# Solution

Find the characteristics of the motor at 10A, 1800rpm first. Look for the following (in order): If, Ia, E, P, w, T.

If = field current = Vf / Rf = 220V/440ohms = 0.5A

Ia = armature current = Il - If = 10A - 0.5A = 9.5A

E = Vt - IaRa = 220V - (9.5V)(0.2ohms) = 218.1V

P = EIa = (218.1V)(9.5A) = 2.07 kW

w = 2(pi)n/60 = 188.5 rad/s

T = P/w = 2.07kW / 188.5rad/s = 11.0 N-m

Now look at the new situation, where torque is 20 N-m. Since the field current is the same, the flux per pole will be the same. Use the torque equation and the power equation. Solve each for T and set the results equal to each other.

T = K1(phi)Ia and also Tw=EIa or T = EIa/w

Thus K1(phi) = E/w = 218.1V / 188.5rad/s = 1.157

Thus Ia = 20N-m/1.157 = 17.3A

Il = Ia + If = 17.3A + 0.5A = 17.8A

E = Vt - IaRa = 220 - (17.3A)(0.2ohms) = 216.54V

w = E/K1(phi) = 216.54V/1.157 = 187.16 rad/s

n = 1787rpm

(The speed changed by only 1%, even though torque doubled. This shows that the shunt motor makes a good constant speed motor.)