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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.

Short Answer

Expert verified

The energy conservation equations are P0-P1-B1-B2-P2-P4-P0,P0-P1-B1-B3-P3-P4-P0 and P2-P3-B3-B2-P0.

Step by step solution

01

Write the given data from the question.

Emf of the battery, V=1.5V

The length of the tungsten filament,L=0.008m.

Radius of filament,r=5×10-6m

Electron mobility of tungsten,μ=1.8×10-3(m/s)/(V/m)

The number of mobile electrons, N=6×1028e/m3

02

Write the statements that helps to write the conservation equation.

The current is always flow from the higher potential to the lower potential and energy remains conserve.

Write the energy conservation equation that must involve a round-trip path that begins and ends at the same location.

03

Write the energy conservation equation.

The energy conservation equation for the loop that starts from P0and go through bulb B1and B2is given as follows.

P0-P1-B1-B2-P2-P4-P0

The energy conservation equation for the loop that starts from P0and go through bulb B1and role="math" localid="1668588087308" B3is given as follows.

P0-P1-B1-B3-P3-P4-P0

The energy conservation equation for the loop that starts from P2and go through bulb B3is given as follows.

P2-P3-B3-B2=0

Hence the energy conservation equations are P0-P1-B1-B2-P2-P4-P0, P0-P1-B1-B3-P3-P4-P0and P2-P3-B3-B2-P0.

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Most popular questions from this chapter

Describe the following attributes of a metal wire in steady

state vs. equilibrium:

Metal Wire

Steady-state

Equilibrium

Location of excess charge

Motion of mobile electrons

inside the metal wire

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?

Suppose that wire A and wire B are made of different metals and are subjected to the same electric field in two different circuits. Wire B has the 6 times the cross sectional area, 1.3 times as many mobile electrons per cubic centimetre and 4 times the mobility of wire A. In the steady state \({\bf{2 \times 1}}{{\bf{0}}^{{\bf{18}}}}\) electrons enters wire A every second. How many electrons enter wire B every second?

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

In a circuit with one battery, connecting wires, and a 12cmlength of Nichrome wire, a compass deflection of 6°is observed. What compass deflection would you expect in a circuit containing two batteries in a series, connecting wires and a36cm length of thicker Nichrome wire (double the cross-sectional area of the thin piece)? Explain.

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