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PH262 AC Circuits: Phase Relationships I

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PH262 AC Circuits: Phase Relationships

I.         BACKGROUND INFORMATION

The rotating phasor representation for series AC circuits should be familiar from textbook and lecture notes.  A brief outline of the essential points is provided here.

 

If a series RLC circuit is connected across a source of emf which is a sinusoidal function of time, then q and all its derivatives will also be sinusoids.  Since all elements in a series circuit share the same current, it is customary to write i = I Cos(ωt) as the reference function against which the others will be compared.  The voltage across each component is then expressed in terms of the current as:

 

v_R = Ri = RICos(ωt) =V_RCos(ωt)

di

v_L = L  =−ωLISin(ωt) =−X_LISin(ωt) =−V_LSin(ωt) 

 

dt

q         1          1

v_C =    =        ∫idt =ωC   ISin(ωt) = X_CISin(ωt) =V_CSin(ωt)

 

 

C        C

    

where lower-case symbols are used for time-varying quantities, and upper-case symbols for the amplitudes.  The ratio of amplitudes V_R/I is seen to be the DC resistance, R.  The ratios of the V to I amplitudes for the other two components, which must also have units of ohms, are given the symbol X and are called "reactance", leading to the name "reactive components" to describe any L or C in a circuit.  

 

Examination of the rightmost terms in each of these equations shows the relative phase of the three voltages.  v_R has the same phase as i [both Cos], v_C [Sin] is 90° behind i, and v_L [Sin] is 90° ahead of i.  These relationships are frequently expressed by phasor diagrams, which show the voltage amplitudes summed with the appropriate phases to equal the amplitude of the driving emf.

 

                          

V

R

V

L

E

φ

L

  

φ

C

E

V

R

V

C

 

Figure 1

 

Above are the voltage phasor diagrams for an RL and an RC circuit.  As is conventional, the phasor rotation is assumed to be counterclockwise.  Thus, it is seen that the external emf will be ahead of the current in the inductive circuit, and behind the current in the capacitive circuit.  Inspection of these diagrams shows that the trigonometric functions of the phase angle can be expressed as ratios of the voltages, for example

 

 

 

It is therefore possible to determine the relative phases of such signals just by making appropriate voltage measurements, which can be done either with an oscilloscope or with an AC voltmeter or a DMM set for AC measurements.  Remember that AC voltmeters and DMM’s usually measure AC voltages in RMS (Root Mean Square) units, not amplitude as seen on an oscilloscope.  RMS voltage is related to amplitude as shown below.

VRMS=Vamplitude / 2

It is also possible, by displaying the signals on an oscilloscope, to measure the relative phase directly by the procedure outlined in the following section.

 

The phase relationships can be expressed in terms of the impedances rather than the voltages.  As indicated by the phasor diagrams, the amplitude of the driving emf is related to the amplitudes across the individual components in a Pythagorean form, eg.

 

????????2 = ????????????????2 + ????????????????2   ????????????????   ????????2 = ????????????????2 + ????????????????2

 

If the amplitude of each phasor is divided by I, one obtains the impedance triangle, with

 

 

where Z= ε/I is called the "impedance" of the entire circuit, X_L is the inductive reactance, and X_C is the capacitive reactance.  The word "impedance" may be used in a more general way to describe the ratio of the voltage amplitude across any two points in a circuit to the current amplitude between those points.  In this usage, the impedance of a resistor is R,  that of a single capacitor is X_C , etc.

 

The trigonometric functions of the phase angle can then be rewritten in terms of the impedance triangle simply by replacing each voltage amplitude by its corresponding impedance term.  The principal point of interest lies in the fact that the reactances are explicit functions of the frequency,

 

 

and consequently both Z and  will also be frequency dependent.  For example:

 

The tangent function is chosen to indicate the nature of the frequency dependence because R is the one parameter that does not depend on frequency, and thus the length of the adjacent side of the triangle is always constant with respect to frequency.  Varying the frequency varies the length of the opposite side (the reactive side), and thereby of the hypotenuse as well.  Note that, through all of this discussion, the relative phase between the resistive and reactive voltages remains ±90°.  It is the angles between the component voltages and the

emf which vary, with the attention usually focused on the V_R phase angle because this is also the relative phase of the current.

 

Finally, let us briefly consider the qualitative reasons for these frequency dependencies.  Since an inductor voltage is proportional to the rate of change of current through the inductor, it stands to reason that no voltage is needed to maintain a non-changing current, ie. the inductor's impedance will be zero at zero frequency ... it is a short circuit. To maintain the same current amplitude as frequency increases will require increasingly rapid rates of change in the alternating current, and therefore increasingly more voltage amplitude across the inductor, ie. its impedance will steadily increase with increasing frequency.

 

A capacitor, on the other hand, allows no current to pass in a constant voltage situation ... it is an open circuit.  It therefore offers infinite impedance at zero frequency.  The magnitude of the capacitor voltage at any instant indicates the charge stored in the capacitor at that instant.  An alternating current shifts charge alternately from one plate to the other.  If frequency is increased, while holding the current amplitude constant, there is increasingly less time for charge to flow before the current reverses and therefore less charge transferred between the plates, ie. the capacitor voltage amplitude, and hence its impedance, will decrease as frequency increases.

 

 

II. EXPERIMENTAL PROCEDURE

Introduction to Phase Measurements

To make a phase measurement requires the use of the dual-trace oscilloscope to display both of the desired signals.  This requires some additional care with grounding and triggering.

 

1. Connect the series circuit under investigation in such a way that the ground is located between the two components to be studied.  For example, if it is desired to measure the phase between the generator and the resistor, then the resistor must be connected to the grounded side of the generator.  In practical usage, this means that one of the signals under study will "always" be the generator voltage. (This condition will be violated at one point during the present experiment in order to demonstrate the constancy of the phase between resistor and reactance, but such a connection would be rare in a "real-life" lab setting).

