AC Circuits Lab Manual

7/23/2019 AC Circuits Lab Manual

http://slidepdf.com/reader/full/ac-circuits-lab-manual 1/18

This lit bulb is part of an AC circuit. Thecoil of wire in the background would have noeffect on a DC circuit, but is a key componentof an AC circuit. It is called an “inductor.”

AC CIRCUITS

Physics 241/261

Fall 2015

1 Introduction

In this lab, we will conclude our study of Alternating Current (AC) electricity by looking indetail at how circuits behave when an AC voltage is applied. The previous labs have taught us thatAC is the natural choice for easy power generation and transmission. Therefore, it is important tounderstand how various types of circuit elements behave in the AC mode.

For our purposes, circuit elements can broadly be classified into three ideal devices: resistors,capacitors, and inductors, denoted by R, C, and L respectively. Many real-world devices can bemodeled as combinations of these. Electric motors, for instance, are mostly inductive in nature,with a non-negligible amount of internal resistance; we can model this as a series combination of an inductor and a resistor. Your toaster oven, on the other hand, is almost completely resistiveand can be modeled simply as a resistor. Finally, a radio has a tuning circuit that is inductive andcapacitive. When you turn to a specific station, you are varying a capacitor in the radio tuning

circuit so that it is in resonance with the broadcast frequency of the station.An important property of resistors is that if an alternating current flows through them, thevoltage drop across them is proportional to the current. In Sections 5-7 you will discover thatthis is not true for capacitors and inductors; while the voltage drop is still a sine wave, it is phase

shifted relative to the current. To better understand the behavior of the AC circuits, we will beusing phasors to represent voltages and currents. While some textbooks may contain a definitionof phasors, the discussion of their use is typically limited, so we have included a discussion of theiruse in Section 2 (note also that conventions may vary between textbooks.)

Many systems have a “natural” frequency at which they tend to oscillate. Examples you have

1



previously encountered in mechanics include a mass on a spring and a child on a swing set. Whensuch a system is driven by some external source at or near its natural frequency, it can absorbenergy very efficiently, causing the amplitude of the oscillations to increase rapidly. This is thephenomenon of resonance, and it is familiar to anyone who has ever made a swing go higher andhigher by moving their legs and body back and forth at just the right rate.

What would it mean for an electrical system to experience resonance? We would expect to

see the “response” of the system increase as the system is driven near its natural frequency by anexternal AC voltage. In experiment 9 we will see that precisely this phenomenon can occur in aseries RLC circuit. To characterize the response of the circuit, we will measure the voltage acrossthe resistor. The voltage is directly related to the power dissipated (P = V 2/R), so a large voltageacross the resistor means that the driver is very efficiently delivering power to the circuit. In thisexperiment, we will use a light bulb as a resistor. Its brightness will give you visible confirmationof resonance.

2 Theory

Before starting this lab, you should be familiar with the following physical concepts. If you needto review them, or if you haven’t yet discussed them in your lecture course, consult the indicatedsection in Young & Freedman, Physics.

• Inductors, §30.2

• Reactance and phasors, §31.1-3

• Resonance in RLC circuits, §31.3,5

• This experiment will use an oscilloscope extensively. If you are even slightly uncomfortablewith oscilloscope measurements, we strongly recommend that you review the principles of thier

operation as described in the Faraday’s Law lab.

2.1 Representing Voltages and Currents with Phasors

In this lab write-up we adopt a convention where the current flowing through an AC circuit atany given time t is

i(t) = I max sin(ωt), (1)

where ω is the angular frequency (in radians per second), which is related to the frequency, f , (inhertz) by ω = 2πf . When you see ω, think “frequency,” since it is angular frequency and is just anotational way of avoiding writing 2πf many times.

In AC, by our convention, the voltage across an inductor, capacitor, or resistor is also a sinefunction and has the form

V (t) = V max sin(ωt+ϕ), (2)

where ϕ is called the phase angle. Since AC voltage and current cycles are continuous oscillationswe arbitrarily set the current to start at zero when t = 0 as in Equation 1. The voltage oscillationmay be at a different point in its cycle so we account for this with a phase angle 1.

