Physics 263 Experiment 1 LRC Transients 1 Introduction In this experiment we will study the damped oscillations and other transient waveforms produced in a circuit containing an inductor, a capacitor, and a resistance which are connected, in series, with a square-wave generator. The square-wave generator output is constant over a specific time interval. The time interval is adjusted so that the entire transient is contained in it. So the equation describing the circuit is one for which there is a constant applied voltage: L d2 q dt + Rdq 2 dt + q C = V 0 (1) Here L is the inductance, R the total resistance, C the capacitance, q the charge on the capacitor, and V 0 the (constant) applied voltage. Making the substitution Q = q CV 0, the equation becomes: L d2 Q dt 2 This equation could also be written + RdQ dt + Q C = 0 (2) m d2 x dt 2 + bdx dt + kx = 0 where we have substituted x for Q, m for L, b for R and k for 1/C. This is an equation for a simple harmonic oscillator with a frictional damping force proportional to the velocity. So we may expect this LRC circuit to have behavior similar to a damped simple harmonic oscillator. A derivation of the solution is given in the Appendix. In this laboratory, we will measure the transient waveforms, and compare the measurements with predictions. In this experiment, we use a signal generator with a rectangular output waveform, an example of which is shown in Figure 1
Figure 1: Output of a square-wave generator. At time t = 0, the voltage has been at -V 0 for a long time; any current from the previous transition has decayed to zero, and the voltage on the capacitor is -V 0. With these two initial conditions, the charge can be shown to be: q = CV 0 2CV 0 e αt (cos ωt + α sin ωt) (3) ω where q is the instantaneous charge, t is time, and ω is the angular frequency of the oscillation. 1 The constant α is α = R 2L (4) For no damping, the oscillation is at the natural angular frequency ω 0 = 1 LC (5) With damping, the angular frequency is less: or [ ] 1 R 2 ω = LC (6) 2L ω = ω 2 0 α 2 (7) Equation 3 describes a sinusoidal oscillation, as long as ω is real. (Remember that the sum of sinusoids, with different amplitudes and phases but the same frequency, is still sinusoidal.) 1 Recall that the frequency, f and the angular frequency, ω are related by ω = 2πf. 2
To keep ω real, Equation 7 tells us that the damping cannot be too large. We must have α < ω 0. For this case we have damped oscillations; an example is shown in Figure 2. Figure 2: Damped oscillations. If α > ω 0, the system is overdamped, and the oscillations disappear. The special case α = ω 0 is called critical damping. Examples of these cases are shown in Figure 3 Figure 3: Critical and overdamped oscillations. It can be shown that for critical damping, the current goes to zero faster than in the overdamped case. It is also possible to show that the current in the circuit is [ α i = 2CV 0 e αt 2 + ω 2 ] sin ωt (8) ω and the magnitude of the voltage difference across a resistor R is just ir. 3
Note that the current is determined entirely by V 0 and the circuit parameters L, R, and C. In this experiment, we will measure V 0, and the time dependent voltage across the load resistor. Using V 0 and the circuit parameters, we predict i with Equation 8, and the voltage across the load resistor, and compare the predicted and measured voltages. 2 Experimental Setup and Data-taking Ch 1 62.5Ω Ch 2 50Ω 0.8H.047µ F 100 Ω Figure 4: Schematic diagram of the apparatus. The 50 and 62.5 Ω resistors are the internal resistances of the generator and the inductor. Set up the circuit shown in Figure 4. The square-wave generator has the same effect as a battery or power supply being switched on and off. In calculating the decay constant α. the resistance R is the total resistance in the circuit, including the resistance of the signal generator, as well as those of the variable load resistor and the inductor. But the voltage drop across the load resistor is just ir load. 2.1 Damped Oscillations 1. Set the signal generator square wave amplitude to about 5.0V. Measure and record its value. Trigger the oscilloscope on Channel 1 (the signal generator output) and look at Channel 2, which is the voltage across the load resistor. 2. Adjust the frequency of the signal generator so that it is low enough that oscillations decay to zero before the voltage makes a transition to the opposite sign. 3. Adjust the timing and vertical gain to fill the display with 6 10 cycles of the oscillation. Then read out the scope with the Open Choice Desktop program. Making sure you have set the tabular display to include time, copy the file to the clipboard, and then paste it into a spreadsheet. 4
4. In the spreadsheet compute the predicted voltage across the load resistor, V Rload, using Equations 8, 6, and 4, and the fact that V Rload = ir load You will need to not include any negative times in order to make correct predictions. Make a plot which shows data and prediction together on the same graph. 5. For the underdamped circuit data, compute the rms deviation of the data from the prediction. This gives a quantitative measure of the agreement between data and prediction. Then make small changes in the parameters R, L, C, and V 0 and observe the resulting changes in the rms deviation. This is reasonable to do, since the nominal values may be off by 5-10%. Find the parameters which minimize the rms deviation. Give these results in your report. 2.2 Critical Damping Replace the load resistor wirh a variable resistor. Increase this resistance until critical damping is reached. Check to see if (R/2L) = (1/LC) Remember to disconnect the variable resistor from the circuit when measuring its resistance. Again, R in the above equation is the total series resistance in the circuit. 2.3 Overdamping Increase the resistance of the variable resistor above its value for critical damping. In your report, describe qualitatively what happens. 5