Three resistors `3 Omega, 6 Omega` and `9 Omega` are connected a battery. In which of them will the power dissipation be maximum if:
(a) They all are connected in parallel
(b) They all are connected in series?
Given reasons.

Explanation

Given `R_(1) = 3 Omega, R_(2) = 6 Omega R_(3) = 9 Omega`
(a) Effective resistance in parallel is given by
`(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) = (1)/(3) + (1)/(6) + (1)/(9)`
`(1)/(R_(P)) = (6 + 3 + 2)/(18)`
`:. R_(P) = (18)/(11) Omega`

`:.` Dissipated power in parallel,
`P_(P) prop (1)/(R_(P)) implies P_(P) prop (1)/(((18)/(11))) :. P_(P) prop (11)/(18) to (1)`
(b) Effective resistance in series is given by
`R_(s) = R_(1) + R_(2) + R_(3) = 3 + 6 + 9 = 18 Omega`
`:.` Dissipated power in series,
`P_(S) prop R_(S) implies P_(S) prop 18 to (2)`
From equations (1) and (2) power dissipation is maximum in series and minimum in parallel.
Reasons :
(1) In series connection, `P prop R` and `V prop R`. Hence dissipated power (P) and potential differenc (V) is more because current is same across each resistor.
In parallel connection, `P prop (1)/(R )` and `I prop (1)/(R )`.
Hence dissipated power (P) and potential difference (V) is less because voltage is same acorss each resistor.