Sol:

From Joule’s Law of heating:

\( \Rightarrow Q = \frac{E^{2}}{R_{1}} * 6 —-[1] \) \( \Rightarrow Q = \frac{E^{2}}{R_{1}} * 3 —-[2] \)

As the amount of heat supplied and applied voltage is the same:

\( \Rightarrow \frac{3}{R_{2}} = \frac{6}{R_{1}} \) \( \Rightarrow R_{1} = R_{2} \)

When the coils are attached in series:

\( \Rightarrow R_{1} + R_{2} \) \( \Rightarrow Q = \frac{E^{2}}{(R_{1}+R_{2})} * t \) \( \Rightarrow Q = \frac{E^{2}}{3R_{2}} * t —-[3] \)

As the amount of heat supplied and applied voltage is the same. Now by comparing both the equations [2] and [3], we get

\( \Rightarrow 3 = \frac{t}{3} \) \( \Rightarrow t = 9 min \)

Therefore, the time taken to boil the water in the kettle is t = 9 min.

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