Shunt motor speed with 75 Ω field resistance added

Given: $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 32$ A, $N_1 = 1500$ rpm. Extra 75 Ω in field circuit, load torque raised by 20%.

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

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

Initial condition

$$\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A} \\ I_{a1} &= 32 - 2 = 30\ \text{A} \\ E_{b1} &= 250 - 30 \times 0.2 = 244\ \text{V} \end{aligned}$$

New field current and flux

$$I_{sh2} = \frac{250}{125 + 75} = 1.25\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.25}{2} = 0.625$$

New armature current

$T_2 = 1.2\, T_1$ and $T \propto \phi I_a$:

$$\begin{aligned} \phi_2 I_{a2} &= 1.2\, \phi_1 I_{a1} \\ I_{a2} &= \frac{1.2 \times 30}{0.625} = 57.6\ \text{A} \end{aligned}$$

New speed

$$\begin{aligned} E_{b2} &= 250 - 57.6 \times 0.2 = 238.48\ \text{V} \\ \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ N_2 &= 1500 \times \frac{238.48}{244} \times \frac{1}{0.625} = 2345.7\ \text{rpm} \end{aligned}$$

Answer: New speed $N_2 \approx 2346$ rpm (armature current rises to 57.6 A).