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KCET2015PhysicsCapacitance

The resistance of the bulb filament is ( 100 ) at a temperature of ( 100^ C ). If its temperature co- efficient of resistance be ( 0.005 ) per ( ^ C ), its resistance will become ( 200 ) at a temperature

Options

  1. A( 300^ C )
  2. B( 400^ C )
  3. C( 500^ C )
  4. D( 200^ C )

Correct answer

A. ( 300^ C )

Step-by-step solution

At T₁=100^ C , resistance, R₁=100 ; temperature coefficient, =0.005 per ^ C Let resistance R₂=200 at temperature T₂ , then R₂=R₁ (1+ (T₂-T₁ ) ) R₂=R₁+ R₁ (T₂-T₁ ) R₂-R₁= R₁ (T₂-T₁ ) (R₂-R₁ ) R₁ =T₂-T₁ T₂=T₁+ (R₂-R₁ ) R₁ Substituting the values, we get T₂=100+ (200-100) 0.005 100 =300^ C

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