91影视

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

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

Resistance of switch when open$R_{open}=鈭�惟$

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

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,

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

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

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

When the switch is closed, equivalent resistance in series of the circuit is given as,

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

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

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

Since$R_{closed}=0$

role="math" localid="1655895714557" $$\begin{gathered} {\mathrm{l}}_{\mathrm{closed}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+0} \\ {\mathrm{l}}_{\mathrm{closed}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}} \end{gathered}$$

Clearly,${\mathrm{l}}_{\mathrm{closed}}>{\mathrm{l}}_{\mathrm{open}}$

Therefore, when the switch is closed, there is some constant value of current in the circuit.