PHY 1402 LAB. REPORT
EXPERIMENT 06
RCL CIRCUITS
NAME:Asmae Aguerd
SECTION:04 .
DATE: .21/11/2022
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1. EXPERIMENTAL PURPOSE:
State the purpose of the experiment. ( 5 points )
The purpose of this experiment is to learn some of basic properties of
alternating current circuits.
2. EXPERIMENTAL PROCEDURES AND APPARATUS:
Briefly outline the apparatus used and the general procedures adopted. (5 points
Apparatus:
The RLC circuit consists of a resistance, a capacitor, an inductance, a
function generator, and an oscilloscope.
Procedures
Connect the two terminals of the audio oscillator across the RC pair. Connect the
oscilloscope as follows: CH1 across the resistor and CH2 across the function generator.
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 sothat 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.
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 tothe positive terminal.
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. Therebyestablishing 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.
 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 ZC as a
function of f. Does equation (2) predict the trend of your graph of ZC?
 Add the two phasors VR and VC to compute V0 (Use equation (4))
 Compare this with the measured value of V0 at 4000 Hz and 10,000 Hz.
 Calculate percent difference.
 Discus this in the space below Table 1 on the worksheet.
 Connect the audio oscillator across RCL. Monitor Vac(t) on CH1 and Vab(t) on CH2
byconnecting the circuit.
 By varying the frequency continuously starting from 1000 Hz, find the
resonance frequencyand compare it to the its theoretical counterpart.
 At the resonance frequency 𝑓0 = 2𝜋𝜔0 in which the impedance Z (𝜔0 ) =
R and consequently, VR=V, 𝜃𝑅 = 0 and 𝑍𝑐 = 𝑍𝐿 . Place your value for f0
in the space provided on the worksheet.
1. RESULTS AND ANALYSIS
TABLE 1 (30 points)
f
V0
(V)
VR
(V)
VC
(V)
4000
5500
7000
8500
10000
11500
5.920
5.760
5.520
5.440
5.280
5.120
2.80
3.40
3.760
4.080
4.160
4.240
5.20
4.640
4.080
3.60
3.120
2.80
𝜃𝑅 − 𝜃0 .𝜃0 − 𝜃𝐶
(degree (degrees)
s)
60.480
27.140
53.40
35.60
47.50
45.30
43.070
45.530
38.490
50.40
29.940
52.940
𝜃𝑅 − 𝜃𝐶
(degrees)
87.620
89.00
92.80
88.60
88.890
82.90
𝑍𝑐
(𝑂ℎ𝑚)
1857,1430
1364,7060
1085,1060
882,3530
750,00
660.3770
Comparison (10 points)
R=1000 Ohm.
𝑉
1. Even though the frequency keeps on increasing, the ratio 𝑉𝐶 increases by a constant, so
𝑅
we can conclude that the graph of Zc vs. frequency will give a linear function.
2. 𝑉0 2 = 𝑉𝐶 2 + 𝑉𝑅 2 + 2. 𝑉𝐶 . 𝑉𝑅 . 𝑐𝑜𝑠(𝜃)
At f = 4000 Hz;
𝑉0 = √5.22 + 2.82 + 2 ∗ 5.2 ∗ 2.8 ∗ 𝑐𝑜𝑠(87.62) = 6.007 V
At f = 10000 Hz;
𝑉0 = √3.122 + 4.162 + 2 ∗ 3.12 ∗ 4.16 ∗ 𝑐𝑜𝑠(88.89) = 5.25 V
3. The value of 𝑉0 at 4000 Hz is 6.007 V, meanwhile it is equal to 5.25 V at 10000Hz.
These two values are approximately equal.
4. 𝑃𝐷(4000 𝐻𝑧) =
6.007−5.92
𝑃𝐷(10000 𝐻𝑧) =
6.007+5.92
2
5.28−5.25
5.28+5.25
2
. 100 = 1.458%
. 100 = 0.569%
5. In this circuit, the voltage changed over time, the phase between the two voltages 𝑉𝐶
and 𝑉𝑅 stayed constant as 90°, this is confirmed by the equation: 𝑉0 2 = 𝑉𝑅 2 + 𝑉𝐶 2
(which means the part 2. 𝑉𝐶 . 𝑉𝑅 . 𝑐𝑜𝑠(𝜃) 𝑖s eliminated) which holds true. The
percentage differences between the phasors are almost zero.
