Question

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.

Short answer

The power dissipated is small because the current is very small.

Step-by-step solution

Definition of the circuit.

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.

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$\mathrm{R}$

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" ${\mathrm{l}}_{\mathrm{open}}$

Effect of the switch when is 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.

Calculating power dissipated 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 $0\mathrm{W}$.