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Chapter 18: Q4Q (page 755)

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.

Short Answer

Expert verified

The two bulbs will glow equally brightly since the current through them is the same.

Step by step solution

01

Given data

Two bulbs are in series in a circuit.

02

Current in a circuit

The current in a circuit where all loads and batteries are connected in series, is equal everywhere.

03

Determination of the glow of the two bulbs

Experimentally it can be checked by connecting two bulbs in a series that they glow equally brightly. Theoretically since the same number of electrons are passing per unit time through the two bulbs in series after steady state is reached, the current through both of them is the same. Hence the glow will not be affected. Electrons don't get used up when they pass through a load. The two bulbs will glow equally brightly.

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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?

Suppose that a wire leads into another, thinner wire of the same material that has only a third the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the drift speedV1¯in the thick wire is 4×10-5ms, what is the drift speed V¯2in the thinner wire?

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.

What is the most important general difference between a system in steady state and a system in equilibrium?

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.
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