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

Given the resistances of `1 Omega, 2 Omega, 3 Omega` how will be 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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It is to be noted that (a) the effective resistance of parallel combination of resistors is less than the individual resistance and (b) the effective resistance of series combination of resistors is more than individual resistance.
Case (i) Parallel combination of `1 Omega` and `2 Omega` is connected is series with `3 Omega`
Effective resistance of `1 Omega` and `2 Omega` in parallel will be given by
`RP = (1 xx 2)/(1 + 2) = (2)/(3) Omega`
`:.` Equivalent resistance of `(2)/(3) Omega` and `3 Omega` in series
`= (2)/(3) + 3 = (11)/(3) Omega` ,
Case (ii) : Parallel combination of `2 Omega` and `3 Omega` is connected in series with `1 Omega` Equivalent resistance of `2 Omega` and `3 Omega` in parallel
`= (2 xx 3)/(2 + 3) = (6)/(5) Omega`
Equivalent resistance of `(6)/(5) Omega` and `1 Omega` in series `= (6)/(5) + 1 = (11)/(5) Omega`
Case (iii) : All the resistance are to be connected in series show
`:.` Equivalent resistance `= 1 + 2 + 3 = 6 Omega`
(Case (iv) : All the resistance are to be connected in parallel
`:.` Equivalent resistance (R ) is given by
`(1)/(R ) = (1)/(1) + (1)/(2) + (1)/(3)`
`= (6 + 3 + 2)/(6) = (11)/(6)` or `r = (6)/(11) Omega`
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