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 Subscribe to view the full document.

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 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 Subscribe to view the full document.

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  • Spring '08
• Covault

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