Series motor speed at 25 A

For a DC series motor below saturation, flux is proportional to the current: $\phi \propto I$. Hence

$$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} = \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2}$$

Given: $V = 230$ V, $I_1 = 100$ A, $N_1 = 1000$ rpm, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $I_2 = 25$ A. Brush drop neglected.

  + o---Rse---Ra---(M)---o -
        I = Ia = Ise

Step 1: Back emf at 100 A

$$\begin{aligned} E_{b1} &= V - I_1 (R_a + R_{se}) \\ &= 230 - 100 \times 0.25 = 205\ \text{V} \end{aligned}$$

Step 2: Back emf at 25 A

$$\begin{aligned} E_{b2} &= 230 - 25 \times 0.25 \\ &= 230 - 6.25 = 223.75\ \text{V} \end{aligned}$$

Step 3: New speed

$$\begin{aligned} N_2 &= 1000 \times \frac{223.75}{205} \times \frac{100}{25} \\ &= 1000 \times 1.0915 \times 4 = 4365.9\ \text{rpm} \end{aligned}$$

Answer: Speed at 25 A $\approx 4366$ rpm.

The speed rises more than four times when current falls to one quarter. This is the typical series characteristic: at light load the flux is weak and the speed becomes very high, so a series motor is always coupled to its load directly or through gears, never through a belt.