An electric tea kettle has two heating coils. When one of the coils is switched on , boiling begins in `6 min`. When the other coil is switched on , boiling begins in `8 min`. In what time will the boiling begin if both coils are switched on simultaneously (i) in series and (ii) in parallel.

Explanation

Let the power of first coil be `P_(1)` and that of the second coil is `P_(2)` . Let `H` be the amount of heat required to boil water. Then `H = P_(1)t_(1) = P_(2)t_(2)` , where `t_(1) = 6 min and t_(2) = 8 min`.
(i) When the coils are connected in series,
`P = (P_(1) P_(2))/(P_(1) + P_(2))`
`t = (H)/(P) = H[ (1)/(p_(1) + (1)/(P_(2)))] = H[ (t_(1))/(H) + (t_(2))/(H)]`
`= t_(1) + t_(2) = 6+8 = 14 min`
When the coils are connected in parallel,
`P = P_(1) + P_(2)`
`t = (H)/(P) = (H)/( P_(1) + P_(2)) = ( H)/((H)/(t_(1)) + (H)/(t_(2)))`
` = (t_(1) t_(2))/( t_(1) + t_(2)) = ( 6 xx 8)/( 6+8) = 3.43 min`