Question

The circuit shown in Figure 18.107 consists of a single battery, whose emf is $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{V}$, and three wires made of the same material but having different cross-sectional areas. Each thick wire has a cross-sectional area $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathbf{m}}^{\mathbf{2}}$and is $\mathbf{25}\mathbf{cm}$long. The thin wire has a cross-sectional area $\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathbf{m}}^{\mathbf{2}}$and is $\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{cm}$long. In this metal, the electron mobility is $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\raisebox{1ex}{$\left({\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}\right)$}\right.$, and there are $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$mobile electrons/m³.

(a) Which of the following statements about the circuit in the steady state are true? (1) At location B, the electric field points toward the top of the page. (2) The magnitude of the electric field at locations F and C is the same. (3) The magnitude of the electric field at locations D and F is the same. (4) The electron current at location D is the same as the electron current at location F . (b) Write a correct energy conservation (loop) equation for this circuit, following a path that starts at the negative end of the battery and goes counterclockwise. (c) Write this circuit's correct charge conservation (node) equation. (d) Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the magnitudes E_Dand E_F of the electric field at locations D and F . (e) Use the appropriate equation(s) to calculate the electron current at location D in the steady state.

Short answer

Statements (1), (2), and (4) are true about the circuit in the steady state condition.

Step-by-step solution

Write the given data from the question.

Emf of battery, $\mathrm{V}=1.8\mathrm{V}$

Cross sectional area of thick wire,${\mathrm{A}}_1=1.4\times {10}^{-6}{\mathrm{m}}^2$

Length of thick wire,${\mathrm{L}}_1=25\mathrm{cm}$

Cross sectional area of thin wire,${\mathrm{A}}_2=5.9\times {10}^{-6}{\mathrm{m}}^2$

Length of thin wire,${\mathrm{L}}_2=6.1\mathrm{cm}$

Electron mobility, $\mu =5\times {10}^{-4}\left(\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\right)/\left(\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.\right)$

Electron density,$n=4\times {10}^{28}\mathrm{e}/{\mathrm{m}}^3$

Determine the formulas to find the correct statement in the steady state of the circuit.

The electric field is defined as the ratio of the voltage and length of the wire.

The expression to calculate the magnitude of the electric field is given as follows.

$\mathbf{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{L}}$

Here, $\mathbf{V}$is the voltage and $\mathbf{L}$is the length of the wire.

Find the correct statement in the steady state about the circuit.

The direction of the electric field is away from the battery's positive terminal and toward the negative terminal. Therefore, the electric field moves from the positive terminal to the battery's negative terminal. So, the direction of the electric field at the location is toward the top of the page.

The electric field depends on the voltage and length of the wire. Because at locations F and C, the voltage and length of the wire are the same. But at the locations F and D length of both the wire are different. Therefore, the magnitude of the electric field is the same at locations F and C but different at locations F and D.

At the steady state condition of the circuit, the electron current at all the locations is the same.

Hence the statement (1), (2), and (4) are true about the circuit in the steady state condition.