1Our timing convention of starting the current on the x axis at t=0 is not universal, in fact, we differ from mosttextbooks here. It is usual to adopt a timing convention setting the voltage at zero when t = 0 and then to putthe phase angle in the expression for the current. Instead, we put the current at zero at t = 0 and then put the

2



x

y

A

θAy

Figure 1: Taking the y axis projection of a vector.

Since the voltage across a circuit element can be out of phase with the current flowing throughit, keeping track of the voltages and currents in AC can be quite a chore. In order to make it easier,the representation of voltages and currents by means of phasors is helpful. A phasor is a vector of constant length, centered on the origin and rotating around the origin at constant speed. For anyvector A, Ay = A sin(θ), where θ is the angle above the x axis made by A, as shown in Figure 1.

If A rotates around the origin at a constant rate then θ = ωt, the projection of A on the y axis is

Ay = Asin(ωt).We can express current as a vector-like quantity (a phasor). In this representation the real

current (physically the rate of charge flow) is the y axis projection of the current phasor. The tail

of the current phasor is fixed at the origin and the head of the phasor rotates counterclockwise with

angular frequency ω. Some “snapshots” of this phasor rotation are shown in Figure 2.We also can express the voltage as a phasor. Both voltage and current phasors rotate around

the origin with the same angular velocity ω. The angle between these is ϕ and it remains constantas the phasors rotate. Note: in this lab, we plot phasors as “snapshots” at t = 0. We now explorethe AC behavior of each of the devices below making use of phasors.

It is important to note that in many textbooks you would represent voltages and currents as

phasors and take the x-axis projection. This is merely a convention and it should not confuse you.We simply use the y-axis projection due to our experimental apparatus. If you are interested inwhy this is necessary, you can ask your instructor (it is due to the triggering mechanism of theoscilloscope).

2.2 Resistors in AC

Resistors under AC behave in the same way as they do under DC—they resist the flow of current,and in so doing, they dissipate some of the electrical energy as heat. A resistor does not care whichway the current happens to be flowing at the moment; only that the current is flowing. Since thevoltage and current in a resistor obey Ohm’s law:

V max = I max R, (3)

V max is the maximum voltage across the resistor, I max is the maximum current flowing through theresistor, and R is the resistance. In a resistor, the applied voltage and the current are in phase,

ϕR = 0. That is, when the applied voltage is a maximum, so is the current; when the voltage is

phase angle in the expression for the voltage. We choose this convention because we are dealing with a series RLCcircuit in which the current is the same for each element. Since the current is the common quantity in this lab, itis straightforward to reference the voltages to it. Switching between the two starting time conventions is simple aslong as you understand each. Refer to a textbook for a complete discussion.

3



Figure 2: Rotation of a phasor that represents the current flowing through some circuit element.The y axis projection of the phasor corresponds to i(t) at that given moment.

4



Figure 3: Phasor diagram for a purely resistive circuit. Notice that the phase difference is 0.

reversed, so is the current; and when the voltage is zero, so is the current. This is shown in a phasordiagram in Figure 3.

In a resistor, power is dissipated by heating. As we have seen in previous labs, for a resistiveelement, the average power dissipated per cycle is

P avg = I RMS V RMS . (4)

This is most conveniently expressed as P = I 2RMS R. Remember that the RMS values for voltageand current are used in AC. RMS values can easily be found from the maximum values, for examplethe RMS voltage is V RMS = V max /

√ 2. Resistors in AC act very much like resistors in DC. They

obey Ohm’s law and the power dissipated is easily obtained as the product of the RMS voltage andRMS current.

