Question

A 230-V DC shunt motor has an armature resistance of 0.25 Ω and runs at 1100 rpm, taking an armature current 40 A. It is desired to reduce the speed to 750 rpm. If the armature current remains the same, find the additional resistance to be connected in series with the armature circuit.
1. 1 Ω
2. 1.5 Ω
3. 1.25 Ω
4. 1.75 Ω

Answer 1

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