Q59P-c A circuit consists of a battery,... [FREE SOLUTION]

Chapter 19: Q59P-c (page 799)

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 脳 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 $(L_{1}, L_{2}, L_{3})$ and diameters $(d_{1}, d_{2})$ 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 $(K, L_{1}, L_{2}, L_{3}, d_{1}, d_{2}, n and u)$ . Explain your work and identify the principles you are using. (d) In the case that $d_{2} 鈮� d_{1}$ , 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 that $d_{2} 鈮� d_{1}$ , what is the approximate voltage shown on the voltmeter, including sign? Express your result in terms of the given quantities $(K, L_{1}, L_{2}, L_{3}, d_{1}, d_{2}, n and u)$ .

Short Answer

Expert verified

(c) The number of electrons that leave the negative end of the battery every second is $\frac{n u 蟺 {(d_{2})}^{2} K}{4 (\frac{2 {(d_{2})}^{2} L_{1}}{{(d_{1})}^{2}} + \frac{{(d_{2})}^{2} L_{3}}{{(d_{1})}^{2}} + 2 L_{2})}$ .

Step by step solution

Given data

A circuit is given which has a battery that has an emf value of K and five Nichrome wires out of 5, three are thick, and two are thin. The mobile-electron density is n, and the electron mobility is u at room temperature for Nichrome. The lengths of the wires are, $L_{1}, L_{2}, L_{3}$ and diameters of the wires are, $d_{1}, d_{2}$ .

Concept

The number of charge carriers moving in the metal wire can be written as,

$n_{e} = n A u E t$ (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.

(c) Calculate the number of electrons that leave the negative end

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

$E_{1} = \frac{A_{2} E_{2}}{A_{1}}$ (2)

Here A₁is the cross-sectional area of the thick wire, E₁is the electric field in the thick wire, A₂is the cross-sectional area of the thin wire, and E₂is the electric field in the thin wire.

Let us apply the circuit law in the loop; we get,

$\begin{array}{rcl} + K + (- E_{1} L_{1}) + (- E_{2} L_{2}) + (- E_{1} L_{3}) + (- E_{2} L_{2}) + (- E_{1} L_{1}) & = & 0 \\ 2 E_{1} L_{1} + E_{1} L_{3} + 2 E_{2} L_{2} & = & K \end{array}$ (3)

Substitute values from equation 2 into equation 3, and we get,

$\begin{array}{rcl} 2 \frac{A_{2} E_{2}}{A_{1}} L_{1} + \frac{A_{2} E_{2}}{A_{1}} L_{3} + 2 E_{2} L_{2} & = & K \\ E_{2} (\frac{2 A_{2} L_{1}}{A_{1}} + \frac{A_{2} L_{3}}{A_{1}} + 2 L_{2}) & = & K \\ E_{2} & = & \frac{K}{(\frac{2 A_{2} L_{1}}{A_{1}} + \frac{A_{2} L_{3}}{A_{1}} + 2 L_{2})} \end{array}$

Substitute this value in equation 2, and we get,

$E_{1} = \frac{A_{2}}{A_{1}} \frac{K}{(\frac{2 A_{2} L_{1}}{A_{1}} + \frac{A_{2} L_{3}}{A_{1}} + 2 L_{2})}$

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

The negative end is connected to the thick wire, and for thick wire, the number of electrons passing for t = 1 s can be written as,

$n_{1} = n A_{1} u E_{1}$ (4)

Cross-sectional areas can be calculated as,

$A_{1} = 蟺 {(d_{1} / 2)}^{2}$

And

$A_{2} = 蟺 {(d_{2} / 2)}^{2}$

Substitute the values in equation 4, and we get,

$\begin{array}{rcl} n_{1} & = & n 蟺 {(d_{1} / 2)}^{2} u \frac{蟺 {(d_{2} / 2)}^{2}}{蟺 {(d_{1} / 2)}^{2}} \frac{K}{(\frac{2 蟺 {(d_{2} / 2)}^{2} L_{1}}{蟺 {(d_{1} / 2)}^{2}} + \frac{蟺 {(d_{2} / 2)}^{2} L_{3}}{蟺 {(d_{1} / 2)}^{2}} + 2 L_{2})} \\ n_{1} & = & n u 蟺 {(d_{2} / 2)}^{2} \frac{K}{(\frac{2 {(d_{2})}^{2} L_{1}}{{(d_{1})}^{2}} + \frac{{(d_{2})}^{2} L_{3}}{{(d_{1})}^{2}} + 2 L_{2})} \\ n_{1} & = & \frac{n u 蟺 {(d_{2})}^{2} K}{4 (\frac{2 {(d_{2})}^{2} L_{1}}{{(d_{1})}^{2}} + \frac{{(d_{2})}^{2} L_{3}}{{(d_{1})}^{2}} + 2 L_{2})} \end{array}$

Thus, the number of electrons that leave the negative end of the battery every second is $\frac{n u 蟺 {(d_{2})}^{2} K}{4 (\frac{2 {(d_{2})}^{2} L_{1}}{{(d_{1})}^{2}} + \frac{{(d_{2})}^{2} L_{3}}{{(d_{1})}^{2}} + 2 L_{2})}$ .