Shunt motor speed with 1 Ω armature resistance added

In a shunt motor the field current, and so the flux, stays constant. Then $T \propto I_a$ and $N \propto E_b$.

Given: $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $I_{L1} = 27$ A, $N_1 = 1200$ rpm. Extra series resistance $R_{ext} = 1\ \Omega$. New torque $T_2 = 1.2\,T_1$.

Step 1: Initial condition

$$\begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{240}{120} = 2\ \text{A} \\ I_{a1} &= I_{L1} - I_{sh} = 27 - 2 = 25\ \text{A} \\ E_{b1} &= V - I_{a1} R_a = 240 - 25 \times 0.4 = 230\ \text{V} \end{aligned}$$

Step 2: New armature current

With constant flux, $T \propto I_a$:

$$I_{a2} = 1.2 \times I_{a1} = 1.2 \times 25 = 30\ \text{A}$$

Step 3: New back emf

The armature circuit resistance is now $R_a + R_{ext} = 0.4 + 1 = 1.4\ \Omega$:

$$\begin{aligned} E_{b2} &= V - I_{a2}(R_a + R_{ext}) \\ &= 240 - 30 \times 1.4 = 198\ \text{V} \end{aligned}$$

Step 4: New speed

With constant flux, $N \propto E_b$:

$$\begin{aligned} N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \\ &= 1200 \times \frac{198}{230} = 1033.04\ \text{rpm} \end{aligned}$$

Assumptions: the field current stays at 2 A (the extra resistance is in the armature only), armature reaction and brush drop are neglected. The 20% rise is taken on the half-load torque.

Answer: New speed $\approx 1033$ rpm (armature current 30 A, back emf 198 V).