Series motor current at 800 rpm

The load condition is not given, so it is assumed that the field is not saturated (Φ ∝ I). Then E_b ∝ I N.

Equivalent circuit

 +--[Ra=0.06]--[Rse=0.04]--(Eb)--+
 |                I               |
 +-------------- 220 V -----------+

R = 0.06 + 0.04 = 0.1 Ω

First condition

E_b1 = 220 − 25 × 0.1 = 217.5 V at 1200 rpm, I₁ = 25 A

Second condition (800 rpm, I₂ unknown)

E_b2 = 220 − 0.1 I₂

E_b2 / E_b1 = (Φ₂ N₂) / (Φ₁ N₁) = (I₂/25) × (800/1200)

220 − 0.1 I₂ = 217.5 × (I₂/25) × (2/3) = 5.8 I₂

220 = 5.9 I₂

I₂ = 37.29 A

Check: E_b2 = 220 − 3.729 = 216.27 V; and 217.5 × (37.29/25) × (2/3) = 216.27 V ✓.

Answer: the motor draws about 37.3 A at 800 rpm (a lower speed means the load torque is higher, so more current is drawn).