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Question
Two identical resistros of `12 Omega` each are connected to a battery of 3V . Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.
Solution
`R_1 = 12 Omega`
`R_2 = 12 Omega`
`V = 3 V`
Minimum resistance ⇒ Parallel combination.
`1/R = 1/R_1 + 1/R_2`
= `12.12/(12 + 12) = 6 Omega`
`6 Omega` = minimum resistance
`therefore P = V^2/R = (3)^2/((6)) = 9/6 = 3/2 = 1.5 "watt"`
for maximum resistance
`R = R_1 + R_2 = 12 + 12 = 24 Omega`
`therefore P_(min) = V^2/R = (3)^2/24 = 9/24 = 3/8 "watt"`
`P_(min)/P_(max) = 3/8 = 1/4`
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