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

Options

A. 100 W
B. 25 W
C. neither
D. both

Answer

B

Explanation

As `R=(("Rated voltage")^(2))/("Rated power")`
`therefore` Resistance of 25 W-220 V bulb is
`R_(1)=((220)^(2))/(25)Omega`
Resistance of 100 W-220 V bulb is
`R_(2)=((220)^(2))/(100)Omega`
When these two bulbs are connected in series, the total resistance is
`R_(s)=R_(1)+R_(2)=(220)^(2)[(1)/(25)+(1)/(100)]=((220)^(2))/(20)Omega`
Current, `I=(440)/((220)^(2)//20)=(2)/(11)A`
Potential difference across 25 W bulb

`=IR_(1)=(2)/(11)xx((220)^(2))/(25)=352V`
Potential difference across 100 W bulb
`=IR_(2)=(2)/(11)xx((220)^(2))/(100)=88V`
Thus, the bulb 25 W will be fused, because it can tolerate only 220 V while the voltage across it is 352 V.