2.3 Capacitors in AC

Capacitors, however, are slightly more complicated. We say that a capacitor reacts to (asopposed to resists) the flow of AC current. Let’s look qualitatively at what happens in the capacitor.When the voltage across the capacitor is zero, there is no electric field between its plates (recallthat for a parallel-plate capacitor, V = Ed), and no charge stored on the plates (Q = CV ). It is atthis point that it is easiest for charges to flow onto (charge) the capacitor’s plates, so we say thatthe current to the capacitor is a maximum. Later, when the voltage across the plates is large, thereexists an electric field against which it is difficult to move more charges from the plates. Whenthe voltage across the capacitor is at its maximum, the current is zero. When the voltage startsto decrease, the stored electric field dissipates, and charge flows (but in the discharging direction).This process repeats with the signs reversed and the AC cycle is completed.

We can restate this result simply: the voltage and current for a capacitor are 90

out of phase,and the current leads the applied voltage. That is, when the applied voltage is going through zerofrom low to high, the current is a maximum; and when the applied voltage is a maximum, thecurrent is zero, but is changing direction. This phase relationship is shown diagrammatically inFigure 4.

In addition to the phase relationship between voltage and current, we need a relationship betweentheir maximum values. It turns out that there is a simple one between the maximum voltage appliedto a capacitor, and the maximum current through the capacitor in AC. It is usually written as:

V max = I max X C , (5)

5



Figure 4: Phasor diagram for a purely capacitive circuit. Notice that the phase difference is −90.

where X C is called the capacitive reactance, and is a function of both the capacitance and thefrequency of the AC power source. Let’s look at the dimensions of X C . Since Equation 5 looksastoundingly similar to Ohm’s law, X C must be in ohms (in fact, it is helpful to think of X C asan “R-like quantity”). Recalling that farads can also be expressed as seconds per ohm, we suspectthat X C has some sort of inverse dependence on capacitance. Given that, we must have an inversedependence on frequency to make the dimensions work.

This makes some intuitive sense when you think about the DC case: a discharged capacitorallows current to flow freely onto its plates, while a fully charged capacitor completely hinders it.If the frequency of the power source were quite high, the capacitor, to an excellent approximation,would be fully discharged. If the frequency were zero, the capacitor would always be almost fullycharged and X C would be infinite. This qualitative argument is, in fact, correct and the exactdependence is

X C = 1

ωC . (6)

Thus, at low frequency a capacitor inhibits the flow of AC current but at high frequency it passesAC.Finally, in a capacitive circuit element, the instantaneous power is the product of the instan-

taneous current and voltage. Unlike a resistive device this product can be positive or negativemeaning that the capacitor can store and return energy to the circuit. This results because thecurrent leads the voltage by 90. For half of the AC cycle, power is supplied to the capacitor, butfor the other half cycle, power is derived from the capacitor. The average power drawn over a fullAC cycle for a capacitor is:

P avg ,C = 0 (7)

This is not surprising since we think of a capacitor as a device by which we can store energy in the

form of an E field and retrieve the energy without losses.

2.4 Inductors in AC

Let’s now consider an inductor in an AC circuit. As you may recall from your text, inductors canstore electrical energy in their magnetic fields. If the strength of this field should change becausethe electrical current through the inductor changes, the inductor will transiently have a potentialdifference across its terminals (this potential difference is called the back-EMF ). Inductors alwaysoppose the change in current in the circuit (Lenz’s law); thus, if a current drops or rises, the inductorsupplies a countering or back-EMF.

6



Figure 5: Phasor diagram for a purely inductive circuit. Notice that the phase difference is +90.

Let’s look qualitatively at what happens to the potential drop across an inductor when an ACcurrent is flowing through it. When the current flowing through the inductor is (instantaneously) atits maximum value, there is no change in the magnetic field (since, at that instant, the current isn’tchanging). By Faraday’s law, this means that there is no potential difference across the inductor’sterminals. As the current gets smaller, the magnetic field within the inductor also gets smaller.This change is opposed, causing a negative potential “drop” across the inductor (i.e., a positivepotential difference) since the inductor will try to maintain the previously higher levels of current.This voltage drop becomes the most negative when the instantaneous current through the inductoris zero. The current (and thus the magnetic field) changes directions. This process then continueswith the signs reversed and the full AC cycle is completed.

