Rectilinear motion Lab



RLC Circuits

OBJECTIVES

[pic]

Figure 1: A series RLC circuit

To study voltage and phase relationships in A.C. circuits with inductors, capacitors and resistors, like the one in Figure 1Figure 1.

APPARATUS

Digital oscilloscope

Differential amplifier box

resistor, inductor, capacitor

Function generator

Cables (3 BNC, 2 banana to alligator, 1 BNC to alligator, 1 two-ended alligator clip)

INTRODUCTION

Root Mean Square

In this lab, we will be using an AC voltage. AC differs from DC in that the voltage, and current, vary with time. When voltages vary with time in a circuit we need a new way of referring to them. If the signal is periodic, like a sine wave, the average value will be zero since equal amounts of time are spent with V > 0 and V < 0. For a periodic wave we can refer to the peak-to-peak value of the voltage (or current) to tell us the magnitude of the signal. Another often used value is the "rms'' value. "rms" is an abbreviation for "root-mean-square.'' This value is useful even when the voltage or current is changing but not as perfect sine waves. For example, the voltage used to provide power for most lights and appliances is roughly Vrms=110 Volts. Although the voltage and current may often be negative, the rms values are always positive. For a pure sine wave voltage, the peak voltage and rms voltage are related by [pic]. Thus the peak to peak voltage swing of a 110 volt AC wire is Vpeak=110([pic]≈155 Volts

[pic]

Figure 2: Phase relationship between I and VR

Phase

In addition to the rms or peak value of the voltage, we have to be concerned about the phase. Imagine a resistor connected to a function generator producing sine waves. When the voltage output is zero, the current will be zero and when the voltage output is at a peak the current through the resistor will also be at a peak (both because of Ohm's law, V=IR). We say that for a resistor, the current and voltage are "in phase." A graphs of the voltage across the resistor, VR, and the current, I, is shown in Figure 2Figure 2. See that the peaks and valleys fall at exactly the same time for the current and the voltage.

An inductor in a circuit is indicated by a symbol. As its symbol indicates, it is often not much more than a coil of wire (often around a core material to enhance the magnetic flux), like a solenoid. When current runs through the coil, a magnetic field is created in the coil and any core material. If a circuit is created with an AC voltage source and an inductor, the current going through the coil will change with the changing AC voltage. Of course that will change the magnetic field which is produced within the coil. According to Faraday’s Law, a changing magnetic flux through a coil will create an emf. The value of that emf will be equal to [pic], where L is a quantity called the inductance (measured in Henries), and it will depend on the geometry of the inductor (diameter, number of coils, whether it has a core). Kirchoff's law tells us that this emf plus the source voltage at any given time must be equal to zero, since these are the only two elements in the circuit loop.

[pic] or [pic]

Equation 1

From this, it seems that the current must be rising the fastest when the voltage is at its positive peak (when V is at its maximum, [pic] must be as well, since L is constant). Similarly, the current is falling the fastest when the voltage is at its negative peak. Finally, when the voltage is zero, the current is not changing ([pic]). Figure 3Figure 3 has a diagram of this behavior. Notice that the current reaches its valleys and peaks one quarter or a cycle after the voltage does. We say that the current in an inductor lags behind the voltage by 90°.

[pic]

Figure 3: Phase relationship between I and VL

[pic]

Figure 44: Phase relationship between I and VC

Finally, a capacitor can be connected to an AC source. The voltage across a capacitor varies with the charge (Q=CV) and the current, I, is [pic]. In this case, when a capacitor is uncharged and the voltage is just increasing from zero we will have the largest current. As the capacitor charges, the current will fall until the voltage changes polarity. This behavior is shown in Figure 4Figure 4. The terminology is that the current in a capacitor leads the voltage by 90°.

Impedance

The impedance of any part of a circuit is the ratio of the rms voltage across that part and the rms current though that part. Because impedance is defined as a ratio of voltage/current (the same as Ohm's law), impedance is measured in ohms. If (=2πf and we measure L and C in henries and farads, respectively, then it can be shown that the impedance XR of an resistance R is R and the impedance XL (in ohms) of an inductance L is XL =(L= 2πfL and the impedance XC (in ohms) of a capacitance C is XC = 1/(C= 1/2πfC. It can also be shown that the impedance Z (in ohms) of an RLC series circuit such as the one shown in Figure 1Figure 1 is

[pic] Equation 2

where, XL and XC are the impedances of the individual inductor and capacitor. The current running through the circuit will be

[pic]

Equation 3

In Equation 3 Equation 3, I and V can be either rms values, or peak-to-peak values. Since XL and XC depend on frequency, the current will also depend on frequency. If you were to measure the voltages across any of the elements in the RLC circuit, you would find:

VR = IR, VL=IXL, VC=IXC

Equation 4

Where I is calculated as in Equation 3Equation 3.

Phasors

[pic]

Figure 55: Phasor diagram

The phase relationship between voltages across the elements of an RLC circuit like that of Figure 1Figure 1 can be conveniently displayed in a phasor diagram. An example is shown in Figure 5Figure 5. In it, the current is represented by a vector, in this case along the positive x axis. The voltage across the resistor, since it is in phase with I, is pointing in the same direction. The counter-clockwise direction represents "leading", so the fact that the voltage in an inductor leads the current by 90° is represented by the VL phasor along the +y axis (it is 90° counter-clockwise from I). Likewise, the current leads the voltage on a capacitor, so the current phasor is 90° counter-clockwise from VC (VC is along the –y axis). The total voltage is the vector sum of VR, VC and VL, in this case the current leads the total voltage as well.

