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- v which of the following equations are valid energy conservation (loop) equations for this circuit? 51 refers to the electric eld in bulb #1, L refers to the length of a bulb lament , etc. Assume that the electric eld in the connecting wires is small enough to neglect. C +2*emf - E1L - EZL = o +2*emf - ElL- EzL - E3L = o E1L - 52L = o +E2L - E3L = o +2*emf - EZL - E3L = 0 51L - E3L = o C C C C C C +2*emf - E1L - E3L = o 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? O i1 = 13 O i1 = 12 O i1 = 12 + 13 Each battery has an emf of 1.4 volts. The length of the tungsten filament in each bulb is 0.015 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? /1 = electrons/s How many electrons per second enter bulb #2? 12 = electrons/s