Question

A dc motor with its rotor and filed coils connected in serieshas an internal resistance of

Answer 1

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