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# An electric kettle has two coils when one of these is switched on the water in the kettle boils in $6\text{ }\min$. When the other coil is switched on, the water boils in $3\text{ }min$. If the two coils are connected in series, the time taken to boil the water in the kettle is:

Last updated date: 20th Jun 2024
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Hint: Using the Joule’s law of Heating, the two coils that are heating the water in series with the voltage passing through the coils, the heating energy required to heat the water is directly proportional to voltage square and time required to heat the water with the two coils separately and two coils combined. The formula for the heat energy separate and combined is:
$H=\dfrac{{{V}^{2}}}{R}t$
where $H$ is the heat energy, $V$ is the voltage required, $R$ is the resistance required, $t$ is the time required to heat the water.

Complete step by step solution:
The heating energy required to heat the water with the second coil is given as:
$H=\dfrac{{{V}^{2}}}{{{R}_{1}}}t$
With the resistance of the coil is given as ${{R}_{1}}$
The voltage that is required to heat the coil is given as $V$
The time required to heat the second coil is $3\text{ }min$
The heat energy required to heat the second coil is:
${{H}_{1}}=\dfrac{{{V}^{2}}}{{{R}_{1}}}3$
The heating energy required to heat the water with the second coil is given as:
${{H}_{1}}=\dfrac{{{V}^{2}}}{{{R}_{1}}}t$ … (1)
With the resistance of the coil is given as ${{R}_{2}}$
The voltage that is required to heat the coil is given as $V$
The time required to heat the second coil is $6\text{ }min$
The heat energy required to heat the second coil is:
${{H}_{2}}=\dfrac{{{V}^{2}}}{{{R}_{2}}}6$ … (2)
Now change the resistance with in terms of the first resistance is given by equating both the energy are:
$\dfrac{{{V}^{2}}}{{{R}_{1}}}3=\dfrac{{{V}^{2}}}{{{R}_{2}}}6$
Hence, the resistance of the second coil in terms of the first resistance is:
$\dfrac{{{V}^{2}}}{{{R}_{1}}}3=\dfrac{{{V}^{2}}}{{{R}_{2}}}6$
$\dfrac{3}{{{R}_{1}}}=\dfrac{6}{{{R}_{2}}}$
${{R}_{2}}=2{{R}_{1}}$
Hence, the heat energy required to heat the water with both the coils are given as:
${{H}_{net}}=\dfrac{{{V}^{2}}}{{{R}_{1}}+{{R}_{2}}}t$
Replacing the value of ${{R}_{2}}=2{{R}_{1}}$
${{H}_{net}}=\dfrac{{{V}^{2}}}{{{R}_{1}}+2{{R}_{1}}}t$
${{H}_{net}}=\dfrac{{{V}^{2}}}{3{{R}_{1}}}t$ …(3)
Now, the time required to heat the water when the coils in series with equation (1) and (3) as:
$\dfrac{{{V}^{2}}}{{{R}_{1}}}3=\dfrac{{{V}^{2}}}{3{{R}_{1}}}t$
$t=9\min$

Therefore, the time required to heat the water in series with the coil is $t=9\min$.

Note: The heat energy or power is the same in this case as both forms voltage that passes current in between the wire and produces heat energy. The heating energy is independent of the current flow meaning even if we don’t know the current flow, the heating energy can be found by voltage and resistance only.