Two electric bulbs marked 20 W- 220 V and 100 W - 220 V are connected in series to 440 V supply. Which bulb will be fused?

Explanation

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
`R=(V^(2))/(P)=(("Rated Voltage")^(2))/("Rated Power")`
For 20W-220V bulb,
`R_(1)=((220)^(2))/(20)Omega=2420Omega`
For 100W-220V bulb,
`R_(2)=((220)^(2))/(100)Omega=484Omega`
Both the bulbs are connected in series. So the current which passes through both the bulbs are same. The current that passes through the circuit, `I=(V)/(R_("tot"))`.
`R_("tot")=(R_(1)+R_(2))`
`R_("tot")=(484+2420)Omega=2904Omega`
`I=(440V)/(2904Omega)=0.151A`
The voltage drop across the 20 W bulb is
`V_(1)=IR_(1)=(440)/(2904)xx2420=366.6V`
The voltage drop across the 100 W bulb is
`V_(2)=IR_(2)=(440)/(2904)xx484=73.3V`
The 20 W bulb will be fused because its voltage rating is only 220 V and 366.6 V is dropped across it.