Two bulbs rated 25 W - 220V and 100 W- 220 V are connected in series to a 440 V supply.Then,

Options

A. 100 W bulb fuses
B. 25 W bulb fuses
C. both the bulb fuses
D. neither of the bulb fuses.

Answer

B

Explanation

Resistance of a bulb `R = (V^(2))/(P)`
`therefore R_(1) = ((220)^(2))/(25) = (484 xx 100)/(25) = 1936 Omega`
`R_(2) = (220)^(2))/(100) = (484 xx 100)/(100) = 484 Omega`
In series `R = 1936 + 484 = 2420 W`
`therefore` Current `I = (V)/(R) = (440)/(2420) = (2)/(11)` amp
`therefore` Potential difference across 25 W bulb = `IR_(1)`
` = 2/11 xx 1936 = 352 V`
Potential difference across 100 W bulb = `IR_(2)`
` = 2/11 xx 484 = 88 V`
Since 25 W can tolerate only 220 V and this combination applies 352 V, the bulb will fuse.