Question

A heater boils a certain amount of water in $15minutes$. Another heater boils the same amount of water in $10minutes$. The time taken to boil the same amount of water when both are used in parallel is?

Answer 1

Step 1: Given

Time taken by the first heater: $t_1=15mins$

Time taken by the second heater: $t_2=10mins$

Resistance of the first heater is $R_1$

Resistance of the second heater is $R_2$

Step 2: Formula Used

Joule’s law of heating,$$H=I^{2}Rt$$
Here,$H$ is the heat output,

$R$ is the resistance,

$I$ is current and $t$ is the time.

$$V=IR$$

$V$ is the voltage

Two resistances $R_1$ and $R_2$ are connected in parallel,

The equivalent resistance is,

$$R_{12}={\displaystyle \frac{R_1{R}_2}{R_1+R_2}}$$

Step 3: Calculating the time

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 current, and$$t_1$$is the time required for the first heater.
According to Ohm’s law,$$V=IR$$
$$\Rightarrow I={\displaystyle \frac{V}{R}}$$
Therefore, we can express the heat supplied by the first heater as,
$$H_1={\left({\displaystyle \frac{V}{R_1}}\right)}^2{R}_1{t}_1$$
$$\Rightarrow H_1={\displaystyle \frac{V^2{t}_1}{R_1}}$$…… (1)
Also, we can express the heat supplied by the second heater as follows,
$$H_2={\displaystyle \frac{V^2{t}_2}{R_2}}$$…… (2)
Here, $$R_2$$is the resistance of the second heater and $$t_2$$is the time required for the second heater.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.
Since the head required to boil the water is same therefore, $$H_1=H_2$$, equating equations (1) and (2), we get,
$${\displaystyle \frac{V^2{t}_1}{R_1}}={\displaystyle \frac{V^2{t}_2}{R_2}}$$
$$\Rightarrow {\displaystyle \frac{t_1}{t_2}}={\displaystyle \frac{R_1}{R_2}}$$
Substituting $15minutes$ for$$t_1$$and $10minutes$ for$$t_2$$in the above equation, we get,
$${\displaystyle \frac{15}{10}}={\displaystyle \frac{R_1}{R_2}}$$
$$\Rightarrow {\displaystyle \frac{R_1}{R_2}}={\displaystyle \frac{3}{2}}$$…… (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,
$$R_{12}={\displaystyle \frac{R_1{R}_2}{R_1+R_2}}$$…… (4)
Let’s express the heat supplied by the combination of heaters as follows,
$$H_{12}={\displaystyle \frac{V^2{t}_{12}}{R_{12}}}$$
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,
$$H_{12}={\displaystyle \frac{V^2{t}_{12}}{{\displaystyle \frac{R_1{R}_2}{R_1+R_2}}}}$$
$$\Rightarrow H_{12}={\displaystyle \frac{\left(R_1+R_2\right)V^2{t}_{12}}{R_1{R}_2}}$$…… (5)
$$H_{12}=H_1$$
$${\displaystyle \frac{\left(R_1+R_2\right)V^2{t}_{12}}{R_1{R}_2}}={\displaystyle \frac{V^2{t}_1}{R_1}}$$
$$\Rightarrow {\displaystyle \frac{\left(R_1+R_2\right)t_{12}}{R_1{R}_2}}={\displaystyle \frac{t_1}{R_1}}$$
$$\Rightarrow {\displaystyle \frac{\left(R_1+R_2\right)t_{12}}{R_2}}=t_1$$
$$\Rightarrow \left({\displaystyle \frac{R_1}{R_2}}+1\right)t_{12}=t_1$$
Substituting $15minutes$ fort $$t_1$$and using equation (3) in the above equation, we get,
$$\left({\displaystyle \frac{3}{2}}+1\right)t_{12}=15$$
$$\Rightarrow \left({\displaystyle \frac{5}{2}}\right)t_{12}=15$$
$$\Rightarrow t_{12}={\displaystyle \frac{30}{5}}$$
$$\therefore t_{12}=6\text{minute}$$
Hence, the total time taken by the heaters when connected in parallel is $6mins$