A and B are two electric bulbs with their ratings respectively 40 W, 110 V and 100 W and 110 V. Find their respective filament resistances. If the bulbs are connected in series with a supply of 220 V, which bulb will fuse ?

Explanation

For A bulb : `P_(1) = 40 Omega, V = 110` V
Suppose resistance of bulb A is `R_(1)` .
` therefore P_(1) = (V^(2))/(R_(1))`
`therefore R_(1) = (V^(2))/(P_(1))`
`therefore R_(1) = (110 xx 110)/(40) = 302.5 Omega`
So current flowing through it,
`I_(1) = (V)/(R) = (110)/(302.5) = 0.3636 ` A
For B bulb = `p_(1) = 1000 Omega,` V = 110 V
Suppose resistance of bulb B is `R_(2)` ,
`therefore P_(2) = (V^(2))/(R_(2))`
` therefore R_(2) = (V^(2))/(P_(2))`
` therefore R_(2) = (110 xx 110)/( 100 ) = 121 Omega`
`I_(2) = (V)/(R_(2)) = (110)/(121) = 0.9091 ` A
Total resistance of circuit when both bulbs are 1V joined in series.
`R = R_(1) + R_(2) = 302.5 + 121 = 423.5 Omega`

Current flowing through both bulbs
I = `(V)/(R)`
`= (220)/(423.5) = 0.52` A
here , I` gt I_(1)` , So bulb A will be fused.
I `lt I_(2) ` , so bulb B will not be damaged.