An eight pole D.C. generator has a simple wave wound armature containing 32 coils of 6 turns each. Its flux per pole is 0.06 wb. The machine is running at 250 rpm. The induced armature voltage is

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ISRO (VSSC) Technical Assistant Electrical 2017 Official Paper
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  1. 96 V
  2. 192 V
  3. 384 V
  4. 768 V

Answer (Detailed Solution Below)

Option 3 : 384 V
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Total No, of conductor = coil × turn × 2 = 32 × 6 × 2 = 384

ϕ = Flux, N = speed, P = Number of poles, A = Number of parallel paths, Z = Number of conductors

Induced armature voltage \(= \left( {\frac{{\phi ZN}}{{60}}} \right)\left( {\frac{P}{A}} \right)\)

\(= \frac{{0.06 \times 384 \times 250}}{{60}} \times \frac{8}{2}\)

(For wave wound A = 2) = 384
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