You only need to know three things 1 The time variation has to look like e tRC

You only need to know three things 1 the time

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the rules for charge on a capacitor without doing any math. You only need to know three things: 1. The time variation has to look like e - t/RC . 2. You need the initial ( t = 0) charge on the capacitor, and 3. You need the final ( t → ∞ ) charge on the capacitor. 83
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Then, you just need to figure out to hook everything together so that you get a formula that obeys the correct boundary conditions. For example, if a problem involves discharging a capacitor, you know that you start out with some initial value Q 0 , and that it must fall towards zero as time passes. The only formula that obeys these conditions and has the correct time variation is Q ( t ) = Q 0 e - t/RC , just what we derived carefully before. If it involves charging up a capacitor, you want a formula that has Q = 0 at t = 0, and that levels off at CV b as t → ∞ . Using 1 - e - t/RC as our “time variation” piece of the solution gets the t = 0 behavior right; multiplying by CV b ensures that we level off at the right value at large t . We end up with Q ( t ) = CV b 1 - e - t/RC · . Again, just what we derived carefully before. If you can remember these physical reasons why the charge behaves as it does, you can save yourself some pain later on. 9.4 Th´ evenin equivalence The RC circuits we’ve looked at so far are quite simple — there’s only one resistor, and it’s hooked up to the capacitor and/or battery in series. How do we handle a more complicated circuit? Consider the following: C V b R R 2 1 How does the charge on the capacitor evolve with time? This problem can be tackled quite simply by using Kirchhoff’s laws: we have multiple loops, so we go around them and write down equations balancing the voltage drops and EMFs; we make sure the currents balance at the junctions; math happens; we find Q ( t ). As an exercise, this is straightforward, but slightly tedious. 84
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There is in fact a very cute way to approach this, based on a theorem proved by L´ eon Charles Th´ evenin 1 : Th´ evenin’s Theorem: Any combination of batteries and resistances with two terminals can be replaced by a single voltage source V OC and a single series resistor R T . For our problem, this means that we can redraw the circuit as V OC R T s C The answer is now obvious! We just steal results from before, and write down Q ( t ) = CV OC 1 - e - t/R T C · . We just need to know how to work out V OC and R T . 1 I tried to find biographical information about this guy and totally failed. Fortunately, an anonymous Spring ’04 student was more persistant than me, and found a web bio: He was a French telegraph engineer, and hence very well versed in practical aspects of electrical circuits. Interestingly, “his” theorem was actually first derived by Hermann Von Helmholtz. 85
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To do this, we take the capacitor out of the circuit and consider what remains as a pair of terminals: V b R R 2 1 If we short circuit these terminals with a wire, a current I SC — the “short circuit” current — will flow. On the other hand, if we leave it open, there will be a potential difference — the “open circuit” voltage, V OC
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  • Spring '08
  • Covault

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