8. FREQUENCY RESPONSE OF THE SERIES R-C NETWORK Objectives:
1. Note the effect of frequency frequency on the impedance of a series R-C network 2. Plot the voltages and current of a series R-C network versus frequency 3. Calculate and plot the phase of the input impedance versus frequency for a series R-C network
List of Equipment and Component:
1. 2. 3. 4. 5. 6.
1 x Resistor 1kΩ (1/4 –W) 1 x Capacitor 0.1μF DMM Oscillator Function generator Frequency counter (if available)
Background:
For series R-L series R-L network, the voltage across the coil increases with frequency since the inductive reactance increases directly with frequency and the impedance of the resistor is essentially independent of the applied frequency (in the audio range). For the series R-C series R-C network, network, the voltage across the capacitor decreases with increasing frequency since the capacitive reactance is inversely proportional to the applied frequency. Since the voltage and current of the resistor continue to be related by the fixed resistance value, the shapes of their curves versus frequency will have the same characteristics. Again, keep in mind that the voltages across the elements in an ac circuit are vectorially related. Otherwise, the voltage readings may appear to be totally incorrect and not satisfy Kirchhoff’s voltage law. The phase angle associated with the input impedance is sensitive to the applied frequency. At very low frequencies the capacitive reactance will be quite large compared to the series resistive element and the network will be primarily capacitive in nature. The result is a phase angel associated with the input impedance that approaches –90 o (v lags i by 90o). At increasing frequencies X C C will drop off in magnitude compared to the resistive element and the network will be primarily resistive, resulting in an input phase angle approaching 0 o (v and i in phase). Caution:Be Caution:Be sure that the ground connections of the source and scope do not short out an element of the network, thereby changing its terminal characteristics
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Procedure: Part 1: VC, VR and I versus Frequency (a) Construct the network of figure 1. Insert the measured value of the resistor R on the diagram.
Figure 1 (b) Maintaining 4 V ( p-p) at the input to the circuit, record the voltage V C (p-p) forthe frequencies appearing in Table 2. Make sure to check continually that E s = 4 V ( p p)witheach frequency change. Do not measure the voltage V R at this point in the experiment. The common ground of the supply and scope will short out the effect of the capacitive element, which may result in damage to the equipment. For each frequency try to read V C to the highest degree of accuracy possible. The higher the degree of accuracy, the better the data will verify the theory to the substantiated. Table 2 Frequency
VC (p‐p)
VR (p‐p)
Ip‐p
0.1kHz 0.2kHz 0.5kHz 1kHz 2kHz 4kHz 6kHz 8kHz 10kHz
(c) Turn off the supply and interchange the positions of R and C in Fig. 1 and measure V R (p-p) for the same range of frequencies with E maintained at 4 V ( p-p). Insert the measurements in Table 2. (d) Calculate I p-p from I p-p = V R (p-p)/ Rmeasured and complete Table 2
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(e) Plot the curve of V C (p-p) versus frequency on Graph 1. Label the curve and clearly indicate each plot point
(f) Plot the curve of V R (p-p) versus frequency on Graph 1. Again, label the curve and clearly indicate each plot point
Graph 1
(g) As the frequency increases, describe in few sentences what happens to the voltage across the capacitor and resistor. Explain why
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(h) At the point where V C=VR , does XC=R? Should they be equal? Why? Record the level of voltage and the impedance of each element below.
VC=VR = XC= R= (i) Determine VC(p-p) and VR(p-p) at some random frequency such as 3.6 kHz from the curves VC(p-p)= VR(p-p)= Are the magnitude such that V C(p-p)+VR(p-p)=E p-p ?
If not, why not? How are they related?
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(j) Plot the curve of I p-p versus frequency on graph 2. Label the curve and clearly indicate each plot point.
(k) How does the curve of Ip-p versus frequency compare to the curve of VR(p-p) versus frequency? Explain why they compare as they do
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(l) At a frequency of 6kHz, calculate the reactance of the capacitor using X C = 1/(2 fC ) and the nameplate capacitance level. Compare with the value obtained from the data of Table 2 using V C ( p p ) X C I p p
XC(calculated)=
XC(from data)=
(m)Use the Pythagorean theorem to determine the voltage V C (p-p) at a frequency of 6 kHz and compare with the measured result of Table 2. Use the peak-to-peak value of V R from Table 2 and E s (p-p) = 4 V.
VC(p-p) (calculated)= VC(p-p) (measured)=
(n) At low frequencies the capacitor approaches a high-impedance open-circuit equivalent and at high frequencies a low-impedance short-circuit equivalent. Do the data of Table 2 and Graph 1 and Graph 2 verify the above statement? Comment accordingly.
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Part 2: ZT versus Frequency
(a) Transfer the result of I p-p from Table 2 to Table 3 for each frequency Table 3 Frequency
0.1 kHz 0.2 kHz 0.5 kHz 1 kHz 2 kHz 4 kHz 6 kHz 8 kHz 10 kHz
E p-p
I p-p
Z T
E p p
Z T
R 2 X C 2
I p p
4V 4V 4V 4V 4V 4V 4V 4V 4V
(b) At each frequency, calculate the magnitude of the total impedance using the equation Z T = E p-p/ I p-p in Table 3. (c) Plot the curve of Z T versus frequency on Graph 3 except for f = 0.1 kHz, which is off the graph. Label the curve and clearly indicate each plot point. (d) For each frequency calculate the total impedance using the equation Z T
R 2 X C 2
and insert in Table 3. (e) How do the magnitudes of Z T compare for the last two columns of Table 3?
(f) On Graph 3, plot R versus frequency. Label the curve.
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(g) On Graph 3, plot X C = 1/2 fC versus frequency. Label the curve and clearly indicate each plot point.
(h) At which frequency does X C = R? Use both the graph and a calculation ( f = 1/2 RC ). How do they compare?
f (graph)=
f (calculation)=
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(i) For frequencies less than the frequency calculated in part 2(h), is the network primarily resistive or capacitive? How about for frequencies greater than the frequency calculated in part 2(h)?
(j) The phase angle by which the applied voltage leads the current is determined by = tan-1(XC/ R) (as obtained from the impedance diagram). The negative sign is clear indication that for capacitive networks, i leads v. Determine the phase angle for each of the frequencies in Table 4. Table 4 Frequency 0.1 kHz 0.2 kHz 0.5 kHz 1 kHz 2 kHz 6 kHz 10 kHz 100 kHz
R (measured)
X C
=
-tan-1(XC/ R)
(k) At a frequency of 0.1 kHz, does the phase angle suggest a primarily resistive or capacitive network? Explain why.
(l) At frequencies greater than 2 kHz, does the phase angle suggest a primarily resistive or capacitive network? Explain why.
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(m)Plot versus frequency for the frequency range 0.2 kHz to 10 kHz on Graph 4. At what frequency is the phase angle equal to –45 o? At –45o what is the relationship between X C and R? Using this relationship, calculate the frequency at which = -45 o.
f ( = -45o) = X C vs. R = f (calculated) = How do the two levels of frequency compare?
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Problems
1. Given the network of fig. 1, calculate f = 1 kHz, calculate the magnitude and phase angle of the input impedance and compare the results to those obtained experimentally in part 2(a) ( Z T = E p-p/ I p-p) and calculated in Table 4.
Calculate: ZT= θ =
Experimental: ZT= θ =
2. Given the network of Fig. 1 with f = 1 kHz, calculate the levels of V C, V R and I (all peakto-peak values) and compare to the measured values of Table 2.
Calculate: VC= VR = I =
Measured: VC= VR = I =
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