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 solve this problem, we need to understand that the speed of a DC shunt motor is inversely proportional to the flux and directly proportional to the electromotive force (E.M.F).

Step 1: Calculate the Back E.M.F at Full Load The back E.M.F (Eb1) can be calculated using the formula: Eb1 = V - Ia1(Ra + Rse), where V is the supply voltage, Ia1 is the armature current, Ra is the armature resistance, and Rse is the series field resistance. In this case, the series field resistance is 0 because it's a shunt motor.

Given that V = 120V, Ia1 = 60A, and Ra = 0.2Ω, we can substitute these values into the formula to get: Eb1 = 120V - 60A * 0.2Ω = 108V.

Step 2: Calculate the Back E.M.F at Half Load At half load, the current is halved, so Ia2 = 60A / 2 = 30A. We can substitute this into the formula to get: Eb2 = 120V - 30A * 0.2Ω = 114V.

Step 3: Calculate the Speed at Half Load The speed at half load (N2) can be calculated using the formula: N2 = N1 * (Eb2 / Eb1), where N1 is the speed at full load.

Given that N1 = 1800 r.p.m., Eb2 = 114V, and Eb1 = 108V, we can substitute these values into the formula to get: N2 = 1800 r.p.m. * (114V / 108V) = 1900 r.p.m.

Therefore, the speed of the motor at half load is 1900 r.p.m.