lab elec 08 - frequency response of the rc network

7/25/2019 Lab Elec 08 - Frequency Response of the RC Network

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8. FREQUENCY RESPONSE OF THE SERIES R-C NETWORK

Objectives:

1.

Note the effect of 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 seriesR-C network

List of Equipment and Component:

1.

1 x Resistor 1kΩ (1/4 –W)

2. 1 x Capacitor 0.1μF

3.

DMM

4.

Oscillator

5.

Function generator

6.

Frequency counter (if available)

Background:

For 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 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 samecharacteristics.

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 –90o (v lags i by 90o).

At increasing frequencies X 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 0o

(v and i in phase).

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 voltageacross the capacitor and resistor. Explain why



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(h)

At the point where VC=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 VC(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

p p

p pC

C I

V X

)(

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? Commentaccordingly.



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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 E p-p I p-p

p p

p p

T I

E

Z

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C T X R Z

0.1 kHz 4 V

0.2 kHz 4 V

0.5 kHz 4 V

1 kHz 4 V

2 kHz 4 V

4 kHz 4 V

6 kHz 4 V

8 kHz 4 V

10 kHz 4 V

(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 equation22

C T X R Z

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 R (measured) X C = -tan-1(XC/ R)

0.1 kHz

0.2 kHz

0.5 kHz

1 kHz

2 kHz

6 kHz

10 kHz

100 kHz

(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 –45o? At –45o what is the relationship

between X C and R? Using this relationship, calculate the frequency at which = -45o.

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: Experimental:

ZT= ZT=

θ = θ =

2.

Given the network of Fig. 1 with f = 1 kHz, calculate the levels of V C , V R and I (all peak-

to-peak values) and compare to the measured values of Table 2.

Calculate: Measured:

VC= VC=

VR = VR =

I = I =

lab elec 08 - frequency response of the rc network

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