There is a voltage across an open switch, such as in Figure 21.43. Why, then, is the power dissipated by the open switch small.

The term "circuit" refers to a set of electrical connections.

Circuits are closed-loop or route systems that consist of a network of electrical components through which electrons can travel.

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$\mathrm{R}$

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

Current in the circuit when the switch is openrole="math" localid="1655897485259" ${\mathrm{l}}_{\mathrm{open}}$

When the switch is open,

Equivalent resistance in series of the circuit given,

$R_{eq}=R+r+R_{open}$

Using Ohm's law, the current in the circuit is given as,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}} \\ {\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+{\mathrm{R}}_{\mathrm{open}}} \end{gathered}$$

$$\begin{gathered} {}_{\mathrm{open}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+\yen } \\ {\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{\yen } \\ {\mathrm{l}}_{\mathrm{open}}=0\mathrm{A} \end{gathered}$$

Therefore, when the circuit is open, there is no current in the circuit.

The power dissipated is given as,

$\mathrm{P}={\left({\mathrm{l}}_{\mathrm{open}}\right)}^2{\mathrm{R}}_{\mathrm{eq}}$

Substituting the values in the above expression, we get,

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

Hence, the power dissipated across the switch when the open comes out to be $0\mathrm{W}$.