Answer:
(a): I = 4.6875 A
(b): P = 562.5 W
(c): P mech = 492.1875 W
Explanation:
Data Given:
V= 120 V
E= 105 V
R= 3.2 Ω
To find:
(a): I = ?
As we know that in a DC series motor the equation to be used will be:
V = Ε + (I) (R)
120 V = 105 V + ( I ) (3.2 Ω),
I= 15/3.2
I= 4.6875 A ans
Now moving towards Power delivered i.e.
(b): P del :
P del = V X I = (120 V) (4.6875 A) = 562.5 W. ans
c) P mech = ?
The mechanical power output is the electrical power input minus the rate of dissipation of energy in the motor’s resistance (assuming that there are no other power losses):
The power P dissipated in the resistance r is
P dsptd = I²r = (4.6875 A)² X(3.2 Ω) = 70.3125 W
P mech = P del - P dsptd
P mech = 562.5 W — 70.3125 W = 492.1875 W Ans
Hope it is clear