We summarize the timing relationship between current and voltage with reference to the phasorsby saying that the current and voltage are 90 out of phase. This is shown in Figure 5. In aninductor, the voltage leads the current, which means that the phase relationship between voltageand current is exactly opposite of the capacitor.

In addition to the phase relationships, there is a relationship between the maximum voltagedrop across the inductor and the maximum current that flows through it in AC. This is usuallywritten:

V max = I max X L, (8)

where X L is known as the inductive reactance, and is a function of both the inductance and thefrequency of the AC power source. Examine the dimensions of X L and note that it varies directlywith the frequency and current and its units are ohms (once again, it is helpful to view X L as an“R-like quantity”).

Here also, we use the limiting cases to see what ought to happen at the extremes of frequency.In the high-frequency case, the magnetic field in the inductor is always changing, meaning that X L

is high. If the frequency is zero the magnetic field doesn’t change, and X L is near zero. This turnsout to be correct and in fact,X L = ωL, (9)

where L is the inductance. Thus, at high frequency an inductor inhibits AC current but at lowfrequency it passes AC.

In terms of AC power, as in a capacitor, an inductor can both receive and deliver energy to andfrom a circuit. Since the voltage leads the current by 90 and the instantaneous power is P = IV ,the power is positive for half the cycle and negative for the other half cycle. The average power

7



over a cycle is thus:P avg ,L = 0 (10)

As we will see in the experiments section, the fact that the average power delivered to an inductorand capacitor is zero will have important consequences for AC circuits composed of series or parallelcombinations of resistors, capacitors, and inductors.

2.5 Phasor Addition in a Series RLC Circuit

We can even keep track of the behavior of the instantaneous voltage and current for an entirecircuit using the phasor representation. For now, we will concentrate on how this is done for a seriesRLC circuit. Since the resistor, the inductor, and the capacitor are in series, Kirchhoff’s law tells usthat the same instantaneous current is flowing through every element of the circuit, meaning thatthe current phasor is at the same angle with respect to the x axis for every element. To find thevoltage phasor for the entire circuit, you add all of the voltage phasors associated with each element

as if they were vectors. (See Figure 6 for an example.) This net voltage phasor will also maintaina fixed phase angle with the current phasor (Why?). We can project this resultant phasor againstthe y axis to obtain the instantaneous voltage across the entire circuit.

Since the sum of the voltage phasors is itself a phasor (refer to your textbook), the magnitudeof this phasor is given as:

V = V 2R + (V L−V C )2. (11)

There is a relationship between the maximum voltage across the series RLC circuit and the current.

V max = I max Z (12)

where Z is the impedance of the series circuit. It is easy to show from current voltage relationshipsabove (Equations 3, 5, and 8) and phasor addition that Z is given by

Z = R2 + (X L−X C )2. (13)

This is the equivalent “R-like quantity” for the entire series RLC circuit. Similarly, the phase anglefor the series circuit is given by

tanϕ = X L−X C

R . (14)

We will refer to these equations as we explore voltage and current relationships for the RLC circuitsin the experiments below.

2.6 Remembering Voltage-Current Phasor Relationships

You must spend some time memorizing which phasor (current or voltage) leads which in eachsituation in order to be completely fluent when discussing AC voltage and current responses. Itmay be helpful to remember the mnemonic “ELI the ICEman ” for this purpose. In the mnemonic,E is the electromotive force, or voltage, and I is the current. E leads I for inductors and I leadsE for capacitors.

8



Vmax, L

( = Imax XL)

Vmax, R

( = Imax XR)

Vmax, C

( = Imax XC )

Vmax, L

Vmax, R

Vmax, C

Vmax ( = Imax Z)!

