Question

A dc series motor of resistance 1 Ω across terminals runs at 1000 rpm at 250 V taking a current of 20 A. When an additional resistance of 6 Ω is inserted in series and taking the same current, the new speed would be
1. 142.8 rpm
2. 166.7 rpm
3. 478.3 rpm
4. 956.6 rpm

Answer 1

Correct Answer - Option 3 : 478.3 rpm

V = 200 V, R₁ = R_a + R_se = 1 ohm, I_a = 15 A

N₁ = 1000 rpm

R₁ = R_a + R_se = 7 Ω

E_b ∝ Nϕ and ϕ ∝ I_a

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

E₁ = 250 – 20 × 1 = 230 V

Given that I_a1 = I_a2

E₁ = 250 – 20 × 7 = 110 V

\(\Rightarrow \frac{{230}}{{110}} = \frac{{1000}}{{{N_2}}}\)

⇒ N₂ = 478.26 rpm