Question

A 120-V, series-wound motor has a field resistance of 80 Ω and an armature resistance of 10 Ω. When it is operating at full speed, a back emf of 75 V is generated. (a) What is the initial current drawn by the motor? When the motor is operating at full speed, where are (b) the current drawn by the motor, (c) the power output of the source, (d) the power output of the motor, and (e) the power dissipated in the two resistances? Options: Option 1: (a) 1.2 A, (b) 0 A, (c) 1200 W, (d) 750 W, (e) 780 W Option 2: (a) 12 A, (b) 6 A, (c) 900 W, (d) 675 W, (e) 880 W Option 3: (a) 8 A, (b) 4 A, (c) 960 W, (d) 720 W, (e) 700 W Option 4: (a) 15 A, (b) 10 A, (c) 800 W, (d) 800 W, (e) 800 W

Answer 1

Option 4 is the correct answer, with the initial current drawn by the motor to be 15 A, the current drawn by the motor at full speed to be 10 A, the power output of the source to be 800 W, the power output of the motor to be 800 W, and the power dissipated in the two resistances to be 800 W.

Explanation:

The given question involves the calculation of the current drawn by the motor, the power output of the source, the power output of the motor, and the power dissipated in the two resistances when the motor is operating at full speed. To determine these values, the equations for the voltage, current, and power in a series-wound motor must be used.

The voltage across the motor is given by the equation E = V - I R, where E is the voltage across the motor, V is the back emf generated at full speed, I is the current drawn by the motor, and R is the total resistance of the motor. In the given problem, V = 75 V, R = 80 Ω + 10 Ω = 90 Ω. Thus, we have E = 75 V - I (90 Ω).

The current drawn by the motor is given by the equation I = V/R, where V is the voltage across the motor and R is the total resistance of the motor. Therefore, the initial current drawn by the motor is I = 75 V/90 Ω = 15 A.

The power output of the source is given by the equation P = V I, where V is the back emf generated at full speed and I is the current drawn by the motor. Therefore, the power output of the source is P = 75 V × 15 A = 1,125 W.

The power output of the motor is given by the equation P = E I, where E is the voltage across the motor and I is the current drawn by the motor. Therefore, the power output of the motor is P = 75 V × 15 A = 1,125 W.

The power dissipated in the two resistances is given by the equation P = I2 R, where I is the current drawn by the motor and R is the total resistance of the motor. Therefore, the power dissipated in the two resistances is P = 152 (90 Ω) = 800 W.

Thus, option 4 is the correct answer, with the initial current drawn by the motor to be 15 A, the current drawn by the motor at full speed to be 10 A, the power output of the source to be 800 W, the power output of the motor to be 800 W, and the power dissipated in the two resistances to be 800 W.