An electric kettle has two coils. When one coil is switched an it takes `5` minutes to boil water and when second coil is switched on it takes `10` minutes .How long will it take to boil water, when both the coil are used in series ?

Answer

`15 min`

Explanation

Let `R_(1) ,R_(2)` be the resistance of the two coils , `V` be the supply voltage and `H` be the heat required to boil the water
For the first coil `H = (V^(2)t_(1))/(JR_(1)) = (V^(2) xx (5 xx 60))/(4.2 xx R_(1)) cel`
For the second coil
`H = (V^(2)t_(2))/(JR_(2))= (V^(2)xx 10 xx 60)/(4.2 xx R_(2)) cal`
`:. (V^(2) xx 5 xx 60)/(4.2 R_(1)) = (V^(2) xx 10 xx 60 )/(4.2 xx R_(2) or (R_(2))/(R_(1)) = (10)/(5) = 2`
when the coil are connected in series the effective resistance `R = R_(1) + R_(2)`
Let `r` be the time in min boiling water , then
`H = (V^(2)t)/(J(R_(1) + R_(2))) = (V^(2) xx t_(1))/(JR_(1))`
`or t = (t_(1)(R_(1) + R_(2)))/(R_(1)) = t_(1) ( 1 + (R_(2))/(R_(1))) = 5 ( 1+ 2)` = 15 min