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Torque and Energy Relation

An energy met...

Question

An energy meter, having meter constant of 1200 rev/kWh, makes 20 revolutions in 30 seconds for a constant load . The load in kW is
2

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Solution

The correct option is A 2
$1200 r e v / k W h, 20 r e v, 30 s e c$

$P_{l o s s} = \frac{20 \times 3600}{1200 \times 30} = 2 k W$

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