Resonance

The impedance Z of the RLC series circuit is a minimum for XL = XC. The frequency for which this occurs is called the resonant frequency. At this frequency, fr, the current through is a maximum, as can be seen in Equation 3Equation 3 above. In an ideal case, I=V/R. At resonance, the voltage VLC across the LC series combination is a minimum (it would be zero for an ideal inductor that had no resistance). Hence one can search for fr by varying the frequency and looking either for a maximum VR, rms signal or a minimum VLC, rms signal.

A search for the minimum has a practical advantage that near the resonant frequency, fr, one can increase enormously the detection sensitivity by going to maximum signal generator amplitude and also by going to higher oscilloscope gain.

Amplifiers

Our oscilloscope allows the simultaneous observation of two voltages. It would be nice to compare the voltage across the resistor and across the inductor/capacitor pair. There is a problem with directly connecting the oscilloscope, though. The [pic] symbol on Figure 1Figure 1 is called the "ground" (you've seen grounds before in the electroscope and Coulomb balance labs), and is connected to the third prong of the power cord of the function generator by the black lead from the generator. Unfortunately, one of the leads from the oscilloscope is also at ground. If it gets connected between the inductor and capacitor, for example, the capacitor will have ground on both sides (one from the generator and one from the oscilloscope). It will then have no voltage across it (the grounds are at the same potential), and will effectively be removed from the circuit ("shorted out").

We can use differential amplifiers to avoid ground problems. The amplifier takes 2 inputs, neither of which is at ground. These amplifiers give an output which is equal to twice the difference of the voltages at the inputs. To measure the voltage across the resistor, connect probes to the two ends of the resistor and plug them into the red and black jacks of an amplifier. Then take a BNC cable and connect the amplifier output to the oscilloscope.

Repeat the process for any other circuit elements you wish to see the voltage across. Then you can compare the signal across the resistor (displayed on one trace) with the signal across other circuit elements (displayed on the other trace). Since across a resistor the voltage and current are in phase, and since the current throughout a series circuit is the same, we can observe the relative phase relationship between the current and any other measured voltage.

PROCEDURE

1. Make sure you have a circuit like that of Figure 1Figure 1. (The function generator supplies the AC voltage.) The companent values should be about R = 5670 Ω (240 and 330 in series), L = 1 mH, C = 0.1µF.

2. Set the generator to the 50 kHz range and leave it there for the whole experiment. Connect the generator to the circuit (using a BNC to alligator clip cable) with the red lead attached to the resistor, and the black one to the capacitor. Set the generator to about 5 kHz.

3. Connect the output of the top amplifier to the oscilloscope channel 1 input using a BNC cable.

4. Find a cable that has one banana plug end and one alligator clip end. Put the banana plug end into the amplifier box. Clip the alligator clips to the same places you connected the function generator. When the amplifier is turned on, you should see the output from the function generator on channel 1.

5. If the signal is too large, the amplifier won’t be able to keep up (particularly at higher frequencies). Change the output of the generator until the voltage on the oscilloscope is about 1.4 V peak to peak or less. You will need to have the –20 dB button pushed in.

6. You will now want to connect the other amplifier up the same way (to channel 2 of the oscilloscope) so that you can use its alligator clips to check the voltages across R, L and C.

7. Use the digital measurements on the oscilloscope to measure Vrms, VR rms, VL rms, VC rms, and VLC rms at frequencies near 5kHz, 10kHz, and 15 kHz Is Vrms=VR rms+VL rms+VC rm? Explain what the relationship between them is.

8. Calculate the frequency for which resonance should occur for the RL and C values you have.

9. Use the oscilloscope to search for the resonant frequency fr by the following techniques and also estimate the uncertainty (fr in each measurement. Use the trigger/TTL output of the signal generator to trigger the oscilloscope.

a. As you vary f, observe the maximum in the current Irms through R. (Since Irms(VR rms, use a differential amplifier with oscilloscope to look at VR rms).

b. Observe the minimum in the voltage VLC rms across the LC combination. (Use the other differential amplifier to input VLC rms to the oscilloscope. To improve sensitivity, turn up signal amplitude and/or scope gain).

c. Produce a Lissajous figure using VR rms and V rms as the X and Y inputs. (Use the X-Y display setting). At resonance the usual elliptical pattern becomes a straight line as VLC rms goes to zero, so V=VR. Again turn up the gains to improve sensitivity.

10. Compare the calculated fr with your observed values. Is there agreement?

11. To measure phase relations, use the oscilloscope (plus a differential amplifier) to observe the voltage across the resistor, VR,. Short the capacitor with a wire (i.e. connect the two ends of the capacitor together with another wire so that there is no voltage across it). Use the second differential amplifier and oscilloscope channel to measure VL. Does the result look like Figure 3Figure 3?

12. Repeat step 11, but short out the inductor and measure VC instead. Does this look like Figure 4Figure 4?

13. Remove the shorting wire and use the oscilloscope to look at the total voltage V across the RLC combination and VR.

a. Set the signal generator near fr. At resonance the two signals should be in phase. Why? (If at f =fr the signals are 180° out of phase, the input polarity is wrong on one differential amplifier; interchange the input leads on one of the differential amplifiers.)

b. Find the resonant frequency by this method. Adjusting the gain and positioning the traces so the signals nearly overlap will help.

c. With f =fr, short out the capacitor with a wire. Note what happens. Unshort capacitor and then short the inductor. Explain results in terms of Figure 5Figure 5. Does current lead or lag the applied voltage in each case?

d. Remove the shorts. Vary the frequency to look at the signals for ffr. Explain the phase behavior.

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