Question

Why is the power dissipated by a closed switch, such as in Figure 21.43, small?

Short answer

Due to the low resistance of a closed switch, the power dissipated is little.

Step-by-step solution

Definition of voltage

Voltage: It's the difference between two points’ electric potentials.

Given

Resistance of the switch when closed$R_{closed}=0\Omega$

Resistance of switch when open$R_{open}=\infty \Omega$

Emf of the battery is$E$

The internal resistance of the battery is$r$

External resistance connected in series with the battery is$R$

Current in the circuit when the switch is closed${\mathrm{l}}_{\mathrm{closed}}$

Current in the circuit when the switch is open${\mathrm{l}}_{\mathrm{open}}$

Finding why the power dissipated by the closed switch is small

The power dissipated is given :

$$\begin{gathered} \mathrm{P}={\left({\mathrm{l}}_{\mathrm{closed}}\right)}^2{\mathrm{R}}_{\mathrm{closed}} \\ \mathrm{P}={\left({\mathrm{l}}_{\mathrm{closed}}\right)}^2\left(0\mathrm{\Omega }\right) \\ \mathrm{P}=0\mathrm{W} \end{gathered}$$

Resistance is directly proportional to power. As a result, the lower the resistance, the lower the power dissipated.

Therefore, the resistance of a closed switch is so low, that the power dissipated is little.