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Initially two bulbs are connected in series with battery because switch shorted the third bulb.

If R is the equal resistance of each bulb , then current I₁ drawn from battery is

I₁ = V / (2R)

Power delivered by battery to two bulbs, P = I₁² (2R) ..................(1)

When switch is open , all three bulbs are in series and connected to battery.

Hence current I₂ drawn from battery is

I₂ = V/(3R) = (2/3) V(2R) = (2/3) I₁

Power delivered by battery to two bulbs, P₂ = I₂² (3R)  = (2/3)² I₁² (3R)

P₂ = (3/2) (2/3)² I₁² (2R) = (2/3) I₁² (2R) = (2/3) P

As the switch S is opened , then power output becomes (2/3)P