A coil takes 15,min to boil a certain amount | vedclass

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(C) Let the heat required to boil the water be $H$. The power of the first coil is $P_1 = \frac{H}{t_1}$ and the power of the second coil is $P_2 = \f

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(C) Let the heat required to boil the water be $H$. The power of the first coil is $P_1 = \frac{H}{t_1}$ and the power of the second coil is $P_2 = \frac{H}{t_2}$.Since $P = \frac{V^2}{R}$,we have $R_1 = \frac{V^2}{P_1} = \frac{V^2 t_1}{H}$ and $R_2 = \frac{V^2}{P_2} = \frac{V^2 t_2}{H}$.When connected in series,the total resistance is $R_S = R_1 + R_2$.The time taken $t_S$ to boil the same amount of water is given by $t_S = \frac{H}{P_S} = \frac{H R_S}{V^2}$.Substituting $R_S$,we get $t_S = \frac{H}{V^2} (R_1 + R_2) = \frac{H}{V^2} (\frac{V^2 t_1}{H} + \frac{V^2 t_2}{H}) = t_1 + t_2$.Given $t_1 = 15\,min$ and $t_2 = 20\,min$,the total time is $t_S = 15 + 20 = 35\,min$.

$A$ coil takes $15\,min$ to boil a certain amount of water,and another coil takes $20\,min$ for the same process. What is the time taken to boil the same amount of water when both coils are connected in series?

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