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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:

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Most popular questions from this chapter

What is the difference between emf and electric potential difference?

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.

Question: Three identical light bulbs are connected to two batteries as shown in Figure 18.106. (a) To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-trip path that begins and ends at the same location. Each segment of the path should go through a wire, a bulb, or a battery (not through the air). How many valid energy conservation (loop) equations is it possible to write for this circuit? (b) Which of the following equations are valid energy conservation (loop) equations for this circuit? E1refers to the electric field in bulb 1; L refers to the length of a bulb filament. Assume that the electric field in the connecting wires is small enough to neglect.

(1) +E2L-E3L=0, (2) E1L-E3L=0, (3)+2emf-E2L-E3L=0, (4)E1L-E2L=0, (5)+2emf-E1L-E2L=0, (6)+2emf-E1L-E3L=0, (7)+2emf-E1L-E2L-E3L=0. (c) It is also necessary to write charge conservation equations (node) equations. Each such equation must relate electron current flowing into a node to electron current flowing out of a node. Which of the following are valid charge conservation equations for this circuit? (1)i1=i3, (2)i1=i2, (3)i1=i2+i3. Each battery has an emf of 1.5V. The length of the tungsten filament in each bulb is 0.008m. The radius of the filament is5×10-6m(it is very thin!). The electron mobility of tungsten is localid="1668588909714" 1.8×10-3(m/s)/(V/m). Tungsten has localid="1668588927161" 6×1028mobile electrons per cubic meter. Since there are three unknown quantities, we need three equations relating these quantities. Use any two valid energy conservation equations and one valid charge conservation equation to solve for localid="1668588943223" E1,E2,i1and localid="1668588965567" i2.

There are very roughly the same number of iron atoms per m3 as there are copper atoms per m3 , but copper is a much better conductor than iron. How does uiron compare with ucopper?

When a single thick-filament bulb of a particular kind and two batteries are connected in series, 3×1018 electrons pass through the bulb every second. When two batteries in series are connected to a single thin-filament bulb, with a filament made of the same material and length as the thick-filament bulb but a smaller cross-section, only 1.5×1018 electrons pass through the bulb every second. (a) In the circuit shown in Figure 18.109, how many electrons per second flow through the thin-filament bulb? (b) What approximations or simplifying assumptions did you make? (c) Show approximately the surface charge on a diagram of the circuit.
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