Lec09 - Scott Hughes Massachusetts Institute of Technology...

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Scott Hughes 9 March 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Lecture 9: Variable currents; Th´evenin equivalence 9.1 Variable currents: Discharging a capacitor Up til now, everything we have done has assumed that things are in steady state: all felds are constant; charges are either nailed in place or else are Fowing uni±ormly. In reality, such a situation is the exception rather than the rule. We now start to think about things that vary in time. Suppose we have a capacitor C charged up until the potential di²erence between its plates is V 0 ; the charge separation is Q 0 = CV 0 . We put this capacitor into a circuit with a resistor R and a switch s : R s C What happens when the switch is closed? Be±ore thinking about this with equations, let’s think about what happens here physically. The instant that the switch is closed, there is a potential di²erence o± V 0 across the resistor. This drives a current to Fow. Now, this current can only come ±rom the charge separation on the plates o± the capacitor: the excess charges on one plate Fow o² and neutralize the defcit o± charges on the other plate. The Fow o± current thus serves to reduce the amount o± charge on the capacitor; by Q = CV , this must reduce the voltage across the capacitor. The potential di²erence which drives currents thus becomes smaller, and so the current Fow should reduce. We expect to see a Fow o± current that starts out big and gradually drops o². To substantiate this, turn to Kirchho²’s laws: at any moment, the capacitor supplies an EM³ V = Q/C . As the current Fows, there is a voltage drop - IR across the resistor: Q C - IR = 0 .
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This is true, but not very helpful — we need to connect the charge on the capacitor Q to the Fow of current I . Thinking about it a second, we see that we must have I = - dQ dt . Why the minus sign? The capacitor is losing charge. A large, positive current reFects a large loss in charge on the capacitor. Putting this into Kirchho±, we end up with a ²rst order di±erential equation: Q C + R dQ dt = 0 . To solve it, rearrange this in a slightly funny way: dQ Q = - dt RC . Then integrate both sides. We use the integral to enforce the boundary conditions : initially ( t = 0), the charge separation is Q 0 . At some later time t , it is a value Q ( t ). Our goal is to ²nd this Q ( t ): Z Q = Q ( t ) Q = Q 0 dQ Q = - Z t t =0 dt RC ln " Q ( t ) Q 0 # = - t RC . Taking the exponential of both sides gives the solution: Q ( t ) = Q 0 e - t/RC . The charge decays exponentially. After every time interval of
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Lec09 - Scott Hughes Massachusetts Institute of Technology...

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