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.

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

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.

During the initial transient leading to the steady state, the electron current going into a bulb may be greater than the electron current leaving the bulb. Explain why and how these two currents come to be equal in the steady state.

Why don’t all mobile electrons in a metal have exactly the same speed?

A Nichrome wire 30 cm long and 0.25 mm in diameter is connected to a 1.5 V flashlight battery. What is the electric field inside the wire? Why you don’t have to know how the wire is bent? How would your answer change if the wire diameter change were 0.35 mm? (Not that the electric field in the wire is quiet small compared to the electric field near a charged tape.)

A steady-state current flows through the Nichrome wire in the circuit shown in Figure 18.90. Before attempting to answer the following questions, draw a copy of this diagram. All of the locations indicated by letters are inside the wire.

(a)On your diagram, show the electric field at the locations indicated, paying attention to relative magnitude.

(b)Carefully draw pluses and minuses on your diagram to show the approximate surface charge distribution that produces the electric field you drew. Make your drawing show clearly the differences between regions of high surface charge density and regions of low surface-charge density. Use your diagram to determine which of the following statements about this circuit are true.

(1) There is some excess negative charge on the surface of the wire near location B.

(2) Inside the metal wire the magnitude of the electric field is zero.

(3) The magnitude of the electric field is the same at locations Gand C.

(4) The electric field points to the left at location G.

(5) There is no excess charge on the surface of the wire.

(6) There is excess charge on the surface of the wire near the batteries but nowhere else.

(7) The magnitude of the electric field inside the wire is larger at location Gthan at location C.

(8) The electric field at location Dpoints to the left.

(9) Because the current is not changing, the circuit is in static equilibrium.

See all solutions

Recommended explanations on Physics Textbooks

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