An electric kettle has two coils. When one of these is switched on, the water in the kettle boils in $6\,\min$ . When the other coil is switched on, the water boils in $3\,\min$. If the two coils are connected in series, the time taken to boil the water in the kettle is ............. $min$
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In series $\frac{1}{{{P_S}}} = \frac{1}{{{P_1}}} + \frac{1}{{{P_2}}}$
$ \Rightarrow \frac{1}{{({H_s}/{t_s})}} = \frac{1}{{({H_1}/{t_1})}} + \frac{1}{{({H_2}/{t_2})}}$
$\because$ ${H_s} = {H_1} = {H_2}$ So ${t_s} = {t_1} + {t_2} = 6 + 3 = 9\,\min$
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