Answer / mubashar ali

total KW * total runing Hours= unit
for example
5 lamp 100 watt each onload for 6 hour
AC 3000W onload for 3 hour
Motor 5HP (5*746=3730W) onload for 2 hour
than

500/1000 * 6= 3
3000/1000 * 3= 9
3730/1000 * 2= 7.46
total unit = 19.46 * Rs.

Answer / ramesh gaur

one unit is equal to one kwh(1000w per hour)

Answer / vinothbabu.k

P=(1.732*VOLTAGE*AMPS)/1000,
FOR UNIT CONSUPTION:

KWATTS*HOURS=UNIT-ITS FOR INDUCTIVE LOAD,
FOR RESISTIVE LOAD THE ABOVE IS CORRECT,LAMP HEATER IS
RESISTIVE LOAD,
IT DEPANDS ON TYPE OF LOAD.
AND FOR CAPACITIVE & INDUCTIVE LOAD NO OF WORKING HOURS
WILL BE VARRY,FOR EXAMPLE TAKE FRIDGE WAS WORKING FOR 2
HOUR IT TAKE 750W MEANS WE SHOULD NOT
CALCULATE .750*2=1.5KWHR/UNIT,FRIDGE IS A CAPACITIVE LOAD
IT WILL ON OPPROXIMATELY 35 MIN FOR EVERY PER HOUR SO WE
SHOULT TAKE FOR 2 HRS IT WILL ON 1.05HRS FOR 2 HRS,
SO 750*1.05=0.787 UNIT

Answer / chandru

power= v*a
unit=kwh

Answer / sivaram senthil.g

p=v*i

Answer / vinothbabu.k

P=(1.732*VOLTAGE*AMPS)/1000

Answer / mad sarfaraj

power(p)=dw/dt(this is the basic formula 4 power)
and it's S.I. unit is joule/second or watt.

Answer / balaji

Power=i^2*R(resistor multiplied with square of current)

Answer / mohanbabublr

Power=(√3xVxIxCosΦ)/1000

Dear all i want to know FEWA Electrical Supervisor Licence test Previous Year Questions ? If any body know please send to me shibin_75@yahoo.com

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