Question

A 120 v, series-wound motor has a field resistance of 74 ω and an armature resistance of 16 ω for a total resistance of 90 ω. when it is operating at full speed, a back emf of 81 v is generated.

what is the initial current drawn by the motor (in a)?

Answer 1

The initial current drawn by a 120 V series-wound dc motor with a total resistance of 90 ω is 1.33 A, calculated using Ohm's Law before any back emf is generated.

Step-by-step explanation:

The question involves a series-wound dc motor operating on a 120 V supply with a field resistance of 74 ω and an armature resistance of 16 ω, totalizing 90 ω. We are asked to find the initial current drawn by the motor when it is just turned on, before back emf comes into play. To find the initial current, Ohm's Law (I = V / R) is applied where V is the supply voltage and R the total resistance of the motor.

Using the given values, the initial current (I) drawn without the back emf can be found with:

I = V / R = 120 V / 90 ω = 1.33 A

This current will decrease when back emf is generated as the motor speeds up to its full operating condition. However, the initial current just after startup will be 1.33 A.