Shunt motor speed with 30 Ω field resistance added

Given: $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 110\ \Omega$, $N_1 = 1400$ rpm at half load, extra 30 Ω in field, load torque +10%.

Note: the current at half load is not given. Since $R_a$ is very small, the $I_a R_a$ drop (about 1-2 V) hardly affects the speed, so the main answer neglects it. A check with an assumed current follows.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Flux ratio

$$\begin{aligned} I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{sh2} = \frac{220}{110 + 30} = 1.5714\ \text{A} \\ \frac{\phi_2}{\phi_1} &= \frac{110}{140} = 0.7857 \end{aligned}$$

New speed (neglecting $I_a R_a$, so $E_{b1} \approx E_{b2}$)

$$N_2 = N_1 \times \frac{\phi_1}{\phi_2} = 1400 \times \frac{140}{110} = 1781.8\ \text{rpm}$$

Check with an assumed half-load line current of 23 A

$$\begin{aligned} I_{a1} &= 23 - 2 = 21\ \text{A}, \quad E_{b1} = 220 - 21 \times 0.08 = 218.32\ \text{V} \\ I_{a2} &= \frac{1.1 \times 21}{0.7857} = 29.4\ \text{A}, \quad E_{b2} = 220 - 29.4 \times 0.08 = 217.65\ \text{V} \\ N_2 &= 1400 \times \frac{217.65}{218.32} \times \frac{1}{0.7857} = 1776.3\ \text{rpm} \end{aligned}$$

The 10% torque increase only raises $I_a$; its effect on speed is under 0.5%.

Answer: New speed $\approx 1782$ rpm (about 1776 rpm if a half-load current of 23 A is assumed).