Given the resistances of 1Omega , 2Omega , 3Omegahow 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?

Given the resistances of 1Omega , 2Omega , 3Omegahow will we combine t

1.	Given the resistances of 1Omega , 2Omega , 3Omegahow 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?
Answer» Solution : (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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Given the resistances of 1Omega , 2Omega , 3Omegahow 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?

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