Simple Nature - Light and Matter

Resonance with damping example 33⊲ What is the amplitude of the current in a series LRC circuit?⊲ Generalizing from example 32, we add a third, real impedance:|Ĩ| = |Ṽ ||Z |==|Ṽ ||R + iωL − i/ωC||Ṽ |√R 2 + (ωL − 1/ωC) 2ag / Example 34.This result would have taken pages of algebra without the complexnumber technique!A second-order stereo crossover filter example 34A stereo crossover filter ensures that the high frequencies go tothe tweeter and the lows to the woofer. This can be accomplishedsimply by putting a single capacitor in series with the tweeter anda single inductor in series with the woofer. However, such a filterdoes not cut off very sharply. Suppose we model the speakersas resistors. (They really have inductance as well, since theyhave coils in them that serve as electromagnets to move the diaphragmthat makes the sound.) Then the power they draw isI 2 R. Putting an inductor in series with the woofer, ag/1, givesa total impedance that at high frequencies is dominated by theinductor’s, so the current is proportional to ω −1 , and the powerdrawn by the woofer is proportional to ω −2 .A second-order filter, like ag/2, is one that cuts off more sharply:at high frequencies, the power goes like ω −4 . To analyze thiscircuit, we first calculate the total impedance:Z = Z L + (Z −1C+ Z −1R )−1All the current passes through the inductor, so if the driving voltagebeing supplied on the left is Ṽ d , we haveand we also haveṼ d = Ĩ L Z ,Ṽ L = Ĩ L Z L .The loop rule, applied to the outer perimeter of the circuit, givesṼ d = Ṽ L + Ṽ R .Straightforward algebra now results inṼ R =Ṽ d1 + Z L /Z C + Z L /Z R.At high frequencies, the Z L /Z C term, which varies as ω 2 , dominates,so Ṽ R and Ĩ R are proportional to ω −2 , and the power isproportional to ω −4 .616 Chapter 10 Fields