Correct Answer - Option 4 : 1.75 Ω

__Calculation:__

Given

V = 230 V,

R1 = Ra + Rse = 0.25 Ω

Ia1 = 40 A

N1 = 1100 rpm,

N2 = 750 rpm

In a DC series motor, Eb = V – IaR

Eb ∝ Nϕ and ϕ ∝ Ia

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{I_{a1}}}}{{{I_{a2}}}} \times \frac{{{N_1}}}{{{N_2}}}\)

E1 = 230 – 40 × 0.25 = 220 V

Given that Ia1 = Ia2

\(\Rightarrow\frac{{{E_1}}}{{{E_2}}} = \frac{{{I_{a1}}}}{{{I_{a1}}}} \times \frac{{{N_1}}}{{{N_2}}}= \frac{{{N_1}}}{{{N_2}}}\)

\( \Rightarrow \frac{{220}}{{{E_2}}} = \frac{{1100}}{{750}}\)

Therefore, E2 = 150 V

150 = 230 – 40 × R2

⇒ R2 = 2 Ω

Therefore, Extra Resistance to be added in Series:

Rex = R2 – R1

= 2 – 0.25 = 1.75 Ω