of 102 Ω. When it is operating at full speed, a back emf of 72 v is generated. (a) What is the initial current drawn by the motor (in A)?

Answer 1

The initial current drawn by the motor is approximately 1.18 A when it is first started, calculated using Ohm's Law with the given resistance values and the applied voltage of 120 V.

Step-by-step explanation:

The initial current drawn by a 120 V, series-wound motor with a field resistance of 84 Ω and an armature resistance of 18 Ω (totaling 102 Ω), when no back emf is generated, can be calculated using Ohm’s Law (I = V/R). The voltage (V) is 120 volts and the total resistance (R) is 102 Ω. Therefore, the initial current (I) can be calculated as follows:

I = V / R

I = 120 V / 102 Ω

I = 1.17647 A (approximately)

This means the initial current drawn by the motor is roughly 1.18 A when it is first started and no back emf is generated.