Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 18: Q15Q (page 755)

Question: Some students intended to run a light bulb off two batteries in series in the usual way, but they accidentally hooked up one of the batteries backwards, as shown in Figure 18.89 (the bulb is shown as a thin filament).

(a)Use+’s and -’s to show the approximate steady-state charge distribution along the wires and bulb.

(b)Draw vectors for the electric field at the indicated locations inside the connecting wires and bulb.

(c)Compare the brightness of the bulb in this circuit with the brightness the bulb would have had if one of the batteries hadn’t been put in backwards.

(d)Try the experiment to check your analysis. Does the bulb glow about as you predicted?

Short Answer

Expert verified

(a) The distribution of charges in the circuit are as follows:

Step by step solution

01

Given data

Two batteries are connected opposing each other in series with a filament.

02

Flow of charges

The negative terminal of the battery forces negative charges to flow from the negative terminal to the positive through the circuit. The positive terminal of the battery forces positive charges to flow from the positive terminal to the negative through the circuit.

03

(a) Determination of the charge distribution in the circuit

The negative terminals of both the batteries are directed towards the circuit and the positive terminal are directed towards each other. Thus there will be a larger concentration of negative charges in the circuit and that of positive charges in between the batteries. The distribution can be depicted as follows:

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a typical drift speed of 5×10-5m/s, an electron traveling at that speed would take about to travel through one of your connecting wires. Why, then, does the bulb light immediately when the connecting wire is attached to the battery?

Since there is an electric field inside a wire in a circuit, why don’t the mobile electrons in the wire accelerate continuously?

State your own theoretical and experimental objections to the following statement: In a circuit with two thick-filament bulbs in series, the bulb farther from the negative terminal of the battery will be dimmer, because some of the electron current is used up in the first bulb. Cite relevant experiments.

In the circuit shown in Figure 18.87, bulbs 1 and 2 are identical in mechanical construction (the filaments have the same length and the same cross-sectional area), but the filaments are made of different metals. The electron mobility in the metal used in bulb 2 is three times as large as the electron mobility in the metal used in bulb 1, but both metals have the same number of mobile electrons per cubic meter. The two bulbs are connected in series to two batteries with thick copper wires (like your connecting wires).

(a)In bulb 1, the electron current is i1and the electric field is E1. In terms of these quantities, determine the corresponding quantities i2and E2for bulb 2, and explain your reasoning.

(b)When bulb 2 is replaced by a wire, the electron current through bulb 1 is i0and the electric field in bulb 1 is E0. How big is i1 in terms of i0? Explain your answer, including explicit mention of any approximations you must make. Do not use ohms or series-resistance equations in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter.

(c)Explain why the electric field inside the thick copper wires is very small. Also explain why this very small electric field is the same in all of the copper wires, if they all have the same cross-sectional area.

(d)Figure 18.88 is a graph of the magnitude of the electric field at each location around the circuit when bulb 2 is replaced by a wire. Copy this graph and add to it, on the same scale, a graph of the magnitude of the electric field at each location around the circuit when both bulbs are in the circuit. The very small field in the copper wires has been shown much larger than it really is in order to give you room to show how that small field differs in the two circuits.

The circuit shown in Figure 18.107 consists of a single battery, whose emf is 1.8V, and three wires made of the same material but having different cross-sectional areas. Each thick wire has a cross-sectional area 1.4×10-6m2and is 25cmlong. The thin wire has a cross-sectional area 5.9×10-6m2and is 6.1cmlong. In this metal, the electron mobility is 5×10-4(ms)(Vm), and there are 4×1028mobile 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 EDand EF 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.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Astrophysics

Read Explanation

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Electric Charge Field and Potential

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free