DC shunt motor speed at reduced voltage and starting resistance

Data: $V_1 = 200$ V, $I_L = 10$ A, $N_1 = 1800$ rpm, $R_a = 0.6\ \Omega$, $R_{sh} = 360\ \Omega$, constant load torque. Assumption: flux is proportional to field current (no saturation).

Initial condition

$$\begin{aligned} I_{f1} &= \frac{200}{360} = 0.5556\ \text{A}, \qquad I_{a1} = 10 - 0.5556 = 9.444\ \text{A} \\ E_{b1} &= 200 - 9.444\times0.6 = 194.33\ \text{V} \end{aligned}$$

(i) Speed at 180 V

The field is across the supply, so the field current (flux) also falls:

$$I_{f2} = \frac{180}{360} = 0.5\ \text{A}, \qquad \frac{\phi_1}{\phi_2} = \frac{0.5556}{0.5} = 1.111$$

Constant torque, $T \propto \phi I_a$:

$$I_{a2} = I_{a1}\frac{\phi_1}{\phi_2} = 9.444\times1.111 = 10.494\ \text{A}$$ $$E_{b2} = 180 - 10.494\times0.6 = 173.70\ \text{V}$$

Since $E_b \propto \phi N$:

$$\begin{aligned} N_2 &= N_1\frac{E_{b2}}{E_{b1}}\cdot\frac{\phi_1}{\phi_2} = 1800\times\frac{173.70}{194.33}\times1.111 \\ &= 1787.7\ \text{rpm} \end{aligned}$$

(ii) Starting resistance for 20 A line current at 180 V

At start $E_b = 0$. Field current $= 0.5$ A, so armature current

$$I_{a,st} = 20 - 0.5 = 19.5\ \text{A}$$ $$\begin{aligned} R_a + R_{ext} &= \frac{180}{19.5} = 9.231\ \Omega \\ R_{ext} &= 9.231 - 0.6 = 8.631\ \Omega \end{aligned}$$

Answer: (i) speed $\approx 1788$ rpm (it falls only slightly because the flux falls with the voltage); (ii) additional series resistance $\approx 8.63\ \Omega$.