EXPERIMENT 10
RCL CIRCUITS
INTRODUCTION:
In this experiment, we will learn some of basic properties of alternating current
circuits.
BACKGROUND
In a direct current (dc) circuit, the electrons always move in the same direction.
Moreover, the emfs and the potential differences are the same value at all times. There are,
however, emfs which vary sinusoidally in time. In fact, the most common source of current, a
wall socket, provides a sinusoidal emf with a frequency of 50 Hz (50 cycles/ second) and an
amplitude of 220 volts. In general, an alternating emf can be written as follows:
Eab (t ) V0 sin(2 f t 0 ) (1a)
or alternatively,
Eab (t ) V0 sin( t 0 ) (1b)
In the above, V0 is called the amplitude (the maximum value of Eab), (t ) t 0 is
called the time dependent phase angle, or simply the phase, f is the frequency measured in
hertz (Hz), or cycles per second, ( 2 f ) is the angular frequency in radians per second,
and 0 is the phase shift. A sketch of Eab(t) versus is given below. We shall use radians
for angular measure ( 90 0 2 , 180
0
, 270 0 3 2, 360 0 2 ).
Remember that Eab(t) is a measure of the potential at point a minus the potential at point b as
a function of time. Suppose, for example, that 0 0 . When t = (1/4)f, Eab = V0, and when t
= (3/4)f, Eab = -V0. At the earlier time, the potential at a is higher than b, while at the later
time, b is at higher potential than a.
Since the imposed emf varies sinusoidally with frequency f, it is not surprising that the
potential differences across the circuit elements as well as the current also oscillate
sinusoidally at the same frequency. This is called an alternating current (AC) circuit.
Describing the potential difference across the elements of an AC circuit is more complex than
the steady state DC circuit. These potential differences are oscillating at the same frequency
as the imposed emf. However, they are not generally in phase with the emf. Phasor
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diagrams, discussed in your text, are a convenient way to represent these oscillating voltages
and currents.
A phasor is a vector whose tail is at the origin. Its length, or amplitude, represents the
maximum value of the voltage across (or current through) a circuit element. The direction of
the phasor represents the phase. For this reason, a phasor is through of as rotating
counterclockwise about the origin at the same frequency as the voltage applied to the circuit.
The projection of the phasor along the y axis gives you the instantaneous value of the voltage
(or current) of the circuit element. For example, in the figure of a phasor diagram on the next
page, we see the projection of the current I on the y-axis. We could do the same for all the
other phasor in the diagram to find their instantaneous values. Since all the phasors drawn for
a single circuit will rotate at the same frequency f, the angle between the phasors will always
remain constant as they rotate.
There are three basic circuit elements: resistors, capacitors and inductors. We need to
know how to draw phasors for the voltage across each of these elements. The amplitude of
the voltage phasor is found by using the equation V = IZ, where V and I are the voltage and
the current, and Z is a quantity called impedance, which acts much like the resistance in a DC
circuit. The impedance of a resistor (ZR) is, in fact, just the resistance R. The impedance of a
capacitor (ZC) or an inductor (ZL) is slightly more complicated. Using V = IZ, we can find the
amplitude of the voltage phasor for a resistor (VR), capacitor (Vc) or inductor (VL):
VR IZR IR
Vc I ZC I (1 c) (2)
VL I ZL I ( L)
Note that the amplitude of VC and VL depend on as well as C and L.
Since phasors are vectors, we also need to know their direction. We know, of course,
that they all rotate with frequency f. We are interested in the angles between the phasors. In a
series circuit, the direction of the phasor VR is always the same as that of the phasor
representing the current in the circuit. This means that the current across a resistor is always
in phase with the voltage across it. The direction of the phasor VC is always 90° less than that
of the current. This means that the voltage across a capacitor lags the current by 90°. The
direction of VL is always 90° greater than that of the current. This means that the voltage
across an inductor leads the current by 90°.
An example of a phasor diagram for an RCL circuit is
voltage s around loop must be zero. This holds true even in
an ac circuit, where the voltage are changing with time. If
we call the voltage that drives the circuit Eab, then
Eab VR VC VL (3)
This means that Eab is the vector sum of the phasors
VR, VC, and VL. This is shown in the figure at right.
We have seen that phasors add like vectors. Thus, we
can use the law of cosines to fined the amplitude of the sum;
of two phasors of arbitrary direction V1 and V2:
Vtot2 V12 V22 2V1V2 COS (4)
where Vtot is the amplitude of the sum of the phasors V1 and V2 and is
the angle between the two phasors. You will use Equation (4) in this experiment to add
phasors.
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EXPERIMENT:
1. APPARATUS:
The RLC circuit consists of a resistance, a capacitor, an inductance, a function
generator, and an oscilloscope. All of these components should be provided in your work
station.
