Physics 241 Lab: RLC Circuit – AC Sourcehttp://bohr.physics.arizona.edu/~leone/ua_spring_2009/phys241lab.htmlName:____________________________ “I’m Nobody! Who are you?”I’m nobody! Who are you?Are you nobody, too?Then there’s a pair of us – don’t tell!They’d banish us, you know.How dreary to be somebody!How public, like a frogTo tell your name the livelong dayTo an admiring bog!-Emily Dickenson “Fire and Ice”Some say the world will end in fire,Some say in ice.From what I’ve tasted of desireI hold with those who favor fire.But if it had to perish twice,I think I know enough of hateTo say that for destruction iceIs also greatAnd would suffice.-Robert FrostImportant:- In this course, every student has an equal opportunity to learn and succeed.- How smart you are at physics depends on how hard you work. Work problems daily.- Form study groups and meet as often as possible.. - Join professional organizations. - Physicists help people: science => technology => jobs.Section 1: The RLC Circuit 1.1. In a circuit where an inductor, resistor and capacitor (RLC) are connected in series and drivenby a sinusoidal voltage source, the properties of the RC circuit and RL circuit that you studiedpreviously combine in a straightforward manner. Let’s summarize the results for the LRC circuit thatyou should already suspect. The voltage across each component oscillates at the same frequency as the driving frequency ofthe source, wdrive. The properties of the inductor and capacitor are frequency dependent, which makesthe circuit respond differently to different driving frequencies.The resistor voltage is chosen as the voltage reference because it is the only Ohmic device andtherefore can be used to determine the current in the circuit:€ V (t)resistor=RZVsourceamplitudesin ωdrivet( ) and therefore € I(t) =1ZVsourceamplitudesin ωdrivet( ), where Z [Ohm] is the total circuit impedance, € Z = R2+ χL− χC( )2. Also, cC [Ohm] is the capacitivereactance € χC=1ωdriveC and cL [Ohm] is the inductive reactance € χL= ωdriveL.Is the current in the circuit increased or decreased when the difference between the capacitivereactance and inductive reactance is increased? Explain your answer. Note that € χL− χC( )2= χC− χL( )2.Your answer and explanation:1.2. The driving source voltage is given with a phase shift with respect to the resistor: € V (t)source= Vsourceamplitudesin ωdrivet + φsource( ),where the source phase shift is given by € φsource= tan−1χL− χCR ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. This shows that the source voltagewill be in phase with the resistor voltage when € χL= χC because tan−10( )= 0. It also shows that if€ χL> χC, then φsource> 0 (the source voltage leads the resistor voltage), and if € χL< χC, then φsource< 0(the source voltage lags the resistor voltage).Is it possible for the source voltage to ever be 180o out of phase with the resistor voltage? Useyour graphing calculator to graph y=arctan(x) and find when y = p or –p. Your answer andreasoning:1.3. The capacitor voltage is phase shifted to lag the resistor voltage by 90o:€ V (t)capacitor=χCZVsourceamplitudesin ωdrivet −π2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟.The inductor voltage is phase shifted to lead the resistor voltage by 90o:€ V (t)inductor=χLZVsourceamplitudesin ωdrivet +π2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟.Sometimes these four boxed (important) voltage equations are rewritten using the fact that€ Iamplitude=Vsource amplitudeZ:€ V (t)resistor= R ⋅Iamplitudesin ωdrivet( ) € V (t)source= Vsourceamplitudesin ωdrivet + φsource( ) (no change)€ V (t)capacitor= χC⋅Iamplitudesin ωdrivet −π2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ € V (t)inductor= χL⋅Iamplitudesin ωdrivet +π2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟If R = 10 W, L = 10 H, C = 10 F, Vsource amplitude = 10 volts and fdrive = 10 Hz, find the voltageamplitudes of the inductor and capacitor. Your work and answers:1.4. It is useful to examine an example graph showing the relationships between each of thecomponent’s voltages:In this example, cC is larger than cL so that Vcapacitor amplitude is larger than Vinductor amplitude. Note that Vresistoramplitude is larger than the other two voltage amplitudes, which indicates that R is larger than cC and cL.This need not be the case since you could always choose to use a smaller resistor in your circuit. Sincethe capacitor voltage lags the resistor voltage by p/2 while the inductor voltage leads the resistorvoltage by p/2, the capacitor and inductor voltages themselves are 180o out of phase (p/2 + p/2 = p).Conservation of energy indicates that the sum of the voltages of the three components at anyinstant of time must equal the voltage of the source at that time. Therefore, if you add the threegraphed voltages, you obtain the source voltage: € Vsource(t) = VL(t) + VR(t) + VC(t). The sum of thethree component voltage functions in the previous graph is shown below with a dotted line.Notice that in this example the capacitor voltage amplitude is larger than the inductor voltageamplitude. This causes the source voltage to reach it’s amplitude a little after the resistor. Recall that€ φsource= tan−1χL− χCR ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ and since cC is larger than cL, you obtain fsource < 0. This means that the sourcevoltage lags the resistor voltage. At driving frequencies where cL > cC, you find that VL > VC, and this causes the source voltageto lead the resistor voltage.Finally, imagine that you are able to adjust the source frequency while leaving Vsource,amplitudeconstant. Since € Iamplitude=Vsource amplitudeZ, the current in the resistor can be maximized by minimizingZ. Remember that Z depends on wdrive since both cC and cL each depend on wdrive. Since € Z = R2+ ΧL− ΧC( )2, Z is minimized when cC is equal to cL so that cL – cC, = 0. This gives Zminimum = R. The driving frequency at which cL – cC, = 0 occurs is called the resonant frequency. This canbe found by setting cC equal to cL so that € 1ωdriveC= ωdriveL. Some algebra (that should appear in yourlab report) gives € ωdriveresonance=1L ⋅C. 1.5. Imagine that you have an RLC circuit being driven sinusoidally at resonance. Assume youhave placed the inductor voltage and capacitor voltage on the two channels of your
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