Boiling water with resistors.

Let the applied voltage be $V$ and the heat required to boil the water be $H$ . So when $R_1$ is used:

$H = I_1^2R_1 t_1$

And when $R_2$ is used:

$H = I_2^2R_2 t_2$

Now, since the source voltage is the same:

$V = I_1R_1=I_2R_2$

Using this result in the Heat expressions:

$H = \frac{V^2t_1}{R_1} = \frac{V^2t_2}{R_2}$ $R_1 = \frac{V^2t_1}{H}$ $R_2 = \frac{V^2t_2}{H}$

Now, if $R_1$ and $R_2$ are connected in parallel, their equivalent resistance is:

$R = \left(\frac{1}{R_1} + \frac{1}{R_2}\right) ^{-1}$

Plugging in expressions for $R_1$ and $R_2$ : $R = \frac{V^2t_1t_2}{H(t_1 + t_2)}$

Rearranging gives:

$H = \frac{V^2}{R} \left(\frac{t_1t_2}{t_1+t_2}\right)$

So, the heat required is the power supplied by the source times the time taken to heat the water. The required time is therefore:

$t =\frac{t_1t_2}{t_1+t_2} = \boxed{6 \ \mathrm{min}}$