if the voltage across the lamp drops by 1%, the power drawn
will be reduced by how many %?
Answers were Sorted based on User's Feedback
Answer / eee
Power = V(square)/R
If voltage is reduced by 1% power will drop by
.99*.99 = .9801
(1-0.9801)*100
Power is reduced by 1.99%.
| Is This Answer Correct ? | 58 Yes | 11 No |
Answer / nirmal maity
Let, voltage across the lamp =100v and resistance R =100 ohm
so, power P =v*I = 100*1=100W {I =V/R =100/100=1 }
If the voltage across the lamp drops by 1 percentage , then present voltage is 99v
Then present power drown by the lamp whose resistance R =100 ohm is P =V^2/R = 98*98/100 = 98.01W .
So power drop of the lamp is 100-98 =2 w
so the power drawn will be reduced by 2percentage
| Is This Answer Correct ? | 29 Yes | 9 No |
Answer / k.prakashchandra
If you connect phase and neutral across the bulb the voltage
should fully drop.1% drop is possible if there are 99
similar bulbs in series with it or a resistance of 99 times
the bulb resistance.
| Is This Answer Correct ? | 1 Yes | 10 No |
Answer / brijesh soni
if you will conect the lamp drop by 1% the power drown by
the bulb will be 1%
| Is This Answer Correct ? | 2 Yes | 17 No |
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