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Two electric bulbs have their resistances in the ratio 2:3 . They are connected (a) first in series and then (b) in parallel across the same voltage . Find the ratio of powers consumed by each of the two bubls in the two combinations. |
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Answer» SOLUTION :`(R_1)/(R_2) = 2/3` a) SERIES combination : power `R= i^2 R` `therefore (P_1)/(P_2) = (i^2 R_1)/(i^2 R_2)` `because ` CURRENT is same in series combination `therefore (P_1)/(P_2) = (R_1)/(R_2) = 2/3` `therefore P_1 : P_2= 2:3` B) Parallel combination `(P_1)/(P_2) = (i_1^2 R_1)/(i_2^2 R_2) = ((V^2)/(R_1^2)R_1)/(((V^2)/(R_2^2)) R_2)"" [ because V= iR]` `=(1//R_1)/(1/R_2) = (R_2)/(R_1) = 3/2 ` `therefore P_1 : P_2 = 3:2` |
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