An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in $$10\,min.$$  When the other coil is used the water boils in $$40\,min.$$  If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

Solution :
Let $${R_1}$$ and $${R_2}$$ be the resistances of the coils, $$V$$ be the supply voltage, $$H$$ be the heat required to boil the water.
For first coil, $$H = \frac{{{V^2}{t_1}}}{{{R_1}}}\,......\left( {\text{i}} \right)$$
For second coil, $$H = \frac{{{V^2}{t_2}}}{{{R_2}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\frac{{{t_1}}}{{{R_1}}} = \frac{{{t_2}}}{{{R_2}}}$$
\[{\rm{i}}{\rm{.e}}{\rm{.}}\,\,\frac{{{R_2}}}{{{R_1}}} = \frac{{{t_2}}}{{{t_1}}} = \frac{{40}}{{10}} = 4\,\,\left[ {\begin{array}{*{20}{l}} {{t_1} = 10\,\min }\\ {{t_2} = 40\,\min } \end{array}} \right]\]
$$ \Rightarrow {R_2} = 4{R_1}\,......\left( {{\text{iii}}} \right)$$
When the two heating coils are in parallel, equivalent resistance is given by
$$\eqalign{ & R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{{R_1} \times 4{R_1}}}{{{R_1} + 4{R_1}}} = \frac{{4{R_1}}}{5} \cr & {\text{and}}\,\,H = \frac{{{V^2}t}}{R}\,.......\left( {{\text{iv}}} \right) \cr} $$
Comparing Eqs. (i) and (iv), we get
$$\eqalign{ & \frac{{{V^2}{t_1}}}{{{R_1}}} = \frac{{{V^2}t}}{R} \cr & \Rightarrow t = \frac{R}{{{R_1}}} \times {t_1}\,\,\left[ {{t_1} = 10\,\min } \right] \cr & \therefore t = \frac{4}{5} \times 10 = 8\,\min \cr} $$