two light bulbs are connected in parallel and then connected

two light bulbs are connected in parallel, and then connected to a battery as shown. you observe that bulb 1 is is twice as bright as bulb 2. assuming that the brightness of a bulb is proportional to the power dissipated in the bulb\'s resistor, which bulb\'s resistor is larger, and by what factor?

Solution

the brightness of a bulb is proportional to the power dissipated in the bulb\'s resistor.

that is,

        brightness = (constant)[ power dissipated in the bulb\'s resistor]

But, P = I^2*R

then,

      P2/P1 = R2/R1

bulb 1 is is twice as bright as bulb 2, that is, P1 = 2P2.

     P2/2P2 = R2/R1

        R2 = (R1)/2

Hence, R1 > R2, therefore, the resistance of the bulb 1 is larger than bulb 2.