Shunt motor speed with 1.2 Ω armature resistance added

For a shunt motor the field is across the constant supply, so flux is constant. Hence $T \propto I_a$ and $N \propto E_b$.

Given: $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $I_{L1} = 30$ A, $N_1 = 1400$ rpm, $R_{ext} = 1.2\ \Omega$, $T_2 = 0.8\,T_1$.

Step 1: Initial condition

$$\begin{aligned} I_{sh} &= \frac{240}{120} = 2\ \text{A} \\ I_{a1} &= 30 - 2 = 28\ \text{A} \\ E_{b1} &= 240 - 28 \times 0.4 = 240 - 11.2 = 228.8\ \text{V} \end{aligned}$$

Step 2: New armature current

$$I_{a2} = 0.8 \times I_{a1} = 0.8 \times 28 = 22.4\ \text{A}$$

Step 3: New back emf

Total armature-circuit resistance $= 0.4 + 1.2 = 1.6\ \Omega$:

$$\begin{aligned} E_{b2} &= 240 - 22.4 \times 1.6 \\ &= 240 - 35.84 = 204.16\ \text{V} \end{aligned}$$

Step 4: New speed

$$\begin{aligned} N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \\ &= 1400 \times \frac{204.16}{228.8} = 1249.23\ \text{rpm} \end{aligned}$$

Even though the load torque falls (which alone would raise the speed slightly), the large drop in the added 1.2 Ω resistance makes the speed fall.

Assumptions: constant field flux, no armature reaction, brush drop neglected.

Answer: New speed $\approx 1249$ rpm.