Three identical light bulbs are connected to two batteries as shown in the diagr
ID: 1428742 • Letter: T
Question
Three identical light bulbs are connected to two batteries as shown in the diagram above.
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? ---Select--- 0 1 2 3 4 5
Which of the following equations are valid energy conservation (loop) equations for this circuit? E1 refers to the electric field in bulb #1, L refers to the length of a bulb filament , etc. Assume that the electric field in the connecting wires is small enough to neglect.
E1L - E3L = 0E1L - E2L = 0+2*emf - E1L - E3L = 0+2*emf - E1L- E2L - E3L = 0+2*emf - E1L - E2L = 0+2*emf - E2L - E3L = 0+E2L - E3L = 0
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?
i1 = i2i1 = i3i1 = i2 + i3
Each battery has an emf of 1.7 volts. The length of the tungsten filament in each bulb is 0.011 m. The radius of the filament is 5e-6 m (it is very thin!). The electron mobility of tungsten is 1.8e-3 (m/s)/(V/m). Tungsten has 6e+28 mobile 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 the following electric field magnitudes:
What is the magnitude of the electric field inside bulb #1?
E1 = V/m
What is the magnitude of the electric field inside bulb #2?
E2 = V/m
How many electrons per second enter bulb #1?
i1 = electrons/s
How many electrons per second enter bulb #2?
i2 = electrons/s
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 = 103.03 V/m
magnitude of electric field inside bulb#1, E1 = 206.06 V/m
magnitude of electric field inside bulb#2, E2 = 103.03 V/m
number of electrons that enter bulb#1 per second = nAmE1 = 175 x 1016 /s
number of electrons that enter bulb#2 per second = nAmE2 = 87.4 x 1016 /s
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