Figure 6: Addition of phasors associated with an ideal resistor, and ideal capacitor, and an idealinductor at a time corresponding to when the shared current phasor is crossing the x axis. (Thecurrent phasor is deleted for the sake of clarity in the figure.)

3 Equipment

• oscilloscope (HP-54603B)

• audio driver (Daedelon model EG-50)

• RLC circuit board (PASCO model CI-6512)

• digital multimeter (DMM) (Meterman 15 XP)

• assorted banana cables

3.1 Initial Setup of the Audio Driver

The audio driver will provide a source of AC voltage in this experiment. It is quite similar tothe function generator that you used in the Faraday’s Law experiment. A principle difference isthat it only provides sine waves, not triangle or square waves.

1. Connect the two outputs on the front of the audio driver (see Figure 8) to points (a) and (d)on the RLC circuit board. In this configuration, the audio driver will provide an alternatingcurrent to the 33 Ω resistor, 8.2 mH inductor, and 10 µF capacitor.

2. Turn on the audio driver, and make sure that the range selector (the knob on the left) is setto 500 Hz.

3. Adjust the frequency selector (the knob on the right) until the audio driver is set to about

850 Hz.

3.2 Initial Adjustment of the Oscilloscope

At this point, turn the oscilloscope on. At present, the oscilloscope isn’t measuring anything, sonow is a good time to make sure that it is set up properly. We’ll do this in two parts. First, we’llmake sure that the oscilloscope is properly grounded and set to trigger:

1. Connect the ground on the oscilloscope (labeled “ ”) to point (a) on the circuit board.

9



Bulb (d)

S t e e l C y l i n d e r

(a)

(b) (c)

(e)

Inductor

Resistor

(f)

C a p a c i t o r

Figure 7: RLC Circuit Board with points (a)–(f) marked and important components labeled. Thewhite lines are printed on the board, and indicate electrical connections inside the board.

Range elector Frequency ontrol

Amplitude Control

Figure 8: Audio driver with controls labeled.

10



2. Connect the oscilloscope terminal labeled “External Trigger” to point (b) on the circuitboard. We will use the voltage across the resistor, which is in phase with the current in theentire circuit, to trigger the oscilloscope.

3. Push the “Setup” key, then the screen key labeled “Default Setup”.

4. Turn the “Volts/Div” knob until input channel 1 is on the 2.00 V scale. For the rest of this

experiment, do not change the “Volts/Div” setting without consulting your GSI! If the twochannels have different settings, your data will be unusable.

5. Turn the “Time/Div” knob until you are on the 500 µs scale (not the 500 ms scale!).

6. Push the “Source” key, then the screen key labeled “Ext.” This tells the oscilloscope to usethe “External Trigger” signal for triggering instead of the signal on Channel 1.

7. Push the “Slope/Coupling” key, then the screen key “Reject HF.” This tells the oscilloscopeto ignore high frequency (HF) noise in the trigger signal.

Your oscilloscope is now configured to display the voltage at various points on your circuit board,

but is does not have a signal to display. We will now connect channels “1 X” and “2 Y” to theRLC circuit board. Figure 9 contains a diagram of the completed setup.

1. Connect the oscilloscope terminal labeled “1 X” to point (b) on the circuit board. You shouldsee a sine wave on the screen. The oscilloscope is plotting the voltage difference between theterminal “1 X” and ground.

2. Adjust the “Volume” knob on the audio driver until the maximum voltage across the resistoris 4 V.

3. Connect the oscilloscope terminal labeled “2 Y” to point (a) on the circuit board. This should

not change your display.

4. Push the “2” key until “On” is highlighted. You should now see a fuzzy horizontal line acrossthe middle of the screen. Turn the “Volts/div” knob until input channel 2 is on the 2.00 Vscale. The oscilloscope is now also plotting the voltage difference between the terminal “2 Y”and ground. In this case, that difference is roughly zero.

