Question

Calculate the amount of heat generated while transforming 90000 coulombs of charge between the two terminals of a battery of 40V in one hour. Also determine the power expended in the process.

Answer 1

Hint: We are given the information on the physical quantities involved in an electric circuit such as the number of charges transformed, the potential applied and the time taken for all these processes. We can easily find the power and heat energy from these relations.

Complete step by step answer:
We know that the current flow in an electric circuit is caused by a driving force due to the applied potential. The heat energy is due to the Joule’s heating due to the resistance offered by the circuit and power is the energy used up in unit time.
Let us consider our present situation. A potential of 40V has caused 90000C to be moved through the circuit in an hour. We can find the current in the circuit as the charge flowing per unit time as –
\[\begin{align}
  & \text{Current, }I=\dfrac{\text{Charge}}{\text{Time taken}} \\
 & \Rightarrow \text{ }I=\dfrac{Q}{t} \\
\end{align}\]
In our situation the charge flowing is 90000C and the time taken is 1hour.
i.e., the current through the circuit is given by –
\[\begin{align}
  & I=\dfrac{Q}{t} \\
 & \Rightarrow \text{ }I=\dfrac{90000}{1\times 60\times 60} \\
 & \therefore \text{ }I=25A \\
\end{align}\]
Now, we can find the heat generated in the circuit by using the Joule’s heating law. According to which –
\[H={{I}^{2}}Rt\]
But we know from Ohm’s law that –
\[V=IR\]
From this we get the heat generated in the circuit over an hour time as –
\[\begin{align}
  & H={{I}^{2}}Rt \\
 & V=IR \\
 & H=VIt \\
 & \Rightarrow H=40V\times 25A\times 3600s \\
 & \therefore H=3.6\times {{10}^{6}}J \\
\end{align}\]
The amount of heat generated by the circuit in one hour is 3600kJ.
Now, we can find the power consumed as the rate of change of energy as –
\[\begin{align}
  & P=\dfrac{H}{t} \\
 & \Rightarrow P=\dfrac{VIt}{t} \\
 & \Rightarrow P=VI \\
 & \Rightarrow P=40V\times 25A \\
 & \therefore P=1000W \\
\end{align}\]
The power dissipated by the circuit is 1kW.
Thus, the heat energy generated is 3600kJ and the power dissipated is 1kW by the circuit element.

Note:
We should understand that the heat energy corresponds to the energy consumption of an electric circuit element. It is therefore, we can find the power dissipated directly from the Joule’s heating law as per the method we have done the exercise.