Question

A heater boils a certain amount of water in 15 minute. Another heater boils the same amount of water in 10 minute. Time taken to boil the same amount of water when both are used in parallel is
A. 25 minute
B. 6 minute
C. 12 minute
D. 12.5 minute

Answer 1

Hint:Use Joule’s law of heating to express the heat supplied by the first heater and then heat supplied by the second heater. Using these two equations, determine the ratio of resistances of the two heaters in terms of time required to boil the water. Express the resistance of parallel combination of the heaters and then use Joule’s law to express the heat supplied by the combination of the two heaters.

Formula used:
Joule’s law of heating, \[H = {I^2}Rt\]
Here, H is the heat output, R is the resistance, I is the current and t is the time.

Complete step by step answer:
We have given that the amount of water is the same for both the heaters. Since the water boils using both the heaters separately, the heat supplied by the heater is the same but the time required to provide that amount of heat is different.

We can use Joule’s law of heating to express the heat supplied by the first heater as follows,
\[{H_1} = {I^2}{R_1}{t_1}\]
Here, \[{R_1}\] is the resistance of the first heater, I is the current, and \[{t_1}\] is the time required for the first heater.

According to Ohm’s law, \[V = IR\]
\[ \Rightarrow I = \dfrac{V}{R}\]
Therefore, we can express the heat supplied by the first heater as,
\[{H_1} = {\left( {\dfrac{V}{{{R_1}}}} \right)^2}{R_1}{t_1}\]
\[ \Rightarrow {H_1} = \dfrac{{{V^2}{t_1}}}{{{R_1}}}\] …… (1)

Also, we can express the heat supplied by the second heater as follows,
\[{H_2} = \dfrac{{{V^2}{t_2}}}{{{R_2}}}\] …… (2)
Here, \[{R_2}\] is the resistance of the second heater and \[{t_2}\] is the time required for the second heater.

Since \[{H_1} = {H_2}\], equating equation (1) and (2), we get,
\[\dfrac{{{V^2}{t_1}}}{{{R_1}}} = \dfrac{{{V^2}{t_2}}}{{{R_2}}}\]
\[ \Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{R_1}}}{{{R_2}}}\]

Substituting 15 minute for \[{t_1}\] and 10 minute for \[{t_2}\] in the above equation, we get,
\[\dfrac{{15}}{{10}} = \dfrac{{{R_1}}}{{{R_2}}}\]
\[ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{3}{2}\] …… (3)

Now, these two heaters are connected parallel to each other and used to boil the same amount of water. We can express the equivalent resistance of the two heaters as,
\[{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\] …… (4)

Let’s express the heat supplied by the combination of heaters as follows,
\[{H_{12}} = \dfrac{{{V^2}{t_{12}}}}{{{R_{12}}}}\]
Here, \[{t_{12}}\] is the time requires to boil the water using the combination of heaters.

Using equation (4) in the above equation, we get,
\[{H_{12}} = \dfrac{{{V^2}{t_{12}}}}{{\dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}}}\]
\[ \Rightarrow {H_{12}} = \dfrac{{\left( {{R_1} + {R_2}} \right){V^2}{t_{12}}}}{{{R_1}{R_2}}}\] …… (5)

Since \[{H_{12}} = {H_1}\], from equation (1) and equation (5), we can write,
\[\dfrac{{\left( {{R_1} + {R_2}} \right){V^2}{t_{12}}}}{{{R_1}{R_2}}} = \dfrac{{{V^2}{t_1}}}{{{R_1}}}\]
\[ \Rightarrow \dfrac{{\left( {{R_1} + {R_2}} \right){t_{12}}}}{{{R_1}{R_2}}} = \dfrac{{{t_1}}}{{{R_1}}}\]
\[ \Rightarrow \dfrac{{\left( {{R_1} + {R_2}} \right){t_{12}}}}{{{R_2}}} = {t_1}\]
\[ \Rightarrow \left( {\dfrac{{{R_1}}}{{{R_2}}} + 1} \right){t_{12}} = {t_1}\]

Substituting 15 minute for \[{t_1}\] and using equation (3) in the above equation, we get,
\[\left( {\dfrac{3}{2} + 1} \right){t_{12}} = 15\]
\[ \Rightarrow \left( {\dfrac{5}{2}} \right){t_{12}} = 15\]
\[ \Rightarrow {t_{12}} = \dfrac{{30}}{5}\]
\[ \therefore {t_{12}} = 6\,{\text{minute}}\]

So, the correct answer is option B.

Note: The voltage supply for the both the heaters will be the same because they are connected parallel to each other. The only factor that determines the power output is the resistance of the heater coils. Note that the resistance decreases in the parallel combination and increases in the series combination.