5. Press the “±” key, then the screen key “1-2”. The oscilloscope is now also plotting the voltagedifference between the terminal “1 X” and the terminal “2 Y.” This is the signal we will usefor the first part of the lab.

6. Push the “1” key until “Off” is highlighted.

7. Push the “2” key until “Off” is highlighted.

8. Have your GSI check your setup!

In this setup we are using the value of “Channel 1” minus “Channel 2” to determine the voltageacross a certain component. In this manner we can move the plugs from one component to anotherand measure their voltage change, regardless of the ground position. Remember, voltages do notneed to be measured relative to ground!

11



Figure 9: After completing 3.2 your setup should be connected like this.

12



resistance

Meter set to measure

(b) and (c)

Probes at points

(!)

Figure 10: Measuring the internal resistance of the inductor.

In the Faraday’s Law lab we used the oscilloscope to plot the voltage difference between one of the oscilloscope inputs and ground. In this experiment, we will plot the voltage difference betweenthe two oscilloscope inputs – you can think of the terminals “1 X” and “2 Y” as being like the twoinputs of a voltmeter.

4 Experiment: Internal Resistance of the Inductor

For our experiment we wish to investigate a resistor, a capacitor and an inductor as ideal devices.Unfortunately, our inductor is not quite ideal. It is primarily inductive but partially resistive, dueto the resistance of the wire windings. It will be important to know the internal resistance of theinductor on the circuit board for later analysis. Measure its resistance before any connections aremade on the board. Set the digital multimeter (DMM) to measure resistance( Ω) and measure theinternal resistance of the inductor (the thing that looks like a coil of wire) by placing the DMMplugs across it, as in Figure 10). Record the resistance on your worksheet.

5 Experiment: AC Voltage and Current in the Resistor

Be sure to go through the steps outlined in Sections 3.1 and 3.2 before beginning the experiments.Make sure the oscilloscope is still connected to measure the resistor (terminal “ 1 X” connected

to (b), terminal “2 Y” connected to (a)), that that the maximum voltage across the resistor isstill set to 4 V, and the audio driver is still set to 850 Hz. It is important that you keep thisfrequency the same for the measurements which follow.

Based on the frequency of the audio driver, calculate ω. Record V R,max and use it to calculateI max. Then sketch the voltage waveform that you see on your oscilloscope. Be sure to use theresistance of the resistor in your calculation, not the resistance of the inductor you measured in theprevious section.

13



V R,max = I maxR (15)

Now, consider Ohm’s Law. What will the current in the resistor (and hence in the whole seriescircuit) look like, given your sketch of V R? On the same graph as your voltage wave, draw what thecurrent looks like, and label the maximum value. Since you are plotting two different values (V Rand I ) you will need two different vertical scales. Clearly indicate each.

Draw a horizontal voltage phasor whose length corresponds to the maximum instantaneousvoltage drop that you read on the oscilloscope across the resistor on your worksheet. We set thevoltage phasor to be horizontal because the projection of this phasor at t = 0 is zero on the y axis(see Section 2 for a discussion of phasors). The current is in phase with the voltage across theresistor, so it is also a horizontal phasor at t = 0.

On all of your phasor diagrams in these experiments, you need to plot your measured values nottheoretical values.

6 Experiment: AC Voltage and Current in the Capacitor

Before changing the oscilloscope connection to view the voltage across the capacitor, checkthat the voltage across the resistor remains at 4.0 V. Store the resistor waveform by pressing the“Autostore” key twice. The stored waveform will be displayed at a lower intensity. Next, move theoscilloscope leads to measure the capacitor voltage, “1 X” to (d) and “2 Y” to (c). Be careful tomake the connections correctly – if you reverse the leads, you will get the wrong phase shift.

