(chapter 18) Answer is provided, provide a step by step solution. Three identica

Question

(chapter 18) Answer is provided, provide a step by step solution.  

Three identical light bulbs are connected to two batteries as shown in the diagram above concervation (loop) equations. Each equation To start the analysis of this circuit you must write through the air). How many valid energy consevation (loop) equations is it possible to writa for this circuit? 3 which of the following equations are valid energy conservation loop equations for this circuit? El e ers t he electric ned i 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 buit #1 L erers to the length or a bu nament etc. Assume that the electric neld in the connecting wres is sm all enough to lect. It is also necessary to write charge conservation equations (node) equations, Each such equation must relate electron current fowing into a node to electron current flowing out of a node. Which of the following are valid charge conservation equations for 1-23 Each battery has an emf of 1.3 volts. The length of the tungsten filament in each bulb i0 012 m cubic meter The adius of the filement is Se 6 m it is very thin! The electron mobility of tungsten is 1.8e-3 m/s)/ V Tungsten has 6e+28 mobile electrons per m Since there are three unknowm quantities, we need three equations relating these quantities. Use any two valid energy conservation equations and one valid charge conservation equation to sclve for the fol lowing electric field magnitudes: What nitude of the electric field inside bulb 1? What e of the electric field inside bulb 2? s per second enter bulb #1? 11.225 18 electrons/s per second enter bulb #2? 2-6.125e17 electrons/s Additional Materials DSection 18.10

Explanation / Answer

We can write 3 valid energy conservation ( loop ) equations.

valid energy conservation ( loop ) equations are

+2*emf - E1L - E3L = 0

+2*emf - E1L - E2L = 0

+E2L - E3L = 0

valid charge conservation equation for this circuit is i1 = i2 + i3

using the above equations

E2L = E3L

since current i = neAVd

n = density of charge carriers

A area of cross-section

Vd drift speed

and Vd = mE

m = mobility

E = Electric field

i =  neAVd = neAmE

since E2L = E3L and E2 = E3

i2 = i3

and since i1 = i2 +i3

neAmE1 = neAmE2 + neAmE3

and E1 = E2 + E3 = 2E2 = 2E3

hence +2*emf = E1L + E2L = 3E2L

E3 = (2 x 1.3)/(0.012 x 3)

E3 = 72.22 V/m

magnitude of electric field inside bulb#1, E1 = 2 E3 = 144.44 V/m

magnitude of electric field inside bulb#2, E2 = E3 = 72.22 V/m

number of electrons that enter bulb#1 per second = nAmE1 = (pi x (5x10-6)2) x 1.8x10-3 x 144.44 x 6x1028 = 1.225 x 1018 /s

number of electrons that enter bulb#2 per second = nAmE2 = 6.125 x 1017/s

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