An electric kettle has two heating coils. Whe | vedclass

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(A) Let $H$ be the heat required to boil the water. The heat produced by a coil is given by $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the

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$3$ $	ext{min}$ $20$ $	ext{sec}$

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(A) Let $H$ be the heat required to boil the water. The heat produced by a coil is given by $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.Since $H$ and $V$ are constant,$t \propto R$,or $R \propto t$.Let $R_1$ and $R_2$ be the resistances of the two coils. Then $R_1 \propto 5$ and $R_2 \propto 10$.When connected in parallel,the equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$.Since $t_p \propto R_p$,we have $\frac{1}{t_p} = \frac{1}{t_1} + \frac{1}{t_2}$.Substituting the given values: $\frac{1}{t_p} = \frac{1}{5} + \frac{1}{10} = \frac{2+1}{10} = \frac{3}{10}$.Therefore,$t_p = \frac{10}{3} 	ext{ min} = 3.33 	ext{ min}$.$0.33 	ext{ min} = 0.33 	imes 60 	ext{ sec} = 20 	ext{ sec}$.Thus,the time taken is $3 	ext{ min } 20 	ext{ sec}$.

An electric kettle has two heating coils. When one coil is used,water in the kettle boils in $5$ minutes,while when the second coil is used,the same water boils in $10$ minutes. If the two coils,connected in parallel,are used simultaneously,the same water will boil in time:

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