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Question
Three resistors of 6Ω, 3Ω and 2Ω are connected together so that their total resistance is greater than 6Ω but less than 8Ω Draw a diagram to show this arrangement and calculate its total resistance.
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Solution
We connect the resistance of 3 Ω and 2 Ω in parallel to each other and then connect this (parallel combination) to the 6 Ω resistance in series. The required arrangement is as shown in given figure. The equivalent resistance of 3Ω and 2 Ω in parallel.
`1/"R" = 1/3 + 1/2 = (2 + 3)/6 = 5/6`
or R = `6/5`Ω = 1.2 Ω
Thus, the total resistance of the set up is (1.2 + 6) Ω = 7.2 Ω
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