- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 1 (28.5 KiB) Viewed 60 times
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- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 3 (39.26 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 4 (29.86 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 5 (38.44 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 6 (38.44 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 7 (49.07 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 8 (34.95 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 9 (69.02 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 10 (31.95 KiB) Viewed 60 times
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- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 13 (25.94 KiB) Viewed 60 times
- Lc And Lcr Circuits Virtual Chalet Mart To The Same The Capacito They Hoc The Ht Pel Padre 21 The The The Other Relati 14 (25.97 KiB) Viewed 60 times
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EXPERIMENT 9. IC AND LCR CTROUITS (VIRTUAL -- it it ojo 14 - JE -HF do - --- PETE PRETI Pure 9.3 The PET LC tanke. Note that is the trio open and S15 is closed during the ring of the cit. The value the capacitors Indoctor en beated by clicking on the compete where the term in the Sit bacut the storie de Subtitate in the list also be (240-6-7)+740) - 46-6)
and LCR Circuits (Virtual) Introduction The analysis of the RC and RL circuits resulted in an exponential transient response followed by a steady state. The LC circuit, on the other hand, demonstrates an oscillating current and charge. In this simulated experiment a LC circuit will be examined and compared to the theoretical results, LC Theoretical Considerations + -- C d those Figure 9.1: The LC tank circuit (parallel). A tank LC circuit is illustrated in Figure 9.1. The capacitor is initially charged. Qat time t0when the switch is first closed. The instant the switch is closed the capacitor "Recall that the charge as a function of time across a capacitor for the series RC combination across ** potential source, Vs was found to be q = cv. (1-0) and for the discharging capacitor acto a resistor wisq = C/RC The current through a RL circuit connected to a source of Em/ is given by i= f(--/. -RC
starts to discharge through the inductor, nis a current. Some time later the capacitor has fully discharged and the potential across both the capacitor and inductor is zero. However, the current through the inductor has generated a magnetic field around the coils. The inductor's magnetic field begins to decrease (once the charge has completely gone through the inductor) but it then induces a current to maintain the magnetic field in the same direction is the original current. This current begins to decrease as the magnetic field decreases, however, this current charges the capacitor and now the right plate is positively charged, + (opposite in polarity as the initial charge). This eyclic process repeats in the opposite direction through the circuit and if there are no resistive losses the cycle will continue indefinitely. To generate the differential equation note that the Kirchhoff's loop rule and node rule imply that +0=0, and (9.1) ic The other relationships that will assist include: di dr > Cuci and di dec di Taking the derivative of the voltage relationship and then substitute in the second set of relationships results in; LC (9.3) The solutions to this differential equation can be written in the form: 1 Icost+o), and -wllesi(1) + dra -0 di di where the One cleant method to solve this equations is to apply the Laplace transform, a method that is beyond the scope of this course. However, I would like to encourage you to evaluate the derivatives of the functions and to verify that they
011 dz --WLI, sin(wt+) where, w to One elegant method to solve this equations is to apply the Laplace transform, a method that is beyond the scope of this course? However, I would like to encourage you to evaluate the derivatives of these functions and to verify that they indeed satisfy the previous equations. The angle o is the phase angle and depends on initial conditions of the system. In the example that was introduced above (and used in this laboratory) incorporated a fully charged capacitor, which implies that o 90° (the current is initially wro while the potential is a maximum). Using some trig identities At the end of this handout a brief derivation of the two equations will be given 50 EXPERIMENT 9. LC AND LCR CIRCUITS (VIRTUAL) one could demonstrate that when = - the current and potential are given by the following functions; i losin(wt), and wll.cos(wt)
Tunctions; Um verwone i = 1, sin(wt), and v=WLI, cos(wt). (9.5) Materials . Computer or Tablet • Phet Simulation (see the Bb site for the appropriate link) Description LC Tank Circuit A schematic of the circuit necessary to demonstrate the resonance in the LC tank circuit is illustrated in Figure 9.2. The circuit consists of a voltage source, V. = 10V, a capacitor, C, an inductor, L, and two switches. SW. and SW2. Construct the tank circuit with the charging battery, as illustrated in Figure 9.3. The voltmeter is connected at the points a and b which is giving the potential across the two devices. The ammeter has a single connection to the left of the inductor. The resulting circuit should be similar to what is illustrated in Figure 9.3. With the two switches open, set the values of C and L to the quantities given in the first trial of Table 9.1. Close SW, and you should see the voltage on the chart go to 10V. Open SW and you might wish to step or toggle (rather than run) the simulation until the current reading is located at one of the grid lines. Close SW2 and allow the simulation to run. You may wish to stop the simulation at a convenient point to read off the munber of cycles during an obvious time interval. In Figure 9.2 illustration you can see that there are two complete cycles in round 2.8 seconds. Dividing the number of cycles by the time interval is the frequency of the signal; cycle +0.714H. The angular frequency would bew=2f = 4.49 Results
the time interval is the frequency of the signal = 0.714H:. The angular frequency would be w = 2*/ = 4.49 2.8 Results Laplace Transform Optional Equation 9.3 is a linear second order differential equation with initial values for the current and the derivative of the current. The solution to these types of equations can be turned into an algebra problem by using what is known as Laplace Transform. The Laplace transform of the current function, i, and its derivatives, 1" and /', can be rewritten in terms of the Laplace transform of the current function, the term, and the
SW Alle V. SW 200 Fre 9.2: The schematie of the LC tanker Trial LID CF Wolbydes C) 1 010 000 2 3 5 T.1: The LC tanker initial conditions (0) and (0), lo this are the two in conditie (0) - do) where the initially on the potential to the device by Solving for the Laplace tom of the terms of the initial conditions and the polysoinsbrale manipulation much partial fraction des the reformeults the desired fet,( Starting with the dieta quos primes to decide the derivatives to simplify the pic d'
Table 9.1: The LC tank circuit results. initial conditions, i(0) and (0). In this particular case, the two initial conditions are (0) - and 7(0) = where the current is initially zero and the potential across the devices is given by: 1 = Solving for the Laplace transform of the current, CE), in terms of the initial conditions and > and then possibly some algebraic manipulation such as, partial fraction decomposition, the inverse transform results in the desired function, i(t). Starting with the differential equation (using primes to denote the derivatives to simplify the process); d + wi-0, di - +21[/"}+wºcx) -0, [vc{1} - wi(0) - (0] + [ wc(0) -
-10 allt to jo1411 2 CCIAC I PRET Figure 0.3: The PET LC tank cretion Notice that the illustration open and SW, is closed during them the circuit The voice of the capacitar and inductor can be adjusted by clicking on the com wure the term in the fint beacket other to be the oldesti? Substitute in the initial conditioned for 2419-o- - 20-6) -) We thereof the function of taken from and
bracket is the substitution for the second derivative Substitute in the initial conditions and solve for C; العة [20-0- Mount + [%c{0} = 0 Z - - Oto sin(wt) - min (1) where, the inverse transform of the function of was taken from standard tables and in this process. The voltage function can be found by di di d lä (sin(t)) Icon(w) = cost) di notice the relationship between the maximum current and the muscum voltage time involves the resoinnt frequency of the circuit, The Laplace transform of the first derivative at the function would be the Bin 2000 from which the second derivative for can be determined