Electrical Analogy For Thermal Conduction MCQ - Practice Questions with Answers

Electrical Conduction

Thermal conduction

1. Natural flow of electric charge is from higher potential to lower potential

1. Heat flows from higher temperature to lower temperature

2. The rate of flow of charge is defined as
 electric current.

i.e., $I=\frac{d q}{d t}$

2. The rate of flow of heat may be called heat current.

i.e., $H=\frac{d Q}{d t}$

3. Similarly, the heat current may be
related to the temperature
difference as$H=\frac{\theta_1-\theta_2}{R}$

where R, is the thermal resistance of
the conductor

4. From the above point the electrical resistance is defined as

$$
\begin{aligned}
& R=\frac{\rho l}{A}=\frac{l}{\sigma A} \\
& \rho=\text { Resistivity }
\end{aligned}
$$

where, $\sigma=$ Electrical conductivity

$$
\frac{d q}{d t}=I=\frac{V_1-V_2}{R}=\frac{\sigma A}{l}\left(V_1-V_2\right)
$$
 

4. Similarly from the above point the thermal resistance may be
defined as $R=\frac{l}{K A}$
where $K=$ Thermal conductivity

$$
\frac{d Q}{d t}=H=\frac{\theta_1-\theta_2}{R}=\frac{K A}{l}\left(\theta_1-\theta_2\right)
$$