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InterviewSolution
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Two elelctric bulbs P and Q have their resistance in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. |
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Answer» Let P=x and Q= 2x Given `P : Q = x : 2x - 1:2` For bulb P power loss `P_(1) = (V_(2))/(R_(1)) = (V_(2))/(P)` `P_(1) = (V^(2))/(x)` Similarly `P_(2) = (V^(2))/(Q) = (V^(2))/(2x)` Ratio of power loss `(P_(1))/(P_(2)) = (cancel(V)^(cancel(2))/(cancel(H)))/((cancel(V)^(2))/(2cancel(H))` `P_(1) : P_(2) = 2:1` |
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