Carefully, measure and record on your worksheet the amplitude voltage across the capacitorfrom your oscilloscope display. We will also need to determine the “phase shift,” ϕ. If we have twoperiodic signals of period T , offset from one another by some ∆t, then the phase shift is defined as

ϕ

360=

∆t

T (16)

Let’s focus our attention on a particular feature of the signals on the oscilloscope: points where thesignal crosses the x-axis with positive slope. The time between two such successive crossings for one

signal is tells us T , while the time between the nearest such crossings for the two different signalstells us ∆t. If you eyeball your signal, you should find that ∆t/T ≈ 1/4, so that ϕ≈ 90 – is thatthe case?

In order to make more precise measurements of ϕ you will need to use a feature on the oscilloscopeknown as cursors. The cursors are dashed lines that you can move around using the knob located justbelow the “Cursors” key. If you place them near interesting features of your signal, the oscilloscopewill report their position with considerable precision. To turn the cursors on, press the “Cursors”key. Now, press the “t1” screen key. The “t1” cursor should be positioned on the y-axis, where it

may be hard to see. We’ll leave it there, since this should be a point where the V R signal crossesthe x-axis with positive slope. Now activate the second cursor by pushing the “t2” key. Use thecursor knob to move this to the next point that the V R signal crosses the x-axis, as shown in Figure11(a) You should now be able to read T from the bottom of the screen.

Next, move the “t2” cursor so that it lines up with the point closest to the origin where theV C signal crosse the x-axis with positive slope, as in Figure 11(b) You can now read ∆t from thebottom of the screen. Since you have T and ∆t, you can now compute the phase shift ϕ usingequation 16.

14



When done, press the “Clear Cursors” screen key to remove the cursors from the display, thendraw a sketch of the display with the stored waveform (resistor) and active waveform (capacitor)clearly marked.

Draw a phasor diagram for the voltage across the capacitor. Remember, your reference for thephase is the voltage across the resistor (which represents the current in the circuit). Make particularnote of the phase of the voltage across the capacitor relative to the current in the capacitor. Does

the current lead or lag behind the voltage applied to a capacitor?With the data taken, you can determine the capacitive reactance from

V C,max = I max X C . (17)

Capacitive reactance is

X C = 1

ωC . (18)

To within your errors, does your calculation of X C based on your measurements agree with thetheoretical value? Finally, what X C phase angle (current to voltage) did you derive for the capacitor?Does it agree with the predicted value?

7 Experiment: AC Voltage and Current in the Inductor

Before changing the oscilloscope connection to view the voltage across the inductor, return theoscilloscope leads to the resistor voltage points, “1 X” to (b) and “2 Y” to (a), and check that thevoltage across the resistor remains at 4.0 V. Next, move the oscilloscope to measure the inductorvoltage, “1 X” to (c) and “2 Y” to (b). Be careful to make the connections correctly – if youreverse the leads, you will get the wrong phase shift.

Carefully, measure and record on your worksheet the amplitude and phase of the voltage acrossthe inductor from your oscilloscope display. Draw a sketch of the display with the stored waveform

(resistor) and continuous waveform (inductor) clearly marked.Draw a phasor diagram for the voltage across the inductor. Does the current lead or lag behindthe voltage applied to an inductor?

We can find the reactance of the inductor, X L, using

V L,max = I max X L, (19)

and the measured voltage across the inductor. Compare the inductive reactance to the resistance youmeasured for the coil in Section 4. Is it reasonable to neglect this resistance in our measurement of the voltage phasor across the inductor? (Even if not, we will neglect it for clarity in this discussion.)Compare the measured value of X L with the relation (which can also be found in your text):

X L = ωL. (20)

Again, to within experimental error, does your value of X L agree with expectations? Finally, towithin experimental error how well does your measured phase angle ϕL agree with the phase angleexpected for an ideal inductor?

15



(a) Since the cursors enclose a single cycle, the period T is reported as “∆t” at thebottom of the oscilloscope display.

(b) ∆t is reported as “∆t” at the bottom of the oscilloscope display.

Figure 11: Using cursors to make measurements. Note that the first cursor “t1” is coincident withthe y-axis, and thus hard to see.

