(**NOTE: The picture is suppose to be connected on the right side as well) Each

Question

(**NOTE: The picture is suppose to be connected on the right side as well)

Each battery has an emf of 1.4 volts. The length of the tungsten filament in each bulb is 0.012 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 = 4 V/m
What is the magnitude of the electric field inside bulb #2?
E2 = 5 V/m
How many electrons per second enter bulb #1?
i1 = 6 electrons/s
How many electrons per second enter bulb #2?
i2 = 7 electrons/s

Explanation / Answer

Whitney,

First let's calculate the resistance of each filament. We start by calculating the material property known as electrical conductivity, and then use the geometry of the filament to calculate its resistance.
Conductivity = qn. q=1.602e-19 coulomb is the charge on an electron. n=electron concentration, mu is the electron mobility.

=1.728e7 per Ohm-metre, and resistivity = 1/ by definition.

Now resisitance=L/A, A is the cross-section of the filament = r^2, r being the radius of the filament, giving (please check my arithmetic) R=8.84 Ohms.

Let J be the current going through the batteries. Let I2 and I3 be the currents through bulbs 2 and 3. Charge conservation now says J=I2+I3 since charge can neither be created nor be destroyed (and this is more subtle: nor can accumulate, under DC conditions) at the branching point.

Let V' be the voltage across B1. The first energy conservation we use is that power generated by battery = power dissipated in resistors.In general, power = accelerating voltage times current. Two batteries in series develop 2.8 Volt between their terminals. That means 2.8*J = V' * J + V'' * I2+ V'' * I3 = (V' + V'')*J using J=I2+I3 (charge conservation stated earlier) giving 2.8 = V' + V''.This is consistent with Kirchoff's voltage law, and ultimately comes from Maxwell's equations (vanishing curl of the electric field for static or zero magnetic fields).

The voltage across the parallel circuit (V'' say) must be the same no matter what path you take (Kirchoff's voltage law again, which is an expression for energy conservation), and the two parallel bulbs are identical, so V''=I2*R=I3*R. That gives I2 = I3 = J/2.

2.8 = V' + V'' = JR + (J/2)*R = (3/2)JR, J=0.21 Amp. We now calculate the voltage across each bulb: V' = 0.21*8.84= 1.86 Volt across B1; V'' = (0.21/2)*8.84= 0.93 Volt across B2 and B3; notice that V'+V'' = 2.8 Volt as required, to the accuracy we are working with. The electric field is given by potential difference / length.

E1 = 155 Volt/metre, E2 = 77.5 Volt/metre.

Finally, i1 = number of electrons per second = current/charge on an electron = J/q = 1.313e18 electrons per sec.

Similarly i2 = I2 / q = (J/2)/q = 6.6e17 electrons per sec.

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