91Ó°ÊÓ

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

Resistance of the switch when closed$R_{closed}=0\mathrm{Î\copyright }$

Resistance of switch when open$R_{open}=∞\mathrm{Î\copyright }$

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

The voltage across the switch role="math" localid="1655896545753" ${\mathrm{V}}_{\mathrm{switch}}$

When the switch is open,

Equivalent resistance in series of the circuit given,

${\mathrm{R}}_{\mathrm{eq}}=\mathrm{R}+\mathrm{r}+{\mathrm{R}}_{\mathrm{open}}$

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

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

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

Using Ohm's voltage across the switch is given:

$$\begin{gathered} {\mathrm{l}}_{\mathrm{open}}=\frac{{\mathrm{V}}_{\mathrm{switch}}}{{\mathrm{R}}_{\mathrm{open}}} \\ 0=\frac{{\mathrm{V}}_{\mathrm{switch}}}{{\mathrm{R}}_{\mathrm{open}}} \\ {\mathrm{V}}_{\mathrm{switch}}=0\mathrm{V} \end{gathered}$$

Hence, the voltage across the switch when open comes out to be $0\mathrm{V}$