Two electric bulbs are connected, one at a time, across a potential difference V, and the powers consumed in them are P₁ and P₂ respectively. Now, if the potential difference V is applied across the series combination of these bulbs, what will be the total power consumed?

Question

Accepted Answer

(3) (P₁P₂)/(P₁+P₂). Idea: each bulb's resistance is fixed, so find the resistances from the powers on V and then use the series resistance. Alone on V: P₁=(V²)/(R₁) and P₂=(V²)/(R₂), so R₁=(V²)/(P₁) and R₂=(V²)/(P₂). In series: R=R₁+R₂=V²(1/(P₁)+1/(P₂)). Total power: P=(V²)/R, so 1/P=1/(P₁)+1/(P₂). P=(P₁P₂)/(P₁+P₂), which is smaller than either P₁ or P₂.

Two electric bulbs are connected, one at a time, across a potential difference V, and the powers consumed in them are P₁ and P₂ respectively. Now, if the potential difference V is applied across the series combination of these bulbs, what will be the total power consumed?

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