An electric bulb is marked with 220 V, 100 W. Find the resistance of the filament of the 14.8 bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hours consumed by the bulb in one month (30 days). Ans. (484 Ω, 15 kWh)



Data Given:


Voltage marked on the Bulb = V = 220 V

Power of the Bulb = P = 100 W

Daily use of the Bulb = t = 5 hrs

Monthly (30 days) use of the bulb = 30 x 5 hrs = 150 hrs


To Find:


Resistance of bulb filament of the bulb R = ?

Energy consumed by the bulb in one month E = ?
 



Solution:


We have the formula for Power

P = I²R 

P = (VR)² xR          I = VR ]

P = V²R² x R

P = V²R

or

R = V²P

by putting value

R = (220V)²100W

R 48400 V²100W

R = 484 Ω  -----------------Ans.


Formula for Energy consumed in KWh (Power (in Watt) and Time (in Hours)

E= PowerxTime1000

By putting values

E = 100Wasx150hrs1000

E = 15 KWh ----------------Ans.


************************************

************************************


Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2020-21 Academic Skills and Knowledge (ASK    

Note:  Write me in the comments box below for any query and also Share this information with your class-fellows and friends.