When a single thick-filament bulb of a particular kind and two batteries are connected in series, $\mathbf{3}\mathbf{\times }\mathbf{10}$¹⁸ electrons pass through the bulb every second. When two batteries in series are connected to a single thin-filament bulb, with a filament made of the same material and length as the thick-filament bulb but a smaller cross-section, only $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{10}$¹⁸ electrons pass through the bulb every second. (a) In the circuit shown in Figure 18.109, how many electrons per second flow through the thin-filament bulb? (b) What approximations or simplifying assumptions did you make? (c) Show approximately the surface charge on a diagram of the circuit.

The number of electrons passes through the single thick filament bulb $\mathrm{n}=3\times 10$¹⁸.

The number of electrons pass through the single thin filament bulb, $\mathrm{n}\text{'}=1.5\times 10$¹⁸.

The relation between the number of electrons of a thin filament bulb and a combination of thin and thick filament bulb is given as follows,

$\frac{\mathbf{N}}{\mathbf{n}}\mathbf{=}{\frac{\mathbf{R}}{\mathbf{R}}}_{\mathbf{eq}}$ ....(i)

Here, $\mathbf{R}$is the resistance of the thick filament bulb, ${\mathbf{R}}_{\mathbf{eq}}$is the equivalent resistance of the circuit, and$\mathbf{N}$ is the number of the electron passes through the thin filament bulb.

The number of electrons passes through the filament bulb is inversely proportional to the filament bulb.

Calculate the resistance of the single thin filament bulb.

$\mathrm{R}\text{'}=\frac{\mathrm{n}}{\mathrm{n}\text{'}}\mathrm{R}$

Substitute$3\times 10$¹⁸ for n and $1.5\times 10$¹⁸ for into the above equation.

$$\begin{gathered} \mathrm{R}\text{'}=\frac{3\times {10}^{18}}{1.5\times {10}^{18}}\mathrm{R} \\ \mathrm{R}\text{'}=\frac{3}{1.5}\mathrm{R} \\ \mathrm{R}\text{'}=2\mathrm{R} \end{gathered}$$

The equivalent resistance of the given circuit can be calculated as,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=\mathrm{R}\text{'}+\frac{\mathrm{R}\times \mathrm{R}}{\mathrm{R}+\mathrm{R}} \\ {\mathrm{R}}_{\mathrm{eq}}=\mathrm{R}\text{'}+\frac{\mathrm{R}}{2} \end{gathered}$$

Substitute for into above equation.

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=2\mathrm{R}+\frac{\mathrm{R}}{2} \\ {\mathrm{R}}_{\mathrm{eq}}=\frac{5}{2}\mathrm{R} \\ {\mathrm{R}}_{\mathrm{eq}}=2.5\mathrm{R} \end{gathered}$$

Calculate the number of electrons passing through the thin filament bulb.

Substitute, localid="1668665406979" $2.5\mathrm{R}$for ${\mathrm{R}}_{\mathrm{eq}}$and $3\times 10$¹⁸ for n into equation (i).

$$\begin{gathered} \frac{\mathrm{N}}{3\times {10}^{18}}=\frac{\mathrm{R}}{2.5\mathrm{R}} \\ \frac{\mathrm{N}}{3\times {10}^{18}}=\frac{1}{2.5} \\ \mathrm{N}=\frac{3\times {10}^{18}}{2.5} \\ \mathrm{N}=1.2\times {10}^{18} \end{gathered}$$

Hence the number of the electron passes through the thin filament bulb is

$1.2\times {10}^{18}$