Speed of 220 V series motor at reduced load current

In a series motor flux is proportional to the series field (line) current, below saturation. So

$$N \propto \frac{E_b}{\phi} \propto \frac{E_b}{I}, \qquad E_b = V - I(R_a + R_{se}) - V_{brush}$$

Given: $V = 220$ V, $I_1 = 38$ A, $N_1 = 600$ rpm, $R_a + R_{se} = 0.4 + 0.2 = 0.6\ \Omega$, brush drop 3 V.

$$E_{b1} = 220 - 38 \times 0.6 - 3 = 194.2\ \text{V}$$

a) Speed at 19 A

$$\begin{aligned} E_{b2} &= 220 - 19 \times 0.6 - 3 = 205.6\ \text{V} \\ N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\ &= 600 \times \frac{205.6}{194.2} \times \frac{38}{19} = 1270.4\ \text{rpm} \end{aligned}$$

b) Speed at 1 A (load removed)

$$\begin{aligned} E_{b3} &= 220 - 1 \times 0.6 - 3 = 216.4\ \text{V} \\ N_3 &= 600 \times \frac{216.4}{194.2} \times \frac{38}{1} = 25\,406\ \text{rpm} \end{aligned}$$

Answer: a) $N \approx 1270$ rpm; b) $N \approx 25\,400$ rpm.

This dangerously high value (about 42 times rated speed) shows why a series motor must never be started or run without load: as current falls, flux falls and speed rises without limit. (In practice residual flux and windage keep it a little lower, but it would still be destructive.)