Interested to Buy Any Domain ? << Click Here >> for more details...
if the voltage across the lamp drops by 1%, the power drawn
will be reduced by how many %?
- 5 Answers
- 24260 Views
- PSPCL, I also Faced
- E-Mail Answers
Answer Posted / 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 |
Post New Answer View All Answers
why Lightning arrestor are made by 2 seperate stack of inslutor but why cvt are in single stack of insulator.
explain remenance . differentiate high remenance type ct & low remenance type ct
What is gross error?
In the Variable Voltage Variable Frequency Drive, how the Voltage and Frequency varied ? Is it varied in the Inverter part of the Drive ? Please give us supporting document if any.
what is the Connected Load (kW) or Contract Demand (kVA)of mahindra nasik plant??
what is your aim
how to set the time in digital timer ?
what are the types of electromagnetic forces developed in practical transformer?
What is servo motor and what are different types of servo motor
difference between power surge and switching transient
If i have two 3000 kW motors and the whole facility requires a 10 MVA transformer. What would be better installing two 5 MVA transformers(one for 1 motor ) or just a single 10 MVA transformer for both of them. Give reasons for your answer.
What is Auto Mains Failure Panel & Auto Synchronizing Panel
HOW TO SELECT THE COPPER CABLE SIZES OF MOTORS,TRANSFORMERS?
what is the difference between single ratio, multiple core and multiple ratio, multiple core ? please explain.
Power factor =1.000 at factory in supply. However, PF = 0.8 at motor end. Current = 100 Amp, Voltage = 400 V. What is the power distribution loss? Can I reduce it by adding capacitor at the load feeder?
