Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse?
Answers
Explanation:
(This problem requires understanding of the given basic formulae :-
V=IR
P=VI
R=V^2/P
Whenever there will be a series connection current I will be constant through out !
Now coming to the question :-
Res. Of 25 w bulb = (V^2/P)= (220^2/25)=1936 ohms. Similarly,
Res of 100 w bulb =(V^2/P)=(220^2/100)=484 ohms
Total resistance= 1936+484=2420 ohms
Given supply = 440 V
Total current in the series Ckt which remains constant through out = 440÷2420= 2/11 A
Voltage drop across 25 w bulb = (2/11)×1936=352V
Given rated voltage for bulb =220 .
Voltage drop is greater than the bulb rating . So it will fuse.
Now let us check Why 100 w bulb will not fuse
Voltage drop across 100w bulb=(2/11)×484=88V
Voltage drop is within the range specified that is less than 220. So it will not fuse !
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(This problem requires understanding of the given basic formulae :-
V=IR - THE OHMS LAW
P=VI
R=V^2/P
Whenever there will be a series connection current IT will be constant through out
Now coming to the question :-
ResISTANCE Of 25 w bulb = (V^2/P)= (220^2/25)=1936 ohms. Similarly,
Resistance of 100 w bulb =(V^2/P)=(220^2/100)=484 ohms
Total resistance of bulbs = 1936+484=2420 ohms
Given supply = 440 V
Total current in the series Ckt which remains constant through out = 440÷2420= 2/11 A
Voltage drop across 25 w bulb = (2/11)×1936=352V
Given rated voltage for bulb =220 .
Voltage drop is greater than the bulb rating . So it will fuse.
Now let us check Why 100 w bulb will not fuse
Voltage drop across 100w bulb=(2/11)×484=88V
Voltage drop is within the range specified that is less than 220. So it will not fuse my friend