The resonance frequency (20 points)
L= 1mH, C =22nF
(f0)experimental =11600 Hz as found when 𝜃𝑅 e q u a l s t o 0
Theoritical resonance frequency
1
1
(f0 )Theoritical = 2𝜋.√𝐿𝐶 = 2𝜋√10.10−3 .22.10−9 = 10730.22 Hz
Comparison between the (f0)experimental and (f0 )Theoritical (10 points)
%𝐷 =
11600 − 10730.22
. 100 = 7.79%
11600 + 10730.22
2
2. CONCLUSIONS: (5 points)
In this experiment, we had a circuit RC at first, we had to check the value of 𝑉𝐶 and
𝑉𝑅 and calculate the angle of phase between them and the 𝑉0, to finally calculate 𝜃𝐶 − 𝜃𝑅
a n d Z c u s i n g t h e f o r m u l a : Zc = (VC VR ) R while changing the frequency. We
plotted a graph the impedance of the capacitor vs. frequency, the function was decreasing and
give us a linear function which means that the impedance of a capacitor is inversely
proportional to frequency.
Using 𝑉0 2 = 𝑉𝐶 2 + 𝑉𝑅 2 + 2. 𝑉𝐶 . 𝑉𝑅 . 𝑐𝑜𝑠(𝜃), and the values of 𝑉𝐶 , 𝑉𝑅 𝑎𝑛𝑑 𝜃 for two
different frequencies, and check between the values of 𝑉0 calculated and those we got from
the experiment, the difference percentage in both cases did not exceed 1.458%. Also, the form
𝑉0 2 = 𝑉𝐶 2 + 𝑉𝑅 2 + 2. 𝑉𝐶 . 𝑉𝑅 . 𝑐𝑜𝑠(𝜃) confirmed that the angle between the phasors of
𝑉𝐶 𝑎𝑛𝑑 𝑉𝑅 is 90° (which means the part 2. 𝑉𝐶 . 𝑉𝑅 . 𝑐𝑜𝑠(𝜃)is eliminated) and it holds because
the percentage difference is near 0%.
The second part of the experiment, we added an inductor to the circuit. We had to find
the resonance frequency 𝑓0 = 2𝜋𝜔0 in which the impedance Z (𝜔0 ) = R and consequently,
VR=V, 𝜃𝑅 = 0 and 𝑍𝑐 = 𝑍𝐿 ,first, through the oscillator varying the frequency starting from
1000 Hz. The resonance frequency (f0) experimental =11600 Hz. The theoretical resonance
1
frequency can be calculated using the formula: (f0 )Theoretical = 2𝜋.√𝐿𝐶 = 10730.22 Hz. The
percentage difference between the two values was about 7.79%.
So, Resonance happens when the impedance of a capacitor and inductor are equal and
cancel out. So only real resistance is there. The energy stored in capacitor and inductor
oscillates between them.
QUESTIONS : (5 points)
1. The impedance of an inductor is 0 when we approach the value of 0. This is because
as inductance reaches 0, the inductor resembles like a straight current carrying
conductor wire which has impedance of 0. By equation impedance 2𝜋𝑓𝐿 which goes
to 0 as goes to 0.
2. Impedance of capacitance goes to infinity as C goes to 0. From equation impedance is
1
. This is because as C goes to 0 we can think the separation between the plates of
2𝜋𝑓𝐿
the capacitance increased to infinity. Hence there is insulator and a gap in circuit in
this limit. Hence impedance is infinity.
GRAPHS: (10 points)
Impedance of the capacitor vs. Frequency
2000.000
Impedance of the capacitor (Ohm)
1800.000
1600.000
1400.000
1200.000
1000.000
800.000
600.000
y = -0.153x + 2285.4
400.000
200.000
0.000
0
2000
4000
6000
8000
Frequency (Hz)
10000
12000
14000