 

 

 Example:

 

 

                    

 

Notice that with the circuit above, you can measure both ε(V_S), the source voltage, and V_R, the voltage across the resistor, since they share a common ground.  Why can’t you connect CH1 of the oscilloscope across the capacitor?

 

2. Connect the probes to the desired components, and set the amplitude and sweep controls to obtain a clean display of both signals.  The “hold off” control may be pushed in for “alternate” or pulled out for “chop” depending on the signal frequency. Low frequencies are usually better observed using “chop;” high frequencies work better on “alternate”. The “coupling source” may be set to either CH1 or CH2, but make sure that the “vertical position” controls for both CH1 and CH2 are pushed in.  If the CH1 “vertical position” control is out, then you will probably see no phase shift 

 

 

Figure 2

 

between the two signals, ε and V_R.  Figure 2 above shows a typical display, triggered when the larger amplitude signal comes up through zero.  Note that, in this picture, the smaller amplitude signal is "leading" the larger, ie. it reaches its maximum earlier, crosses the axis earlier, etc.  The time interval ?t corresponds to the relative phase in the same ratio as T corresponds to 2π radians (or 360°), ie;

 

 

                                                                                        

 

????????

????????

=

1

Time period = 1ms

Time bar scale

5.6

cm = 1ms

So:

1

0

mm = 0.178 ms

?

????????

mm

=10

 

In “Every Circuit” APP the time bar scale is given on x-axis and it represents time period T. Set the time period 1ms.  Adjust the waveform in such a way that the sinosidual wave of the function generator starts from one side of the time bar scale and ends on the other side of it.

Measure the length of the time bar scale and time interval ????????? carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Now you have T and ????????? from simulationscreen, compute  as above. 

 

Experiment

 

Phase Measurements 

1. Connect an RC series circuit across the sine generator in such a way that the generator voltage and the resistor voltage can be simultaneously displayed.  Try to find a frequency setting for the generator such that the amplitude of the V_R signal is approximately 1/2 of the generator amplitude.  According to the trigonometric relations above, this should correspond to a phase of approximately 60° (or π/3 radians).  Verify this by a direct measurement of the relative phase of the two signals.

 

2. The impedance relations predict that Tan ???????? in a capacitive circuit should vary as . 

Keep in mind that the simulation provides you frequency . As you know that .

Go to the tool and select the frequency (????????). Rotate the flywheel to adjust the frequency of oscillation. Use the dual trace feature of the oscilloscope to make phase measurements over a sufficiently wide range of frequencies to plot a good graph of Tan  vs 1/ω, and extract the value of RC from your graph. Compare with the actual values of the components. For each measurement of the phase you should also measure the voltage for the signal generator (ε or V_S), V_C, and V_R from oscilloscope of simulation screen. Place all of your phase, amplitude and frequency data in a neat table.

 

Slect the

frequency

????????

in the

tool bar

????????

=

2

????????

????????

             

Adjust the

flywheel to

the change

the frequency

 

3. Use your voltage data to calculate the phase using 

cos    = V_R / ε

         and compare your phase () results to the direct measurement of    made with the dual trace oscilloscope.

 

4. Use your voltage data to show that the pythagorean relation

ε 2 = VR2 + VC2

      holds for each set of voltages.

 

5. Construct an LR circuit using the 22mH inductor and an R equals 1500 Ω. With this circuit tan  (phase angle between R and ε), tan  _L= V_L / V_R, can be monitored by measuring V_L and V_R (from oscilloscope). The waveform in the screenshot below is of the circuit on the right. R is grounded. CH1 is connected function generator and CH 2 across resistor. Calculating the ratio V_L / V_R. Collect a set of data for tan  vs ω.  Plot the appropriate graph, analyze the slope in terms of L and R, and compare with the actual component values.

 

????????

????????

????????

?

????????

 

         The waveform in the screenshot below is of the circuit on the left. Inductor (L) is grounded. CH1 is connected across function generator and CH 2 connected across resistor. 

 

 

????????

????????

????????

?

????????

 

 

6. Construct an LR circuit using the 22 mH inductor and an R that is  1500 Ω.  But this time we wish to measure V_L and V_R simultaneously with the dual trace oscilloscope. Notice that you cannot do this with a simple series L-R since the oscilloscope ground can’t be in two different places at the same time.  So we have to be a bit clever as we were before in Lab #10.

 

     

 

         Construct the L-R circuit as shown above.  Make sure that both inductors and resistors are as close to the same values (22 mH, 1500 Ω) as possible.  Notice that you can now use the oscilloscope to measure V_L and V_R at the same time, with no ground conflict.  

????????????????????????????????

????????

????????????????????????????????????????

????????

5

cm

=100

µ

s

?

????????

Signal

across V

R

and V

L

 

 

         Find a frequency such that the two signals have approximately the same amplitude. Measure the relative phase between the two signals on the screen. Note that the relative phase between the resistor and the inductor is, as predicted, ~90º, and will remain constant independent of frequency. Explain why the angle you measure is not exactly 90º.

 

 

 

 

III. CONCLUSION/QUESTIONS

    

1. Starting with the defining equations for R and C, show that RC has units of time.

 

2. Starting with the defining equations for L, R, and C, show that ωL and 1/ωC both have units of ohms.

 

3. Show that taking the time derivative of any sinusoidal function, such as cos (ωt + ), has the effect of advancing its phase by π/2.

 

4. If the internal resistance of an inductor is not negligible, how does it change the relative phase of voltage and current for the inductor ?  Is this effect more important at high or at low frequency ?