Physics 241 Lab: RLC Circuit – AC Source http://bohr.physics.arizona.edu/~leone/ua_spring_2009/phys241lab.html Name:____________________________ “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 frog To tell your name the livelong day To 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 desire I hold with those who favor fire. But if it had to perish twice, I think I know enough of hate To say that for destruction ice Is also great And would suffice. -Robert Frost Important: • 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 driven by a sinusoidal voltage source, the properties of the RC circuit and RL circuit that you studied previously combine in a straightforward manner. Let’s summarize the results for the LRC circuit that you should already suspect. The voltage across each component oscillates at the same frequency as the driving frequency of the source, ωdrive. The properties of the inductor and capacitor are frequency dependent, which makes the circuit respond differently to different driving frequencies. The resistor voltage is chosen as the voltage reference because it is the only Ohmic device and therefore can be used to determine the current in the circuit: ! V (t)resistor=RZVsourceamplitudesin"drivet( ) and therefore ! I(t) =1ZVsourceamplitudesin"drivet( ), where Z [Ohm] is the total circuit impedance, ! Z = R2+"L#"C( )2. Also, χC [Ohm] is the capacitive reactance ! "C=1#driveC and χL [Ohm] is the inductive reactance ! "L=#driveL. Is the current in the circuit increased or decreased when the difference between the capacitive reactance and inductive reactance is increased? Explain your answer. Note that ! "L#"C( )2="C#"L( )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"drivet +#source( ), where the source phase shift is given by ! "source= tan#1$L#$CR% & ' ( ) * . This shows that the source voltage will be in phase with the resistor voltage when ! "L="C because tan#10( )= 0. It also shows that if ! "L>"C, then #source> 0 (the source voltage leads the resistor voltage), and if ! "L<"C, then #source< 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? Use your graphing calculator to graph y=arctan(x) and find when y = π or –π. Your answer and reasoning:1.3. The capacitor voltage is phase shifted to lag the resistor voltage by 90o: ! V (t)capacitor="CZVsourceamplitudesin#drivet $%2& ' ( ) * + . The inductor voltage is phase shifted to lead the resistor voltage by 90o: ! V (t)inductor="LZVsourceamplitudesin#drivet +$2% & ' ( ) * . Sometimes these four boxed (important) voltage equations are rewritten using the fact that ! Iamplitude=Vsource amplitudeZ: ! V (t)resistor= R " Iamplitudesin#drivet( ) ! V (t)source= Vsourceamplitudesin"drivet +#source( ) (no change) ! V (t)capacitor="C# Iamplitudesin$drivet %&2' ( ) * + , ! V (t)inductor="L# Iamplitudesin$drivet +%2& ' ( ) * + If R = 10 Ω, L = 10 H, C = 10 F, Vsource amplitude = 10 volts and fdrive = 10 Hz, find the voltage amplitudes of the inductor and capacitor. Your work and answers: 1.4. It is useful to examine an example graph showing the relationships between each of the component’s voltages: In this example, χC is larger than χL so that Vcapacitor amplitude is larger than Vinductor amplitude. Note that Vresistor amplitude is larger than the other two voltage amplitudes, which indicates that R is larger than χC and χL. This need not be the case since you could always choose to use a smaller resistor in your circuit. Since the capacitor voltage lags the resistor voltage by π/2 while the inductor voltage leads the resistor voltage by π/2, the capacitor and inductor voltages themselves are 180o out of phase (π/2 + π/2 = π).Conservation of energy indicates that the sum of the voltages of the three components at any instant of time must equal the voltage of the source at that time. Therefore, if you add the three graphed voltages, you obtain the source voltage: ! Vsource(t) = VL(t) + VR(t) + VC(t). The sum of the three component voltage functions in the previous graph is shown below with a dotted line. Notice that in this example the capacitor voltage amplitude is larger than the inductor voltage amplitude. This causes the source voltage to reach it’s amplitude a little after the resistor. Recall that ! "source= tan#1$L#$CR% & ' ( ) * and since χC is larger than χL, you obtain φsource < 0. This means that the source voltage lags the resistor voltage. At driving frequencies where χL > χC, you find that VL > VC, and this causes the source voltage to lead the resistor voltage. Finally, imagine that you are able to adjust the source frequency while leaving Vsource,amplitude constant. Since ! Iamplitude=Vsource amplitudeZ, the current in the resistor can be maximized by minimizing Z. Remember that Z depends on ωdrive since both χC and χL each depend on ωdrive. Since ! Z = R2+ "L# "C( )2, Z is minimized when χC is equal to χL so that χL – χC, = 0. This gives Zminimum = R. The driving frequency at which χL – χC, = 0 occurs is called the resonant frequency. This can be found by setting χC equal to χL so that ! 1"driveC="driveL. Some algebra (that should appear in your lab report) gives ! "driveresonance=1L # C. 1.5. Imagine that you have an RLC circuit being driven sinusoidally at resonance. Assume you have placed the inductor voltage and capacitor voltage on the
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