Shunt motor speed with series armature resistance

The flux is constant (shunt field across the supply), so torque $\propto I_a$ and speed $\propto E_b$.

Field current: $I_f = 240/120 = 2$ A.

Half-load condition:

$$I_{a1} = 27 - 2 = 25\ \text{A}, \qquad E_{b1} = 240 - 25(0.4) = 230\ \text{V}, \qquad N_1 = 600\ \text{rpm}$$

With the extra 1 Ω, the armature circuit resistance is $0.4 + 1 = 1.4\ \Omega$.

(i) Load torque constant at half-load value

The torque is unchanged, so $I_{a2} = I_{a1} = 25$ A.

$$E_{b2} = 240 - 25(1.4) = 205\ \text{V}$$ $$N_2 = 600\times\frac{205}{230} = 534.8\ \text{rpm}$$

(ii) Load torque increased to full-load torque

Full-load torque is twice the half-load torque, so

$$I_{a2} = 2\times 25 = 50\ \text{A}$$ $$E_{b2} = 240 - 50(1.4) = 170\ \text{V}$$ $$N_2 = 600\times\frac{170}{230} = 443.5\ \text{rpm}$$

Answer: (i) $534.8\ \text{rpm}$; (ii) $443.5\ \text{rpm}$.