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An electric kettle has two heating coils. When one of the coils is connected to a.c. source, the water in the kettle boils in 10 minute. When the other coil is used the water boils in 40 minute. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be:
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The correct answer is:
8 minute
Here $Q=\frac{V^2}{R_1} \times t_1=\frac{V^2}{R_2} \times t_2$
$$
\begin{aligned}
& =\frac{V^2}{R} \times t \\
& \therefore \quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \quad \frac{Q}{V^2 t}=\frac{Q}{V^2 t_1}+\frac{Q}{V^2 t_2} \\
& \Rightarrow \quad \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_4} \\
& \Rightarrow \quad t=\frac{t_1 t_2}{t_1+t_2} \\
& =\frac{10 \times 40}{10+40}=8 \mathrm{~min} \\
&
\end{aligned}
$$
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