An electric kettle has two coils. When one coil is connected to the AC mains the water in the kettle boils in 10 minutes. When the other coil is used the same quantity of water takes 15 minutes to boil. How long will it take for the same quantity of water to boil if the two coils are connected in parallel?

6 min

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12 min

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18 min

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24 min

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Solution

The correct option is A 6 min
Let H be the amount of the heat energy needed to boil the given quantity of water. If $R_{1}$ and $R_{2}$ are the resistances of coils and V is the applied voltage then
$H = \frac{V^{2} t_{1}}{R_{1}} = \frac{V^{2} t_{2}}{R_{2}}$
$\frac{R_{2}}{R_{1}} = \frac{t_{2}}{t_{1}} = \frac{15}{10} = \frac{3}{2}$
when the coils are connected in parallel the combined resistance is given by
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{R_{1} + R_{2}}{R_{1} R_{2}}$
$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
If the water takes t minutes to boil then
$\frac{v^{2} t}{R} = \frac{V^{2} t_{1}}{R_{1}}$
$t = t_{1} \times \frac{R}{R_{1}} = t_{1} \times \frac{R_{1} R_{2}}{R_{1} + R_{2}} \times \frac{1}{R_{1}} = t_{1} \times \frac{1}{(\frac{R_{1}}{R_{2}} + 1)} = \frac{10 \times 1}{(\frac{2}{3} + 1)} = 6$ min

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