Question

An electric tea kettle has two heating coils. When one of the coils is switched on, boiling begins in 6 min. When the other coil is switched on, boiling begins in 8 min. In what time will the boiling begin if both coils are switched on simultaneously (a) in series and (b) in parallel.

Answer 1

(a) 14 min (b) $\frac{24}{7}$ min

Answer 2

Let $Q$ be the amount of heat required to boil the water. Let $t_1 = 6$ min and $t_2 = 8$ min be the times taken by the first and second coils, respectively. The powers of the coils are $P_1 = \frac{Q}{t_1} = \frac{Q}{6}$ and $P_2 = \frac{Q}{t_2} = \frac{Q}{8}$.

To simplify, let $Q = \text{lcm}(6, 8) = 24$ units of heat. Then, $P_1 = \frac{24}{6} = 4$ units of power, and $P_2 = \frac{24}{8} = 3$ units of power.

(a) Coils in series: The equivalent power for series combination is given by: $\frac{1}{P_{series}} = \frac{1}{P_1} + \frac{1}{P_2} = \frac{1}{4} + \frac{1}{3} = \frac{3+4}{12} = \frac{7}{12}$ So, $P_{series} = \frac{12}{7}$ units of power. The time taken $t_{series}$ is: $t_{series} = \frac{Q}{P_{series}} = \frac{24}{\frac{12}{7}} = 24 \times \frac{7}{12} = 2 \times 7 = 14 \text{ min}$

(b) Coils in parallel: The equivalent power for parallel combination is given by: $P_{parallel} = P_1 + P_2 = 4 + 3 = 7 \text{ units of power}$ The time taken $t_{parallel}$ is: $t_{parallel} = \frac{Q}{P_{parallel}} = \frac{24}{7} \text{ min}$