Worksheet #12 - PHY102 (Spr. 2008)AC Circuits : Transients and resonanceDue Friday April 11th 6pmIn earlier worksheets we have studied the behavior of damped, mass-spring systems. We also took a brief look at the linear and non-linear pen-dulum problems. The equations describing these problems also describe thebehavior of linear and non-linear circuits and there are many analogies be-tween mechanical and electrical problems that are often used in the physicsdiscussions. For example the concepts of underdamped, critically dampedand overdamped apply to electrical problems as well as to mechanical prob-lems. Similarly the concept of resonance applies to both mechanical andelectrical problems. In circuits, a resistor absorbs energy and acts as thedamping. In mechanical problems energy is transferred between potentialenergy and kinetic energy. In electrical problems energy is transferred be-tween energy stored in the magnetic field and energy stored in the electricalfield. In the case of circuits energy is transferred between capacitors andinductors.The relation between the magnitudes of the voltage and current in circuitelements L,R,C are as follows:VR= IR, VL= LdIdt, VC=1CZt0Idt (1)Kirchhoff’s voltage law states that the sum of the voltage drops around aseries LRC circuit must sum to zero, yielding the following equation,Vs− VR− VL− VC= 0 (2)Using the relations above between the current and voltages for L,R,C givenabove we find,Vs− IR − LdIdt−1CZt0Idt = 0 (3)Taking a derivative of this equation yields,dVsdt− Ld2Idt2− RdIdt−IC= 0 (4)1Recall that a damped mass-spring system is described by Fs− bv − kx = ma,so that,−Fs+ md2xdt2+ bdxdt+ kx = 0 (5)where Fsis the driving force on the mass-spring system. The analogies be-tween the mass-spring system and the electric circuit are evident from theseequations, that is, L acts like the mass, R like the damping and 1/C like thespring constant. The derivative of the source voltage acts like the drivingforce.Problems1. (Transients) Consider a series LRC circuit connected to DC voltageof 1V. A switch in the circuit is initially open and the capacitor is ini-tally uncharged. Consider closing the switch at time zero. Using DSolve,find the current in the RLC circuit as a function of time after the switchis closed. The natural frequency when R = 0 is ω0= 1/(LC)1/2. UsingL = 10mH, C = 1µF On the same graph, plot the behavior of i(t) forR = 50, 199, 500, to illustrate the cases of underdamped, critically dampledand overdamped circuits.2. (Resonance) Consider an LRC circuit driven by a steady sinusoidalAC source with amplitude 1V. Use the same values of L, C as in the firstproblem. Also set R = 10. Use DSolve to find i(t) when an AC voltagecos(ωt) is applied to the circuit, instead of the constant voltage of problem1. In this case, at long times, the current oscillates at the same frequency asthe applied voltage. The amplitude and phase of the current depends on thenatural frequency of the circuit and the applied frequency. Resonance occurswhen they are the same. Make a plot of the current as a function of timefor the case ω = 50s−1. Note the time at which the steady state behaviorsets in. By using “Table” to make a list of the values of the current in thesteady state regime, and then using “Max” to find the maximum current inthis regime, plot the dependence of the maximum current in the steady stateregime as a function of applied frequency. It is nicer to normalize this to thecurrent at resonance, forming the ratio imax(ω)/imax(ωR). It is easy to findthe maximum current at resonance as there, only the resistive part of theimpedance remains, so that imax= V0/R and the phase angle is
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