An electric kettle heater has two coils. When one coil is switched ON, the water in the kettle begins to boil after 15 minutes, If the other coil is switched on, the water boils in 30 minutes. Which of the following statement is correct.

Water boils in 60 minutes if the coils are connected in series

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Water boils in 10 minutes if the coils are connected in parallel

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Water boils in 22.5 minutes if the coils are connected in series

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Water boils in 20 minutes if the coils are connected in parallel

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Solution

The correct option is B Water boils in 10 minutes if the coils are connected in parallel
Let $R_{1}$ and $R_{2}$ are the resistances of the first and the second coils and V be mains voltage.
As, $Q = \frac{V^{2} t_{1}}{R_{1}} = \frac{V^{2} t_{2}}{R_{2}} \Rightarrow \frac{R_{1}}{R_{2}} = \frac{t_{1}}{t_{2}} = \frac{15}{30} = \frac{1}{2}$
or $2 R_{1} = R_{2}$
In series, $R_{e q} = R_{1} + R_{2}$
$\Rightarrow Q = \frac{V^{2} t_{3}}{R_{e q}} = \frac{V^{2} t_{1}}{R_{1}} \Rightarrow t_{3} = 3 t_{1} = 45 m i n u t e s$
In parallel, $Q = \frac{V^{2} t_{4} (R_{1} + R_{2})}{R_{1} R_{2}} = \frac{V^{2} t_{1}}{R_{1}}$
$\Rightarrow t_{4} = \frac{R_{2}}{R_{1} + R_{2}} t_{1} = 10 m i n u t e s$
Alter : $R = \frac{V^{2}}{P} t$
Hence : $\frac{R_{1}}{t_{1}} = \frac{R_{2}}{t_{2}} \frac{R_{1} + R_{2}}{t_{3}} = \frac{R_{1} R_{2}}{(R_{1} + R_{2}) t_{4}} = k (s a y)$
$R_{1} = 15 k R_{2} = 30 k$
$R_{1} + R_{2} = k \times t_{3}$
$45 k = k t_{3} \Rightarrow t_{3} = 45 m i n$
$\frac{R_{1} R_{2}}{(R_{1} + R_{2})} = k t_{4}$
$\frac{15 \times 30}{45} k = k t_{4} \Rightarrow t_{4} = 10 m i n$

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