Three identical light bulbs are connected to two batteries as shown in the diagr

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? 3 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 +2 emf E1L E2L 0 +2 emf E1L- E2L E3L O +2 E1L E3L E2L E3L E1L E2L E E2L E3L. 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? 2 Each battery has an emf of 1.7 volts. The length of the tungsten filament in each bulb is 0.008 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? V/ E1 What is the magnitude of the electric field inside bulb #2? V/m E2 How many electrons per second enter bulb #1? electrons/s How many electrons per second enter bulb #2? electrons/s

Explanation / Answer

Hi,

In this case the problem is almost solve, because the suppositions that were made at the beginning are right. The only thing to do is use the equations provided with the values given in order to find the things they ask.

In the case of the electric fields, we use the following system of linear equations:

2 emf - E1L - E2L = 0 ::::: 3.4 - 0.008*E1 - 0.008*E2 = 0

2 emf - E1L - E3L = 0 ::::: 3.4 - 0.008*E1 - 0.008*E3 = 0

E2L = E3L :::::::::::::::::::::: E2 = E3

When we solve this system we get the following values:

E1 = 268.2 V/m

E2 = 156.8 V/m (which is equal to E3, so E3 = 156.8 V/m)

In the case of the currents (I1 and I2), we find the value of the current 1 and thanks to the equations we know that the current 1 is two times the value of current 2 or 3.

To find current 1, in terms of the electrons, we use the following equation:

I = n*v*A ; where n is the density of mobile electrons in the material, v is the average mobility of the electrons and A is the cross sectional area of the material.

A = *r2 = *(5*10-6 m)2 = 7.85*10-11 m2

I1 = (6*1028 electrons/m3)*(1.8*10-3 m/s)*(7.85*10-11 m2) = 8.48*1015 electrons/s

I2 = (1/2) I1 = 4.24*1015 electrons/s

I hope it helps.

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