Series motor current at 800 rpm

In a series motor the field current is the armature current, so (unsaturated) the flux is proportional to the current: $\Phi\propto I$. The back emf then follows $E_b \propto \Phi N \propto I N$.

Total resistance $R = R_a + R_{se} = 0.06 + 0.04 = 0.1\ \Omega$.

Condition 1: 1200 rpm, 25 A

$$E_{b1} = V - I_1R = 220 - 25(0.1) = 217.5\ \text{V}$$

Condition 2: 800 rpm, current $I_2$

$$E_{b2} = 220 - 0.1I_2$$

Using $\dfrac{E_{b2}}{E_{b1}} = \dfrac{\Phi_2N_2}{\Phi_1N_1} = \dfrac{I_2N_2}{I_1N_1}$:

$$\frac{220 - 0.1I_2}{217.5} = \frac{I_2}{25}\times\frac{800}{1200}$$ $$220 - 0.1I_2 = 5.800\,I_2 \;\Rightarrow\; I_2 = \frac{220}{0.1 + 5.800} = 37.29\ \text{A}$$

Check: $E_{b2} = 220 - 0.1(37.29) = 216.27\ \text{V}$.

Answer: the motor draws $I_2 = 37.29\ \text{A}$ at 800 rpm (assuming an unsaturated magnetic circuit).