Question

two electric bulbs, rated at 220V,25W and 220V,100W are connected in series across a 220V, voltage source. if the 25w and 100w bulbs draw powers p1 and p2 respectively, then

Answer 1

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 !