Question

Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse?

A) 25 WB) BothC) 100 WD) Neither

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Answer 1

Answer: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 !

Answer 2

Answer: A) 25 W

Explanation:

Resistances of both the bulbs are

R1=V^2/P1 = 220^2/25

R2 = V^2/P2 = 220^2/100

Hence R1gt ; R2

When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V devides in such a way that voltage across 25 w bulb will be more than 220 V. Hence 25 w bulb will fuse.