An electric kettle has two heating coils. Whe | vedclass

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

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$8$

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(C) Let $Q$ be the heat required to boil the water. The heat produced by a coil is given by $Q = \frac{V^2}{R} 	imes t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.For the first coil: $Q = \frac{V^2}{R_1} 	imes t_1$,so $\frac{1}{R_1} = \frac{Q}{V^2 t_1}$.For the second coil: $Q = \frac{V^2}{R_2} 	imes t_2$,so $\frac{1}{R_2} = \frac{Q}{V^2 t_2}$.When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.The heat equation for the parallel combination is $Q = \frac{V^2}{R_{eq}} 	imes t$,which gives $\frac{1}{R_{eq}} = \frac{Q}{V^2 t}$.Substituting these into the parallel resistance formula: $\frac{Q}{V^2 t} = \frac{Q}{V^2 t_1} + \frac{Q}{V^2 t_2}$.This simplifies to $\frac{1}{t} = \frac{1}{t_1} + \frac{1}{t_2}$.Substituting the given values $t_1 = 10$ and $t_2 = 40$: $\frac{1}{t} = \frac{1}{10} + \frac{1}{40} = \frac{4+1}{40} = \frac{5}{40} = \frac{1}{8}$.Therefore,$t = 8 \; min$.

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