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# Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio a) 1 : 2 b) 1 : 1 c) 2 : 1 d) 1 : 4

Saurabh Koranglekar
one year ago
Vikas TU
14149 Points
one year ago
The equation for power in a resistance R, connected to a voltage V, is P = V^2/Rn. where Rn is the resistance of light bulb N.
Let R1 - resistance of Bulb 1 and R2 = resistance of Bulb 2. Since the bulbs are connected in parallel, then V1 = V2 = V. Given that R2 = 2٠R1, then P1 = V^2/R1, and P2 = V^2/(2٠R1). so the ratio of the power dissipated in the two bulbs is P1/P2 = (V^2/R1)/[V^2/(2٠R1)] = (2٠R1)/R1 = 2 (or 1/2, depending on which way you write the ratio).