A 120 V d.c. shunt motor has an armature resistance of 0.2 Ω and a field resistance of60 Ω. The full load line current is 60 A and the full load speed is 1800 r.p.m.. If thetotal brush contact drop is 3 V, find the speed of the motor at half load

To find the speed of the motor at half load, we first need to understand the relationship between the speed and the load of a DC shunt motor. The speed of a DC shunt motor is inversely proportional to the load. This means that as the load decreases, the speed increases.

Here are the steps to find the speed of the motor at half load:

1. Calculate the back emf (Eb1) at full load: Eb1 = V - Ia1(Ra + Rse) Where: V = Supply voltage = 120V Ia1 = Armature current at full load = Line current - Field current = 60A - (120V/60Ω) = 58A Ra = Armature resistance = 0.2Ω Rse = Series field resistance = 0 (since it's a shunt motor) So, Eb1 = 120V - 58A*0.2Ω = 103.6V

2. Calculate the armature current at half load (Ia2): Since the torque is proportional to the armature current, the armature current at half load will be half of the full load armature current. So, Ia2 = Ia1/2 = 58A/2 = 29A

3. Calculate the back emf at half load (Eb2): Eb2 = V - Ia2(Ra + Rse) So, Eb2 = 120V - 29A*0.2Ω = 114.2V

4. Calculate the speed at half load (N2): The speed of a DC motor is directly proportional to the back emf and inversely proportional to the flux. Since the motor is a shunt motor, the flux remains constant. Therefore, the speed at half load can be calculated using the formula: N2 = N1 * (Eb2/Eb1) Where: N1 = Speed at full load = 1800 rpm So, N2 = 1800 rpm * (114.2V/103.6V) = 1983 rpm

Therefore, the speed of the motor at half load is approximately 1983 rpm.