Answer

Verified

54.6k+ views

**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.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main