1.

Two identical resistors each of resistance 12 Omega are connected (i) in series (ii) in parallel, to a battery of 6 V. Calculate the ratio of power consumed in the combination of resistors in the two cases.

Answer»

Solution :`R_(1)=R_(2)=12Omega`
V=6V
`R_(s) = R_(1)+R_(2)`
`=(12+12)Omega= 24 Omega`
`(1)/(R_(p))=(1)/(12)+(1)/(12)`
or `"" (1)/(R_(p))=(2)/(12)`
`:. R_(p)=6Omega`
Power consumed by series combination
`=P_(s)`
`=(V^(2))/(R_(s))= (6xx6)/(24)`
=1.5 W
Power consumed by parallel combination
` =P_(p)=(V^(2))/(R_(p))=(6xx6)/(6)`
= 6 W
Ratio =` (P_(s))/(P_(p)) = (15)/(6) = (1)/(4) = 1:4`


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