An electric bulb rated 100W 40 V has to be operated at 50V 50Hz AC supply. What is the value of inductance of inductor in the bulb?

Asked by joshimohit1050 | 22nd Jul, 2021, 07:12: PM

Expert Answer:

Power rating of bulb, P = 100 W
 
Voltage rating of bulb , V = 40 V
 
Current I drawn by bulb , I = 100 / 40 = 2.5 A
 
Resistance of bulb , R = V / I = 40 / 2.5 = 16 Ω
 
To get 2.5 A current in 50V , 50Hz ac power supply , required impedence Z is given as
 
Z = Vm / Im = 50 / 2.5 = 20 Ω
 
When an inductance L is in series with bulb of resistance R , impedence Z is given as
 
begin mathsize 14px style z space equals space square root of R squared plus L squared omega squared end root space equals space 20 space capital omega end style  ............................. (1)
where ω is angular frequency that is given as ω = 2π f , where f = 50 Hz is frequency of alternating voltage 
 
if we substitute R = 16 Ω in above eqn.(1) , then we get Lω = 12 Ω 
 
Hence , inductance L = 12 / ω = 12 / ( 2π f )  = 12 /[( 2π (50) ] =  ( 0.12 / π )  H
 
 
 

Answered by Thiyagarajan K | 22nd Jul, 2021, 07:58: PM

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