Question

An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in $10$ minutes. When the other coil is used the water boils in $40$ minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

• A. 8 minutes
• B. 4 minutes
• C. 25 minutes
• D. 15 minutes

Answer 1

Answer: (A)

Let $R_1$ and $R_2$ be the resistance of the two coils and $V$ be the voltage supplied.
Effective resistance of two coils in parallel $=\frac{R_1{R}_2}{R_1+R_2}$
Let $H$ be the heat required to begin boiling in kettle.
Then $H=$ Power $\times$ time $=\frac{V^2{t}_1}{R_1}=\frac{V^2{t}_2}{R_2}$
For parallel combination, $H=\frac{V^2\left(R_1+R_2\right)t_p}{R_1{R}_2}$
$\Rightarrow \frac{1}{t_p}=\left(\frac{t_2+t_1}{t_2{t}_1}\right)$
$\therefore t_p=\frac{t_1{t}_2}{t_1+t_2}=\frac{10\times 40}{10+40}=8$ minute.