Question

A dc series motor with a resistance between terminals of 1 Ω, runs at 800 rpm from a 200 V supply taking 15 A. If the speed is to be reduced to 475 rpm for the same supply voltage and current the additional series resistance to be inserted would be approximately
1. 2.5 Ω
2. 3 Ω
3. 4.5 Ω
4. 5 Ω

Answer 1

Correct Answer - Option 4 : 5 Ω

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

N₁ = 600 rpm, N₂ = 475 rpm

In a DC series motor, E_b = V – I_aR

E_b ∝ Nϕ and ϕ ∝ I_a

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

E₁ = 200 – 15 × 1 = 185 V

Given that I_a1 = I_a2

\( \Rightarrow \frac{{185}}{{{E_2}}} = \frac{{800}}{{475}}\)

Therefore, E₂ = 110 V

110 = 200 – 15 × R₂

⇒ R₂ = 6 Ω

Therefore, Extra Resistance to be added in Series:

R_ex = R₂ – R₁

= 6 – 1 = 5 Ω