Problem 2:

1. What is the initial current supplied by the battery just after the switch is closed?
2. What is the current supplied by the battery a long time after the switch is closed?
3. What is the charge on each capacitor and the voltage across each capacitor a long time after the switch is closed?

1. Many people assumed that this was a complicated "Kirchoff's laws" problem. But we did not discuss how to apply Kirchoff's laws to circuits with capacitors (and notice as well that there is only one battery). Instead, this is exactly the same kind of problem as example 23-8 in Tipler, as the suggested problems 23-39 and 23-41, and as an example I did in the problem session, and does not require the full "machinery" of Kirchoff's laws.

   Let's ask ourselves what the capacitors are doing at time t=0 (the switch is just closed). Well, they are initially uncharged, which tells us that the charge happily wants to run onto them, so all the current will pass through them initially - they act as a short-circuit. Thus no current flows through the 30 ohm resistor at first! (as an aside: this can also be seen by noting that if Q=0 on the capacitors, then V=0, since Q=CV; thus the total V across the capacitors is zero to start. This must be the same as the total voltage across the 30 ohm resistor, as they are connected by bare wires. If the voltage across a resistor is zero, then V=IR says the current is zero).

   Well, if the capacitors can be replaced by a short-circuit (bare wire) and the 30 ohm resistor can be ignored this is just a battery in series with two resistors! Thus:

      I = V/R_eq = (30 V)/(5 ohm + 25 ohm) = 1 A

2. After a "long time" (i.e. when the capacitors are fully charged) they will no longer conduct current. Thus they now appear as an "open circuit", and all of the current will go through the 30 ohm resistor. The circuit is then a simple loop of three resistors in series and

      I = V/R_eq = (30 V)/(5 ohm + 30 ohm + 25 ohm) = 0.5 A

3. What about the charge on the capacitors? A depressing number of people decided that the voltage across each capacitor was 30 V (not true even if there were no resistors in the circuit, and certainly not true where there is voltage drop across each resistor...), and also managed to find different charges on the two capacitors. The two capacitors are in series with each other and therefore must have the same charge (as an aside: consider the "I" shaped piece of the circuit made from one plate of each capacitor and the wire joining them - it is isolated from the rest of the world, and so the charge on each plate must be equal and opposite - there is no way for extra charge to get onto this...). So, we recognize that for two capacitors in series we have

      1/C_eq = 1/C₁ + 1/C₂ = 1/3µF + 1/6µF

      C_eq = 2µF

   Now, how to find the voltage across the equivalent capacitance? It is just the same as the voltage across the 30 ohm resistor (look at the circuit again) which we can get from:

      V = I R = (0.5 A)(30 ohm) = 15 V

   so we have

      Q = C_eq V = (2µF)(15V) = 30µC

   This is the charge on each capacitor. The voltages on each are found from

      V₁ = Q/C₁ = 30µC/3µF = 10 V

      V₂ = Q/C₂ = 30µC/6µF = 5 V

   (of course the voltages add up to 15 V, as they should...

   Problem 3
   Test 2
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   College of William and Mary,
   Dept. of Physics
   armd@physics.wm.edu
   last updated: March 25 1998