lec09

lec09 - Scott Hughes 8 March 2005 Massachusetts Institute...

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Unformatted text preview: Scott Hughes 8 March 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 9: Variable currents; Th´evenin equivalence 9.1 Variable currents 1: Discharging a capacitor Up til now, everything we have done has assumed that things are in steady state: all fields are constant; charges are either nailed in place or else are flowing uniformly. In reality, such a situation tends to be 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 difference between its plates is V ; the charge separation is Q = CV . 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? Before thinking about this with equations, let’s think about what happens here physically. The instant that the switch is closed, there is a potential difference of V across the resistor. This drives a current to flow. Now, this current can only come from the charge separation on the plates of the capacitor: the excess charges on one plate flow off and neutralize the deficit of charges on the other plate. The flow of current thus serves to reduce the amount of charge on the capacitor; by Q = CV , this must reduce the voltage across the capacitor. The potential difference which drives currents thus becomes smaller, and so the current flow should reduce. We expect to see a flow of current that starts out big and gradually drops off. To substantiate this, turn to Kirchhoff’s laws: at any moment, the capacitor supplies an EMF V = Q/C . As the current flows, there is a voltage drop- IR across the resistor: Q C- IR = 0 . 80 This is true, but not very helpful — we need to connect the charge on the capacitor Q to the flow 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; more accurately, the charge separation is being reduced. In this situation, a large, positive current reflects a large reduction in the capacitor’s charge separation. Putting this into Kirchhoff, we end up with a first order differential 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 . At some later time t , it is a value Q ( t ). Our goal is to find this Q ( t ): Z Q = Q ( t ) Q = Q dQ Q =- Z t t =0 dt RC → ln " Q ( t ) Q # =- t RC . Taking the exponential of both sides gives the solution: Q ( t ) = Q e- t/RC ....
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This note was uploaded on 07/19/2010 for the course 8 8.022 taught by Professor Scotthughes during the Spring '10 term at MIT.

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lec09 - Scott Hughes 8 March 2005 Massachusetts Institute...

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