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Chapter 18: Q10Q (page 755)

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?

Short Answer

Expert verified

The bulb lights immediately because mobile electrons are present throughout the circuit that start moving simultaneously when switched on.

Step by step solution

01

Given data

Drift speed of electrons = 5×10-5m/s

02

Electrical force on electrons in a circuit

The force due to electric field applied on a circuit affects all mobile electrons simultaneously.

03

Determine reason why a light bulb turns on immediately when switched on

There are mobile free electrons present throughout the circuit wire. When switched on, the applied electric field from the battery acts simultaneously on all electrons throughout the circuit and they start moving instantaneously. Thus a bulb located anywhere in the circuit is lit immediately.

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Most popular questions from this chapter

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.

Inside a chemical battery it is not actually individual electrons that are transported from the + end to the – end. At the + end of the battery an “acceptor” molecule picks up an electron entering the battery, and at the – end a different “donor” molecule gives up an electron, which leaves the battery. Ions rather than electrons move between the two ends to support the charge inside the battery.

When the supplies of acceptor and donor molecules are used up in a chemical battery, the battery is dead because it can no longer accept or electron. The electron current in electron per second times the number of seconds of battery life, is equal to the number of donor molecules in the battery.

A flashlight battery contains approximately half a mole of donor molecules. The electron current through a thick filament bulb powered by two flashlight batteries in series is about 0.3 A. About how many hours will the batteries keep this bulb lit?

The emf of a particular flashlight battery is 1.7 V. If the battery is 4.5 cm long and radius of cylindrical battery is 1 cm, estimate roughly the amount of charge on the positive end plate of the battery.

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.

Question: A circuit is constructed from two batteries and two wires, as shown in Figure 18.104. Each battery has an emf of 1.3V. Each wire is26cmlong and has a diameter of 7×10-4m. The wires are made of a metal that has7×1028mobile electrons per cubic meter; the electron mobility is 5×10-5(m/s)/(V/m). A steady current runs through the circuit. The locations marked by ×and labeled by a letter are in the interior of the wire. (a) Which of these statements about the electric field in the interior of the wires, at the locations marked by ×'s, are true? List all that apply. (1) The magnitude of the electric field at location G is larger than the magnitude of the electric field at location F. (2) At every marked location the magnitude of the electric field is the same. (3) At location B the electric field points to the left. (b) Write a correct energy conservation (round-trip potential difference) equation for this circuit, along a round-trip path starting at the negative end of battery 1 and traveling counterclockwise through the circuit (that is, traveling to the left through the battery, and continuing on around the circuit in the same direction). (c) What is the magnitude of the electric field at location B? (d) How many electrons per second enter the positive end of battery 2? (e)If the cross-sectional area of both wires were increased by a factor of 2, what would be the magnitude of the electric field at location B? (f) Which of the diagrams in Figure 18.105 best shows the approximate distribution of excess charge on the surface of the circuit?
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