Starting current; speed and % drop of shunt motor

Why the starting current is high

Reason: the armature current of a DC motor is

$$I_a = \frac{V - E_b}{R_a}$$

At the instant of starting the armature is at rest, so the back emf $E_b = \dfrac{\Phi ZNP}{60A} = 0$. The current is then limited only by the very small armature resistance:

$$I_{a(start)} = \frac{V}{R_a}$$

For example, a 220 V motor with $R_a = 0.5\ \Omega$ would draw $220/0.5 = 440$ A, about 15–20 times its full-load current (say 25 A).

Speed at 51 A

The field is across the supply, so $I_{sh} = 250/250 = 1$ A and the flux is constant. Then $N\propto E_b$.

Condition 1 (1000 rpm, 2 A):

$$I_{a1} = 2 - 1 = 1\ \text{A}, \qquad E_{b1} = V - I_{a1}R_a = 250 - 1(0.2) = 249.8\ \text{V}$$

Condition 2 (51 A):

$$I_{a2} = 51 - 1 = 50\ \text{A}, \qquad E_{b2} = 250 - 50(0.2) = 240.0\ \text{V}$$ $$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \;\Rightarrow\; N_2 = 1000\times\frac{240.0}{249.8} = 960.8\ \text{rpm}$$

Percentage speed drop

$$\frac{N_1 - N_2}{N_1}\times 100 = \frac{1000 - 960.8}{1000}\times 100 = 3.92\%$$

Answer: speed at 51 A $= 960.8\ \text{rpm}$; speed drop $= 3.92\%$.