lec18

lec18 - Scott Hughes 21 April 2005 Massachusetts Institute...

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Unformatted text preview: Scott Hughes 21 April 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 18: AC Circuits. Power and energy; Resonance. Filters. 18.1 AC circuits: Recap and summary Last time, we looked at AC circuits and found that they are quite simple to analyze provided we follows some simple rules: 1. Work with complex valued voltages and currents. Our driving AC EMF is usually some- thing like E ( t ) = E cos ωt ; replace this with ˜ E ( t ) = E e iωt (so that E ( t ) = Re[ ˜ E ( t )]). 2. The voltage drop across any circuit element obeys a generalized, complex version of Ohm’s law: ˜ V X = ˜ I X Z X where Z X is the impedance of circuit element X . Impedance works just like resistance, but is complex and frequency dependent: Z R = R Z C = 1 / ( iωC ) Z L = iωL . 3. Analyze the circuit as though it were a simple DC circuit containing only resistors as circuit elements. You may find phasor diagrams helpful for making sure that you get the magnitudes and phases correct. 4. Take the real part at the end of the day. 168 18.2 Power delivered to an AC circuit Let’s look again at our prototypical driven RLC circuit: C L R Assume that the EMF supplied is E ( t ) = E cos ωt . In the complex representation, this becomes ˜ E ( t ) = E e iωt . We can now solve for the complex current in this circuit: ˜ E = ˜ V R + ˜ V L + ˜ V C = ˜ I • R + iωL + 1 iωC ‚-→ ˜ I = ˜ E R + i [ ωL- 1 / ( ωC )] . We now substitute ˜ E ( t ) = E e iωt , ˜ I ( t ) = I e- iφ e iωt . Then, we find I = E q R 2 + [ ωL- 1 / ( ωC )] 2 ≡ E / | Z tot ( ω ) | where | Z tot ( ω ) | is the magnitude of the total impedance. The phase is defined by tan φ =- Im[ ˜ I ] Re[ ˜ I ] = ωL R- 1 ωCR . The real current flowing in this circuit is thus given by I ( t ) = I cos( ωt- φ ) 169 where I = E | Z tot ( ω ) | tan φ = ωL R- 1 ωCR . How much power is delivered to this circuit? From the definition of EMF as work per unit charge, and from the definition of current as charge per unit time, we know that the power delivered to this circuit must be P ( t ) = I ( t ) E ( t ) . Let’s plug in the results we have for I ( t ) and E ( t ): P ( t ) = E 2 | Z tot ( ω ) | cos ωt cos( ωt- φ ) = E 2 | Z tot ( ω ) | [cos ωt cos ωt cos φ + cos ωt sin ωt sin φ ] ....
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lec18 - Scott Hughes 21 April 2005 Massachusetts Institute...

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