An electric kettle has two independent heating coils. When connected individually to a 220 V supply, the first coil boils a certain mass of water in 10 minutes , while the second coil takes 40 minutes to boil the same

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 25 : 4

Let the heat required to boil the water be

H

and the supply voltage be

V

.

For the first coil:

H = \dfrac{V^2}{R_1} t_1 \Rightarrow R_1 = \dfrac{V^2}{H} t_1

For the second coil:

H = \dfrac{V^2}{R_2} t_2 \Rightarrow R_2 = \dfrac{V^2}{H} t_2

When connected in series, the equivalent resistance is

R_S = R_1 + R_2

.

Time taken in series

t_S = \dfrac{H}{V^2} R_S = \dfrac{H}{V^2} (R_1 + R_2) = t_1 + t_2

t_S = 10 + 40 = 50\text{ minutes}

When connected in parallel, the equivalent resistance is

R_P = \dfrac{R_1 R_2}{R_1 + R_2}

.

Time taken in parallel

t_P = \dfrac{H}{V^2} R_P = \dfrac{H}{V^2} \left( \dfrac{R_1 R_2}{R_1 + R_2} \right) = \dfrac{t_1 t_2}{t_1 + t_2}

t_P = \dfrac{10 \times 40}{10 + 40} = \dfrac{400}{50} = 8\text{ minutes}

The ratio

t_S : t_P = 50 : 8 = 25 : 4

.

Answer:

25 : 4

An electric kettle has two independent heating coils. When...

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