1 Ω के टर्मिनलों के बीच प्रतिरोध वाला एक dc श्रेणी मोटर 20 A लेते हुए 250 V आपूर्ति से 1000 rpm पर चलता है। जब 6 Ω के अतिरिक्त प्रतिरोध को श्रेणी में डाला जाता है और समान धारा ली जाती है, तो नयी गति क्या होगी?  

This question was previously asked in
ESE Electrical 2014 Paper 2: Official Paper
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  1. 142.8 rpm
  2. 166.7 rpm
  3. 478.3 rpm
  4. 956.6 rpm

Answer (Detailed Solution Below)

Option 3 : 478.3 rpm
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Detailed Solution

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V = 200 V, R1 = Ra + Rse = 1 ओम, Ia = 15 A

N1 = 1000 rpm

R1 = Ra + Rse = 7 Ω

Eb ∝ Nϕ और ϕ ∝ Ia

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

E1 = 250 – 20 × 1 = 230 V

दिया गया है कि Ia1 = Ia2

E1 = 250 – 20 × 7 = 110 V

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

⇒ N2 = 478.26 rpm
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