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 |
What Should be the rating of transformer used in 33KV substation?
why the field winding in dc motor is not short circuited eventhough dc supply is applied across the winding
limitation of conductor height on ground when we consider EMC
can v rotate single phase induction motor in both the direction...?
why we dont store the ac power in battery?
why the vfd motor take current high in vfd panel then mcc panel.current in vfd panel=160A in 93%.and mcc panel it shows 130A. (132kw motor)
How shall i select the value of capacitor or inductor in a system?
why are maintine power factor 0.99 in Eb line? why are used capacitor bank in Eb line? what is the purpose capacitor bank?
4 Answers TNEB Tamil Nadu Electricity Board,
our plant maximum demand has 500 kva.need to suggest me to maintained pf 0.98.how much capacitor bank will be connect to the supply line.
State superposition theorem?
What is the necessary and sufficient condition for stability.
What is the difference between shortcircuit current, rated current of cable & MCB with example? What will be the size of cable required for 5.5 KW motor (Full load current 10 Amps approx.), Starting current with DOL is 120 Amps. Aprrox. Short circuit current with LLLG fault is 2000 Amps Approx.