Show how you would connect three resistors, each of resistance `6Omega` so that combination has a resistance of
(i) `9Omega` (ii) `4Omega`

Explanation

Parallel combination of `6 Omega` and ` 6 Omega`
`(1)/(R_(p)) = (1)/(6) + (1)/(6) = (1+1)/(6) = (2)/(6) = (1)/(3)`
Resultant `3Omega + 6 Omega= 9Omega`
(ii)
R = `9 Omega`
`₹ = 12 Omega `
Resultant `implies (1)/(12) +(1)/(6) = (1+2)/(12) =(3)/(12) =(1)/(4)`
`R= 4 Omega `