Final answer:
The initial current drawn by a 120 V series-wound motor with a total resistance of 90 Ω is calculated using Ohm's Law. It comes out to be approximately 1.33 A, using the full source voltage without considering the back emf.
Step-by-step explanation:
The question relates to a 120 V series-wound motor with field and armature resistances resulting in a total resistance of 90 Ω. To find the initial current drawn by the motor, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R): I = V/R. Since we want to find the initial current, we need to consider the full source voltage without the back emf. Therefore, the initial current drawn by the motor can be calculated using the total resistance and the source voltage without the back emf effect, as the back emf opposition only develops when the motor has already started operation.
Using the given total resistance of 90 Ω and the source voltage of 120 V, the initial current (I) drawn by the motor is:
I = 120 V / 90 Ω = 1.33 A (approximately).
The initial current is crucial for the motor as it is the current at the instant the motor starts before any back emf is generated. As the motor reaches its operating speed, the back emf increases and the actual running current will decrease accordingly from this initial value.