Starting current and starting resistance of shunt motor

At starting, speed is zero so back emf $E_b = 0$; only $R_a$ limits the armature current.

Given: $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 100\ \Omega$.

Part 1: Starting current at normal voltage

$$\begin{aligned} I_{a,st} &= \frac{V}{R_a} = \frac{220}{0.08} = 2750\ \text{A} \\ I_{sh} &= \frac{220}{100} = 2.2\ \text{A} \\ I_{L,st} &= 2750 + 2.2 = 2752.2\ \text{A} \end{aligned}$$

Part 2: Starting resistance

"Reduced by 80%" is read as: new starting armature current = 20% of 2750 A. Supply voltage is 10% lower, $V' = 0.9 \times 220 = 198$ V.

$$I'_{a,st} = 0.2 \times 2750 = 550\ \text{A}$$ $$\begin{aligned} I'_{a,st} &= \frac{V'}{R_a + R_{st}} \\ R_a + R_{st} &= \frac{198}{550} = 0.36\ \Omega \\ R_{st} &= 0.36 - 0.08 = 0.28\ \Omega \end{aligned}$$

(If line current is used instead: $I'_L = 0.2 \times 2752.2 = 550.44$ A, $I_{sh} = 1.98$ A, $I'_a = 548.46$ A, giving $R_{st} = 0.281\ \Omega$, practically the same.)

Answer: Starting current $\approx 2750$ A in the armature (2752.2 A from the line); starting resistance required $R_{st} \approx 0.28\ \Omega$.