2. PROCEDURE
I. RC Circuit
Connect the two terminals of the audio oscillator across the RC pair as Shown in the figure.
Connect the oscilloscope as follows: CH1 across the resistor and CH2 across the function
generator as shown in figure 1 (a).
Note that it is very important that both black terminals be connected to a common point since
each is grounded inside the oscilloscope.
This presents no problem since it is only potential differences being measured. On the other
hand, if the black terminals were connected at two different points in the circuit, then in fact
all points between them, including the circuit elements, would be maintained at ground
potential contrary to our intentions.
CH2 CH2
CH1 CH1
B1&B2 B1&B2
(b) (a)
Figure 1: RC circuit
CH1 now monitors VR and CH2 monitors V0. The mode switch should be set to BOTH and
the scale dials set on 2 volts/div. Set the oscillator frequency at f = 4000 Hz and adjust its
output to that the amplitude V0 = 6 volts (or 12 V peak to peak). V0 will have to be set to 6
volts at each frequency. Also make sure the calibration knobs are set all the way clockwise so
that the scaling (fudge) factor is 1.0. Change the MODE switch to CH1, and measure and
record the value of VR in table 1. Change the MODE to CH2, and measure and record the
value of V0 in table 1. To measure R 0 , change the MODE to BOTH. Put trigger to
CH2. Adjust the sweep time so that two or three cycles at most appear on the screen. By
using the cursor find the period T and then find the number X (time per division) between
corresponding phase (e.g. = 0) on the V0 and VR traces. The final result is
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X
( )360 R 0 (in deg rees )
T
where X and T are the distances shown in the figure.
T
Figure 3: tow signal phase shifted
Now, switch the leads connecting the RC components to the function generator, so that the
capacitor is connected to the negative terminal of the genrator and the resistor is connected to
the positive terminal. Folowing the same procedures,
determine 0 C
Determine the algebraic sign of R - C : Make sure you know which trace is VC and which
is VR. Simply use the ground switch at the input terminals to eliminate one of traces. Thereby
establishing the identity of the two. If VR is leading VC then the sign is (+).
Now repeat the procedures outlined above by seeping the frequency range from 4000
Hz to 11,500 Hz in increments of 1500 Hz, each time adjusting V0 to 6 volts. Again record
the results in table 1 on the worksheet.
RC CIRCUIT CALCULATIONS
From equation (2), one can show that the impedance
Zc (VC VR ) R.
1. Compute ZC in units or the resistance R for the various frequencies and record the results
in table 1 on the worksheet. Draw a graph of Z C as a function of f. Does equation (2)
predict the trend of your graph of ZC?
2. Add the two phasors VR and VC to compute V0 (Use equation (4))
3. Compare this with the measured value of V0 at 4000 Hz and 10,000 Hz.
4. Calculate percent difference.
5. Discus this in the space below Table 1 on the worksheet.
II. RCL Circuit; Resonance
Connect the audio oscillator across RCL. Monitor Vac(t) on CH1 and Vab(t) on CH2 by
connecting the circuit as shown in the figure below.
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CH2 CH1
B1&B2
Figure2: RLC circuit
By varying the frequency continuously starting from 1000 Hz, find the resonance frequency
and compare it to the its theoretical counterpart.
At the resonance frequency f 0 2 0 , the impedance Z ( 0 ) R . Consequently, VR=V
and R =0. Also, ZC=ZL. Place your value for f0 in the space provided on the worksheet.
QUESTIONS
(1) What is the impedance of an inductor when we approaches 0 (i.e., a dc circuit)? Explain
why your answer is reasonable. (HINT: Look at Equation 2 and your graph of ZL.)
(2) What is he impedance of a capacitor when we approaches 0? Explain why your answer is
reasonable. (HINT: Look at Equation 2 and your graph of ZC.
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PHY 1402 LAB. REPORT
EXPERIMENT 10
RCL CIRCUITS
NAME: . . DATE: . .
SECTION: . .
***
1. EXPERIMENTAL PURPOSE:
State the purpose of the experiment.( 5 points )
2. EXPERIMENTAL PROCEDURES AND APPARATUS:
Briefly outline the apparatus used and the general procedures adopted. (5 points )
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3. RESULTS AND ANALYSIS
TABLE 1 (30 points)
f V0 VR VC R .
0 0 C . R C . ZC
(degrees) (degrees) (degrees)
4000
5500
7000
8500
10000
11500
Comparison (10 points)
The resonance frequency (20 points)
(f0)experimental = Hz as found when R
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Theoritical resonance frequency
(f0 )Theoritical= Hz
Comparison between the (f0)experimental and (f0 )Theoritical (10 points)
4. CONCLUSIONS: (5 points)
QUESTIONS: (5 points)
GRAPHS: (10 points)
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