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.