Armature resistance to reduce shunt motor speed to 1250 rpm

This is armature-resistance speed control. For a shunt motor with constant flux, $N \propto E_b$, and with constant load torque the armature current also stays constant ($T \propto I_a$).

Given: $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_L = 30$ A, $N_1 = 1500$ rpm, $N_2 = 1250$ rpm, torque constant.

Step 1: Armature current and back emf at 1500 rpm

$$\begin{aligned} I_{sh} &= \frac{250}{125} = 2\ \text{A} \\ I_a &= 30 - 2 = 28\ \text{A} \\ E_{b1} &= 250 - 28 \times 0.2 = 244.4\ \text{V} \end{aligned}$$

Step 2: Back emf needed at 1250 rpm

Torque constant and flux constant, so $I_{a2} = I_{a1} = 28$ A.

$$\begin{aligned} E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \\ &= 244.4 \times \frac{1250}{1500} = 203.67\ \text{V} \end{aligned}$$

Step 3: Series resistance

$$E_{b2} = V - I_a (R_a + R)$$ $$\begin{aligned} R_a + R &= \frac{V - E_{b2}}{I_a} = \frac{250 - 203.67}{28} = 1.655\ \Omega \\ R &= 1.655 - 0.2 = 1.455\ \Omega \end{aligned}$$

Check: $E_{b2} = 250 - 28 \times 1.655 = 203.67$ V, and $1500 \times 203.67/244.4 = 1250$ rpm.

Power wasted in the added resistor $= 28^2 \times 1.455 \approx 1141$ W, which is why this method is inefficient.

Answer: A resistance of about $1.455\ \Omega$ must be connected in series with the armature.