We know that electric heater contain coils which have some resistances. So let R_{1} and R_{2} be the resistances of the two heaters respectively. The amount of heat required is the same in all the three cases given in the problem as it is required to heat the same amount of water i.e 1 kg. Let t be the time required to heat the water with the combination of heaters connected in parallel.

We know from **Joule’s Law of Heating**

**Heat (H) = V**^{2} x time(t) / R

So,

H1 = V^{2} x t_{1}/ R_{1}

H2 = V^{2} x t_{2}/ R_{2}

H1 = H2

Equating H1 to H2 we get the relation R_{2} = t_{2}R_{1} / t_{1}.

Parallel combination of heaters: H_{3} = V^{2} x (R_{1} + R_{2}) x t / R_{1}R_{2}

Equating H_{3} to H_{1} and substituting for R_{2} in terms of R_{1}, we get

=> V^{2} x (R_{1} + R_{2}) x t / R_{1}R_{2} = V^{2} x t_{1}/ R_{1}

=> (R_{1} + R_{2}) x t / R_{2} = t_{1}

=> (R_{1} + t_{2}R_{1}/t_{1}) x t / (t_{2}R1 / t_{1}) = t_{1}

=> R_{1}(1 + t_{2}/t_{1}) x t t_{1}/ (t_{2}R1) = t_{1}

Solving we get :

**t = t**_{1} t_{2 }/ (t_{1} + t_{2})

**So, If both the heaters are connected in parallel, the combination will boil the water in time**

**t = t**_{1} t_{2 }/ (t_{1} + t_{2})