An electric kettle has two heating coils. Whe | vedclass

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Question

(A) Let $V$ be the voltage of the supply and $R_1, R_2$ be the resistances of the two coils.The heat required to boil the tea is $H = \frac{V^2}{R_1}

Vedclass · Accepted Answer

$14$ minutes

Vedclass · Answer

(A) Let $V$ be the voltage of the supply and $R_1, R_2$ be the resistances of the two coils.The heat required to boil the tea is $H = \frac{V^2}{R_1} 	imes T_1 = \frac{V^2}{R_2} 	imes T_2$.Given $T_1 = 6$ minutes and $T_2 = 8$ minutes, we have $R_1 = \frac{V^2 T_1}{H}$ and $R_2 = \frac{V^2 T_2}{H}$.When connected in series, the total resistance is $R_{eq} = R_1 + R_2$.The time taken $T$ to boil the tea in series is $H = \frac{V^2}{R_{eq}} 	imes T$.Substituting $R_{eq}$, we get $H = \frac{V^2}{R_1 + R_2} 	imes T$.$T = H 	imes \frac{R_1 + R_2}{V^2} = H 	imes \left( \frac{V^2 T_1 / H + V^2 T_2 / H}{V^2} ight) = T_1 + T_2$.Therefore, $T = 6 + 8 = 14$ minutes.

An electric kettle has two heating coils. When the first coil is switched on, the tea boils in $6$ minutes. When the second coil is switched on, it boils in $8$ minutes. If both coils are connected in series and switched on, how long will it take for the tea to boil?

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