Physics 217, Fall 2002 4 December 2002(c) University of Rochester 14 December 2002 Physics 217, Fall 2002 1Today in Physics 217: AC circuits Finish resistive circuit problem from last time The LRC circuit Energy in LRC, and the meaning of Quality4 December 2002 Physics 217, Fall 2002 2The third simplest AC circuit: LRCSuppose that, in this simple circuit, the charge on the capacitor is q0, and the currentI = 0, at t = 0. What is the chargeon the capacitor at later times?Set up first: Let upper plate have positive charge initially, and define Iflowing into this plate, so that With this choice of direction for I, the polarities for the potential differences across L and R are as shown. The magnitudes of the potential differences are LCRI+++.Idqdt=, , and (back EMF!). CR LVqCVIR VLdIdt== =---4 December 2002 Physics 217, Fall 2002 3LRC (continued)Now apply Kirchhoff’ssecond rule:and define two useful new quantities:so thatLCRI+++---220, or0,qdIIR LCdtqdqdqRLC L dtdt++ =++=00(natural frequen1,cy) (quality,)LQLC Rωω≡ =220020.dq dqqQdtdtωω++=Not to be confused with charge!Physics 217, Fall 2002 4 December 2002(c) University of Rochester 24 December 2002 Physics 217, Fall 2002 4LRC (continued)Many of you will recognize this as the equation of motion of a damped harmonic oscillator, and will be able to recite the solution without my help. Too bad, here it is anyway. Use an exponential trial solution:Thus2222,,.tttqAedqAe qdtdqAe qdtρρρρρρρ=====22000.Qωρρω++=4 December 2002 Physics 217, Fall 2002 5LRC (continued)This is just a quadratic equation, and its solutions areThe general solution to the differential equation is thus22200 0001141.22 2 2iQQ Q Qωω ωρωω =− ± − =− ± −  ()ωωωω=−+−+−− −2002001exp 1221exp 1 .22tqt A i tQQtBitQQ4 December 2002 Physics 217, Fall 2002 6LRC (continued)Now write , and apply the initial conditions, () ()==00 and 00:dqqqdt()()()()ωωωωωωωωωωω=+==− + +− − =−+ +−− −==+000000000000000,00, so220,222, and...2qABqdqiDA iDBdt Q QiDA iDq AQQiDA iDqQ()=−2112DQPhysics 217, Fall 2002 4 December 2002(c) University of Rochester 34 December 2002 Physics 217, Fall 2002 7LRC (continued)So the general solution becomes =− =−=+  0001, 1.22 22qqiiABqAQD QD()ωωωωωωω− =− − +   ++ − −     =− − ++      00000000001exp22 21exp22 211exp 1 1 .222 22iDt iDtqtiqt i DtQD QqtiiDtQD QtiiqeeQQD QD4 December 2002 Physics 217, Fall 2002 8LRC (continued)Rearrange the complex exponentials and you get sines and cosines:()ωωωωωωωω−−=− ++− =− + 000000000011exp22 211 122 21exp cos sin .22iDt iDtiDt iDttqt q e eQeeQD i itqDtDtQQD4 December 2002 Physics 217, Fall 2002 9LRC (continued)It is useful to consider the limit of high quality, Now use and note that since  1:Q=− ≅2111 1 to first order in ,2DQQ()ωωω =− + 00001so exp cos sin .22tqt q t tQQ()αβ α β α β−= +cos cos cos sin sin ,()()≅≅ 1, sin 1 2 1 2 and cos 1 2 1 to first order :QQQQ()ωω =− −  0001exp cos .22tqt q tQQPhysics 217, Fall 2002 4 December 2002(c) University of Rochester 44 December 2002 Physics 217, Fall 2002 10LRC (continued)This is a damped oscillation: less strongly damped the larger Q (or the smaller R) is. Its period is t0q−0q0Tω−02tQe()qtπω=02.T4 December 2002 Physics 217, Fall 2002 11Energy in LRC, and the significance of qualityAt the extrema of q, the current is zero, so the total energy of the circuit is stored in the capacitor, as  Between extrema, the peak charge decreases by a factor ofso the energy decreases by the square of this factor, or Energy is conserved, so the energy dissipated during one period of the oscillation is=2max2.WqCωπ−−=02,TQ Qeeπ−2.Qe()π−∆= −2max1.QWW e4 December 2002 Physics 217, Fall 2002 12Energy in LRC, and the significance of quality (continued)If thenThus an interpretation of Q emerges: 1,Qπππ−=≅ =∆−−−max211.22111QWQWeQππ==∆=maxmaximum stored energy22energy dissipated per periodmaximum stored energy.energy dissipated per