Two electric bulbs marked 25W-220V and 100W . . .

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


Doubt by Sonali

Solution : 

Method I : Long Method

Generally the bulb got fused when its filament gets melted due to too much heating and too much heating happened when the current exceed its maximum value.

Here
For First Bulb
P1 = 25W
V1 = 220V

We know, 
P1=V12/R1
R1=V12/P1
R1 = (220×220)/25
R1 = 44×44
R1=1936 Ω

Also, 
P=VI
I = P/V

I1 = P1/V
I1 = 25/220
I1 = 0.11 A


For Second Bulb
P2 = 100W
V2 = 220 V
Similarly 
R2=(220×220)/100
R2 = 484 Ω

Similarly 

I2 = P2/V
I2 = 100/220
I2 =0.45 A

Both the bulb are connected in series. So net resistance of the circuit will be 
R = R1+R2
= 1936 Ω + 484 Ω
= 2420 Ω

V = 440 V

Current drawn from the battery 
I =V/R
I = 440/2420
I = 0.18 A

∵ I1 = 0.11 A & I2 = 0.18 A

But 0.18 A of current will pass through both of them. Clearly, first bulb of 25W will fuse.

Method II : Short Method (Recommended)

We know, 
P=V2/R
R = V2/P
When Applied Voltage (V) across both the bulbs will be same then 
R∝1/P — (1)

When both the bulbs are connected in series then current through both of them will remain same. 
So, 
H = I2Rt
For constant I and t
H∝R — (2)

Using (1) and (2)
H∝1/P

i.e. more heat will be produced in a bulb having low power. 

Hence, 25 W will be produce more heat and will be fused.
Similar Question : Two 120 V light bulbs, one of 25 W and other of 200 W were connected in series across a 240 V line. One bulb burn out almost instantly. Which one was burnt and why?
Short Answer : The bulb of 25 W will burns out.