Question

A circuit consists of a battery, whose emf is K, and five Nichrome wires, three thick and two thin as shown in Figure 19.78. The thicknesses of the wires have been exaggerated in order to give you room to draw inside the wires. The internal resistance of the battery is negligible compared to the resistance of the wires. The voltmeter is not attached until part (e) of the problem. (a) Draw and label appropriately the electric field at the locations marked × inside the wires, paying attention to appropriate relative magnitudes of the vectors that you draw. (b) Show the approximate distribution of charges for this circuit. Make the important aspects of the charge distribution very clear in your drawing, supplementing your diagram if necessary with very brief written descriptions on the diagram. Make sure that parts (a) and (b) of this problem are consistent with each other. (c) Assume that you know the mobile-electron density n and the electron mobility u at room temperature for Nichrome. The lengths $\left(L_1,L_2,L_3\right)$and diameters$\left(d_1,d_2\right)$ of the wires are given on the diagram. Calculate accurately the number of electrons that leave the negative end of the battery every second. Assume that no part of the circuit gets very hot. Express your result in terms of the given quantities$\left(K,L_1,L_2,L_3,d_1,d_2,n\text{and}u\right)$ . Explain your work and identify the principles you are using. (d) In the case that${\mathit{d}}_{\mathbf{2}}\mathbf{\ll }{\mathit{d}}_{\mathbf{1}}$ , what is the approximate number of electrons that leave the negative end of every second? (e) A voltmeter is attached to the circuit with its + lead connected to location B (halfway along the leftmost thick wire) and its - lead connected to location C (halfway along the leftmost thin wire). In the case thatrole="math" localid="1663035964741" ${\mathit{d}}_{\mathbf{2}}\mathbf{\ll }{\mathit{d}}_{\mathbf{1}}$ , what is the approximate voltage shown on the voltmeter, including sign? Express your result in terms of the given quantitiesrole="math" localid="1663036061574" $(K,L_1,L_2,L_3,d_1,d_2,n\text{and}u)$ .

Short answer

(e) In the case that $d_2\ll d_1$, the approximate voltage shown on the voltmeter is $\frac{K}{4}$and, the sign is positive as the positive terminal is connected to the higher potential.

Step-by-step solution

Given data

A circuit is given with a battery with an emf value K and five Nichrome wires in a number of 5, three out of five are thick, and two are thin. The mobile-electron density is denoted as n, and the electron mobility is denoted as u at room temperature for the Nichrome metal wire. The lengths of the wires are given as $L_1,L_2,L_3$, and the diameters of the wires are given in the question as $d_1,d_2$.

Concept

The number of charge carriers that are moving between two terminals in a conductor can be measured as,${\mathit{n}}_{\mathbf{e}}\mathbf{=}\mathit{n}\mathit{A}\mathit{u}\mathit{E}\mathit{t}$ (1)

Here n is the charge density of the charge carrier; A is the cross-sectional area of the conductor, E is the electric field in the conductor, u is the mobility of the charge carriers, and t is the time.

(e) In the case d2≪d1, calculation of the approximate voltage shown on the voltmeter

The voltmeter is connected to two points. Point B is connected to the positive terminal, and point C is connected to the negative terminal of the voltmeter.

Point B is in the middle of the thick wire, and Point C is in the middle of the thin wire.

Thus, we can calculate the voltage shown in the voltmeter as,

The potential difference between the two terminals that are connected to the points on the wire will be shown on the voltmeter. That difference can be written as,

$\Delta V=E_2\frac{L_2}{2}$

Here E₂ is the electric field in the thin wire.

(Refer to SID:875865-19-59 P-c)

The expression for the electric field in the thin wire is,

$E_2=\frac{K}{\left(\frac{2A_2{L}_1}{A_1}+\frac{A_2{L}_3}{A_1}+2L_2\right)}$

Substitute this value in the above expression, and we get,

$\Delta V=\frac{K}{\left(\frac{2A_2{L}_1}{A_1}+\frac{A_2{L}_3}{A_1}+2L_2\right)}\frac{L_2}{2}$ (2)

Here A₁ is the cross-sectional area of the thick wire, and A₂is the cross-sectional area of the thin wire.

Cross-sectional areas can be calculated as,

$A_1=\pi {\left(d_1/2\right)}^2$

And

$A_2=\pi {\left(d_2/2\right)}^2$

Now we can write equation 2 as,

$$\begin{gathered} \Delta V=\frac{K}{\left(\frac{2\pi {\left(d_2/2\right)}^2{L}_1}{\pi {\left(d_1/2\right)}^2}+\frac{\pi {\left(d_2/2\right)}^2{L}_3}{\pi {\left(d_1/2\right)}^2}+2L_2\right)}\frac{L_2}{2} \\ \Delta V=\frac{K}{\left(\frac{2{\left(d_2/2\right)}^2{L}_1}{{\left(d_1/2\right)}^2}+\frac{{\left(d_2/2\right)}^2{L}_3}{{\left(d_1/2\right)}^2}+2L_2\right)}\frac{L_2}{2} \end{gathered}$$ (3)

If

$$\begin{gathered} d_2\ll d_1 \\ \frac{d_2}{d_1}\ll 1 \end{gathered}$$

Then the term $\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}$will be written as,

$\frac{2{\left(d_2\right)}^2{L}_1}{{\left(d_1\right)}^2}+\frac{{\left(d_2\right)}^2{L}_3}{{\left(d_1\right)}^2}\approx 0$.

The above expression 3 will be written as,

$$\begin{gathered} \Delta V=\frac{K}{\left(0+2L_2\right)}\frac{L_2}{2} \\ \Delta V=\frac{K}{\left(2L_2\right)}\frac{L_2}{2} \\ \Delta V=\frac{K}{4} \end{gathered}$$

Thus, in the case that$d_2\ll d_1$, the approximate voltage shown on the voltmeter is $\frac{K}{4}$and, the sign is positive as the positive terminal is connected to the higher potential.