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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