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The RC Series AC Circuit
Purpose

To investigate the effects produced by an AC voltage applied to an RC series circuit.
The circuit will be analysed using phasors.
Introduction and Theory
As we saw in “Discharging and Charging a Capacitor” lab, there exists a current “through” the
capacitor during the charging and discharging processes. If we connect a capacitor to an AC
source, there will be an alternating yet continuous current “through” the capacitor, as shown in
Figure 1. The current is not in phase with the applied voltage: it is leading the voltage by 90*.
The magnitude of the current depends on how fast the charging process is, which depends the
frequency of the AC source and the capacitance.
V  Vmax sin t
I
C
V
I
Figure 1a
Figure 1b
Figure 1. Voltage and current of a capacitor connected with an AC power source
Basic rules for capacitor in AC circuit

I is leading V by 90;

I max 
Vmax
1
1
where X C 

is called capacitive reactance.
XC
C 2fC
We will not test the above theory directly, as it is impossible for us to find a circuit without any
resistance. Our AC source, the function generator, has an internal resistance of 50 . Instead,
we will test RC series circuit.
*
The phrase “ELI the ICE man” is a handy way to remember the relations between the
voltages of RC and RL circuits. The current is I and the voltage is E. For inductance L, current
I lags E by 90 so “ELI”, and for capacitance C, current leads E by 90 so “ICE”.
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Phasor Diagram
V  Vmax sin t
R
I
VR
C
Figure 2a
V

t
VC
Figure 2b
Figure 2. Phasor diagram for a RC series AC circuit
Consider the circuit in Figure 2a. A resistor and capacitor are in series with a sinusoidal voltage
source of amplitude Vmax and angular frequency.
For series connection, the current I through each element will be the same. The voltage across
the resistor VR and the voltage across the capacitor VC will combine to balance the applied
voltage V. Because the power is an AC source rather than a DC source, it is the instantaneous
voltage that will add up rather than the maximum voltage:
V (t )  VR (t )  VC (t ) or Vmax sin t  VR max sin(t  R )  VC max sin(t  C )
Let’s examine the phases. For the resistor, the voltage is in phase with the current. For the
capacitor, the voltage lags the current by 90. Therefore, the voltage across the resistor always
leads the voltage across the capacitor by 90:
Vmax sin t  VR max sin(t   )  VC max sin(t    90 )
Above relation can be graphically presented by Figure 2b, where each voltage is represented
by a vector rotating about the origin at an angular frequency ωt. This is called a phasor
diagram.
The phasor diagram contains all the information necessary to analyse the circuit: the
magnitude of the “vector” is the maximum of the voltage, while the “y-component” gives the
instantaneous voltage value. The angle between the current I and the overall voltage V is called
the phase angle.
From the phasor diagram, we know
Vmax  VR max  VC max
2
Because Vrms 
1
2
2
Vmax , it follows that Vrms  VRrms  VCrms
2
2
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We also get from the phasor diagram that
tan  
VC max
VR max
Because VR max  RImax and VC max  X C I max , we have tan  
XC
R
The impedance of the whole circuit will depend on both the capacitor and the resistor:
VR max  VC max
V
Z  max 
Imax
I max
2
2

(RI max )2  ( X C I max )2
I max
 R 2  XC
2
We will test these relations in our lab.
Phase angle measurement
Phase angle can be measured using two different methods, the peak shift method and the
Lissajous method.
The Peak Shift Method
We can measure the phase angle directly using the dual-trace oscilloscope. Since the current
is leading the total in phase in time, their sine waves will appear shifted horizontally. Therefore,
if we connect the total voltage to CH1 and the voltage across the resistor (which is directly
proportional to the current) to CH2, we can measure the time difference directly. This shift is
proportional to the phase difference between the signals.
Figure 3. The peak shift method
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The Lissajous Method
B
A
Figure 4. The Lissajous Method
Again we will connect the total voltage to CH1 and the resistor voltage to CH2, but this time we
set the oscilloscope to x-y mode. If the x and y signals are in phase, the resulting trace on the
screen will be a slanting line. If they’re out of phase, the resulting trace will be a slanted ellipse.
If we compare the maximum total voltage (A, above) to the total voltage when the resistor
voltage is reading 0 (B, above), we can calculate that the x-signal leads the y-signal by a phase
angle of
B
  sin 1
A
And the meanings of B and A are shown. Note that for our oscilloscopes, channel 2 is the
x-signal.
A feature about the Lissajous method is that the x and y scales do not need to be the same,
since we are only comparing the maximum total voltage with the total voltage at another time.
Apparatus
BK Precision 2120 dual-trace oscilloscope, BK Precision 4011 function generator, two Fluke
73III multimeters, BK Precision 875B LCR meter, decade resistance box, 0.047 µF and 0.22 µF
capacitors.
Record the following:






Internal resistance of the function generator:
Accuracy of the Fluke multimeter:
Accuracy of the LCR meter:
Accuracy of the decade box:
Actual capacitance (and uncertainty) of the 0.047 µF capacitor:
Actual capacitance (and uncertainty) of the 0.22 µF capacitor:
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Data
Connect the circuit shown in Figure 2a.
Part A: Testing Vrms  VRrms  VCrms  (1)
2
2
Set the output of the function generator to sine wave with f = 200 Hz and Vrms = 2.0 V (using the
multimeter). Use the 0.047 µF capacitor and set the decade resistance box to 10 k.
Measure Vrms, VRrms and VCrms using the multimeter. Change the values of f, C, and R and
repeat measurements 2 or 3 times. In the space below, list following results in a table: f, R, C,
VRrms, VCrms, Vrms calculated from Eq. (1) and Vrms from the multimeter measurement. Verify
Eq. (1) for each case by comparing the percentage difference with the percentage uncertainty.
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Part B: Testing the capacitive reactance X C 
1
 (2)
2fC
Set f = l00 Hz, R = 1.8 k, and C = 0.22 µF.
If Eq. (2) holds, a graph of XC vs l/f should be a straight line. Increase the generator frequency
from 100 Hz to 1000 Hz in steps of 100 Hz. For each frequency, measure Irms and VCrms using
two separate Fluke multimeter and record the results below.
Using Excel, calculate XC by its definition, XC=VCrms/Irms, and plot a graph of XC versus l/f.
Sketch the graph below. Is it a straight line? In other words, does your result agree with Eq. (2)?
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Part C: Testing the impedance: Z  R 2  X C  (3)
2
Set f = 200 Hz, R = 1.8 k and C = 0.22 µF.
Using the multimeter, measure and record Irms, Vrms, f, R and C.
Calculate XC using Eq. (2). Calculate Z using its definition Z=Vrms / Irms. List results and explain
whether they support Eq. (3).
Change to another set of f, R and C and repeat above process. List results below.
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Part D: Testing the phase angle between the current and the applied voltage:
X
tan   C  ( 4)
R
Set f = 400 Hz, C = 0.22 µF and R = 1.8 k.
With the oscilloscope, measure  using both the peak-shift and Lissajous methods. Change to
a different set of f, C, and R and repeat measurements for a different value of .
Compare your two values with the values obtained by Eq. (4), calculating XC using Eq. (2).
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