Question

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 ?

Answer 1

Answer:

$302.5\ \Omega$, $121\ \Omega$

the bulb of resistance $121\ \Omega$ will be fused.

Explanation:

Given,

power rating of bulb 1, P1 = 40 W

power rating of bulb 2, P2 = 100 W

voltage rating of each bulb ,  = 110 v

hence, resistance of bulb 1 is given by

$R1\ =\ \dfrac{(V)^2}{P1}$

    $=\ \dfrac{110^2}{40}$

    $=\ 302.5\ \Omega$

  resistance of bulb 2 is given by  

$R2\ =\ \dfrac{(V)^2}{P2}$

     $=\ \dfrac{110^2}{100}$

      $=\ 121\ \Omega$

Since, both bulbs are to be connected in series through the voltage of 220 V. Since , the applied voltage is more than the rating voltage of the bulb so the bulb will get fused. So , the bulb of less resistance will fuse first and break the circuit so the remaining bulb will be saved. So, the bulb of resistance$\ 121\ \Omega$ will be fused.