16



8 Experiment: AC Voltage and Current in a Series RLCCircuit

In this final measurement you will repeat the measurement for all three components together.Next, move the oscilloscope to measure the voltage across the entire circuit, “1 X” to (d) and “2 Y”to (a). Be careful to make the connections correctly – if you reverse the leads, you will get the

wrong phase shift.Draw the phasor diagram for the complete RLC circuit. Be sure to clearly label the phase

angle and the magnitude of the voltage vector. Compare this to what you expect from adding theindividual R, L and C phasors.

First, draw the phasor which corresponds to V R (and thus total current) along the positivex-axis. Next, based on your responses above, draw the theoretical phasors for V C and V L on thepositive/negative y-axis. Next, perform vector addition to add all three phasors on your diagram.Label your resultant phasor V RLC on a new diagram (also draw V R on the new diagram). Usingthe coordinates for the V RLC vector, determine the phase shift between voltage and current in thecircuit.

We can now determine the impedance of the entire circuit. It is

Z =

(R+RL)2 + (X L−X C )2 (21)

where RL represents the resistance of the inductor (measured in Section 4). Using

V max = I max Z, (22)

determine the maximum voltage provided by the audio driver and compare that with what youmeasured.

9 Experiment: Resonance in RLC Circuits (Frequency De-pendence of Z )

You do not need to use the oscilloscope in this experiment, so you can disconnect it from theRLC circuit board. Construct an RLC circuit using the 10 µF capacitor, the inductor, and thelight bulb:

• Connect one output of the audio driver to the 10 µF capacitor (d).

• Connect a jumper from the inductor (b) to the connector just above the light bulb (f).

• Connect the other output of the audio driver to the connector just below the light bulb (e).

To measure the voltage across the light bulb, simply connect your multimeter to the two connectorsabove and below the bulb. Be sure that your multimeter is set to measure AC volts ( V ). Recallthat in this configuration, the meter reports V RMS = V Max/

√ 2

Begin with the audio driver frequency set to 100 Hz, and adjust the Volume knob until thevoltage reads approximately 0.2 V. The bulb will probably not light. Now measure V bulb as afunction of frequency for 10 values ranging from 100 Hz to about 1500 Hz. (For frequencies above1000 Hz, you will need to adjust the range selector on the audio driver to 5000 Hz.) Evenly-spacedfrequencies are not necessarily best—use fine measurements (50 Hz spacing) when the voltage is

17



changing more slowly and a coarser scale when it is changing more rapidly. Above 1000 Hz, it issufficient to use a 200 Hz spacing. Record and plot your data on your worksheet. Be sure to findthe resonant frequency, f r, at which V bulb reaches its maximum value.

The circuit can absorb the most energy from the audio driver when its impedance Z is at aminimum. The impedance of the circuit is given by

Z = R2

+ (X L−X C )2

= R2 + (ωL−1/ωC )2.

Because the inductive and capacitive reactance terms appear with opposite signs, and depend indifferent ways on frequency, they can cancel each other out when X L =X C . This leads to a minimumin the impedance at the resonant frequency,

f r = 1

2π√ LC

.

Compute the resonant frequency for your circuit and compare it to your observations. Do they

agree?At the resonant frequency, use your multimeter to measure the voltage drop across the capacitor?

Is it zero? What is the power dissipated in the capacitor? What about the inductor?

Measuring the Inductance of an Unknown Inductor

Adjust the frequency on the audio driver so that your RLC circuit is at resonance again. Nowtake the steel cylinder from your circuit board and insert it into the inductor. What happens tothe bulb?

Scan the frequency on the audio driver until you find the new resonant frequency of the circuit.Based on this result, what is the inductance of the inductor when the steel cylinder is inserted?

Increasing the inductance by inserting a magnetic material into a coil is analogous to increasingcapacitance by adding a dielectric. For this reason, you will often see inductors wound on an ironcore.

AC Circuits Lab Manual

Documents

Transcript of AC Circuits Lab Manual