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 = `(\frac{V}{R})²` ｘR          [ I = `\frac{V}{R}` ]

P = `\frac{V²}{R²}` ｘ R

P = `\frac{V²}{R}`

or

R = `\frac{V²}{P}`

by putting value

R = `\frac{(220 V)²}{100 W}`

R = `\frac{48400 V²}{100 W}`

R = 484 Ω  -----------------Ans.


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

E= `\frac{Power ｘ Time}{1000}`

By putting values

E = `\frac{100 Wattsｘ 150 hrs}{1000}`

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


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