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Given the resistances of 1Omega , 2Omega...

Given the resistances of `1Omega , 2Omega , 3Omega ` how will we combine them to get an equivalent resistance of (i) `(11/3) Omega, (ii) (11/5) Omega, (iii) 6 Omega , (iv) (6/11) Omega` ?

Text Solution

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(i) By connecting parallel combination of `2Omega` and `1 Omega` in series with 3 `Omega ` we have
Total resistance ` = 3 + (1)/(1/1 + 1/2) = 11/3 Omega`
(ii) By connecting parallel combination of 2 `Omega ` and 3 `Omega` in series with 1 `Omega` , we have
Total resistance = `1 + (1)/( (1/2 + 1/3)) = 11/5 Omega`
(iii) By connecting all the three resistances in series
Total resistance `= 1 + 2 + 3 = 6.Omega`
(iv) By connecting all resistances in parallel
Total resistance ` = (1)/(1/2 + 1/2 + 1/3) = 6/11 Omega`
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