Adc motor with its rotor and filed coils connected in serieshas an internal resistance of
3.2î©. when running at full load on a 120-v lined, the emf inthe rotor is 105v.
a)what is the current drawn by the motor from the line?
b)what is the power delivered to the motor?
c) what is the mechanical power developed by this motor?

(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