An electric kettle used for heating water has two heating elements. One raises the temperature of water by `Delta theta` in time `t_(1)` and another in time `t_(2)`. Assuming no loss of heat , find the time taken to raise the temperature by `Delta theta`.
(i) When heating elements are connected in series.
(ii) When heating elements are connected in parallel.

Explanation

Suppose 'H' is heat required to raise the temperature of water by `Delta theta` , then for first coil
`H = (V^(2))/(R_(1)). t_(1)implies R_(1) = (V^(2)t_(1))/(H)`
and for second coil
`H = (V^(2))/(R_(2)). t_(2)implies R_(2) = (V^(2)t_(2))/(H)`
(i) When both filaments are connected in series.
`H = (V^(2) .t)/( R_("total")) = (V)/((R_(1)+R_(2)))t`
`H = (V^(2).t)/((V^(2))/(H)t_(1)+(V^(2))/(H)t_(2))impliest=t_(1)+t_(2)`
(ii) When both the filaments are conneced in parallel.
`H = (V^(2))/( R_("total")).t`
`H- (V^(2))/(R_(1)R_(2)).t`
`=((V^(2).t)/(1))/((1)/(R_(1))+(1)/(R_(2)))`
`=((V^(2).t)/(1))/((H)/(V^(2)t_(1))+(H)/(V^(2)t_(2)))`
`H = V^(2) .t [(H)/(V^(2)t_(1) )+ (H)/(V^(2)t_(2))]`
`(1)/(t)= (1)/(t_(1))+(1)/(t_(2))`