PSU-Physics
PH-315
Andrés La Rosa
RLC SERIES CIRCUIT RESONANCE
(Complex impedance)
_______________________________________________________________________________
PURPOSE
To observe the frequency-dependence of the impedance in an AC circuit and the
resonance frequency.
We will use the resonance frequency to determine the inductance of a coil. The
oscilloscope will be used to measure the phase lags between the voltage and the
current.
THEORETICAL CONSIDERATIONS
We will use the following notation:
vA, i, z complex quantities.1
VoA, Io, Zo constant real quantities.
j2 = -1
In the circuit shown the three elements, C, L and R connected in series.
The current i is the same across each element.
The input voltage is given by,
vA  VoA e j t
(Driving voltage)
(1)
Here VoA is a positive real number; the input voltage controlled by the user.
A
i
C
VoA
vA
t
L
vA
R
Fig. 1 RLC series circuit.
1
Fig.2 Phasor vA , of amplitude VoA rotating
with angular velocity 
For ab rief description on complex variable see the companion file Complex Variable available online at the PH315 webpage.
http://www.pdx.edu/nanogroup/sites/www.pdx.edu.nanogroup/files/NOTES_2013__COMPLEX_NUMBERS_for
_Exp_RLC_SERIES.pdf
1
Since the driving voltage is changing harmonically with angular frequency , we assume that
the steady current also changes harmonically with the same frequency , except with an
eventual phase difference.
i  Io e
j ( t -  )
(2)
The peculiar characteristic is that I o  I o () and    ()
Given C, L and R, as well as  and VoA, we have to figure out the values of Io and .
Capacitor complex impedance
Since the current i is the same across each element, we will express their drop of voltage in
terms of that current.
vR  R i

Voltage across the resistor:

Voltage across the capacitor:
On one hand:
vC 
d vC
dt

(3)
q
, which implies
C
i
C
(i)
On the other hand: The voltage across the capacitor should be changing also
harmonically at the frequency  ; like vC  ( vC ) o e j t .

This implies
d vC
 j ( vC ) o e j t
dt
d vC
 j vC
dt
(ii)
Equating the two expressions (i) and (ii), one obtains,
i
;
j vC 
C
Or, equivalently,
 1
vC  
 j C

i

(4)
 1 - j  /2) 

e
 i
 C
Voltage across the capacitor is
lagging 90o with respect to the
current.
The expression
2
ZC 
1
(5)
j C
is called the capacitance complex impedance.
vC  Z C i
(6)
Inductor complex impedances
Voltage induced across the inductor:
di
On one hand:
vL  L
dt
On the other hand: Since the current is assumed to be changing according to
i  I o e j ( t - )
Then
vL  L
di
 L j I o e j ( t - )
dt
 L j i
vL   jL i
Thus,
(7)
 [Le j  /2) ] i
The expression
Z L  j L
Voltage across the inductor
leading 90o with respect to the
current.
(8)
is called the inductance complex impedance.
vL  Z L i
(9)
Kirchhoff law for the RLC circuit
Kirchhoff law states that, at any instant of time, the addition of the three voltages vR , vC ,
and vL in circuit 1 above should be equal to vA..
vA  vR +
 Ri
+
vL
+
jL i
+
vC
1
j C
3
i
1
vA  [ R  j L 
j C
(10)
]i
More explicitly,
VoAe j t  [ R  j L 
1
j C
] I o e j ( t -  )
Cancelling the factor e j t ,
VoA  [ R  j L 
VoA  [ R ] I o e - j  
1
j C
] I o e- j 
[ j L ] I o e - j 
(11)

[
1
j C
] I o e- j 
(12)
where I o  I o () and    ()
Expression (12) is very convenient for answering the questions that appear in the experimental
section below [in particular see question e) in that section]. For a given , you have to measure
both Io and , and subsequently verify that those experimental values fulfill expression (12).
But, how to calculate or measure Io and  ?
First, we have to calculate Io and  in terms of R, L, C, and VoA .
 Impedance
Let's identify first the total complex impedance of the RLC series circuit. In expression (10),
1
vA  [ R  j L 
] i , we identify such total impedance as,
j C
z  R  j L 
1
j C
 R  j ( L 
1
)
ωC
Total impedance RLC series circuit
(13)
That is, z has the form
z  Z o e j
where
(14)
1

 ( L   C )
  arctan 
R







and
Z o  R 2  ( L 
1 2
)
C
4

L

R
L – 1/(C)
1/(C)
Fig. 3 Representation of the impedance in the complex plane.

Solving for Io and  from expression (11)
VoA  [ R  j L 
1
j C
] I o e- j 
Using (13) and (14),
VoA  [ z ] I o e -j 
VoA  [ Z o e j ] I o e -j 
Since VoA is a real number, the right side of the last expression has to be real as well. This
requires that  has to be equal to . The latter also implies that IoA = VoA / Zo. That is,
V
I o ( )  oA =
Zo
VoA
1 2
R  (L 
)
C
2
1

 (L   C )
and  ( )    arctan 
R







(15)
In summary, for the RLC-series circuit in Fig. 1 above,
i  Io e
where
j ( t -  )
Io 
(16)
VoA
1 2
R  (L 
)
C
 I o ( )
2
1

 ( L   C )
 ( )  arctan 
R


5





Notice,
at very low frequencies:
 0  -  (current I leads the voltage v)
1
we have:= 
LC
at   o 
at very high frequencies

v
 
  (current i lags the voltage v)

i

v
v
i


i
i
~ 0
  ∞

v
~ 0
Fig. 4 Phasor
v
and phasor i rotate with the same frequency. Their
phase difference changes as the frequency  varies. Here
v  VoA e j t

A similar diagram is shown in Fig. 5.

v
i
se is measured relative to

v
i
Case: <o
Case:  >o
< 0
> 0
Fig. 5 The phasor voltage and the phasor current rotating
counterclockwise. Left: The current lags the voltage. Right: The
current leads the voltage.
Notice in (14) and (15) that when L  1 /(C) ,
The impedance is minimum
The current is maximum
Hence
  o 
1
LC
(17)
6
is called the resonance frequency.




Fig. 6 Resonance curve (from expression (16)..
Example (Taken from James Brophy, “Basic Electronics for Scientists,” McGraw Hill). Calculate
the current in the circuit and the voltages across each component for the circuit in Fig. 7.
Consider a) the case of resonance, and b) a case out of resonance.
Io (mA)
i
A
C= 0.1 F
L= 250 mH
10V
R= 100 
Fig. 7 RLC series circuit.
The input voltage is
Fig. 8 Resonance curve of circuit 7.
vA  10 Volts e j t
VoA
After substituting the values for R, C and L into Eq. (16), I o ( ) 
1 2
R  (L 
)
C
found that the current in the circuit changes with frequency as illustrated in Fig. 8.
The current maximum is I0.max = 10/100 = 0.1 Amps at the resonance frequency,
2
fo 
1
2 LC

1
6.28 250  10-3  0.1  10-6
a) Case of resonance
7
 1006 Hz
, it is
The voltage drop across each element at resonance illustrates an important feature of
alternating currents. Let’s see.
Current at resonance:
I = 0.1 Amp
Voltage amplitude across the resistor:
VR = R I = 100   0.1 A = 10 Volts
Voltage amplitude across the capacitor: VC =
1
1
I =
 0.1 A = 158 Volts
2  1006 Hz  10-7 F
C
Voltage amplitude across the inductor: VL =  L I = 2  1006 Hz  0.25 H  0.1 A = 158 Volts
It is evident the amplitude voltage drops around the circuit do not add up to zero.
Notice however that,
According to expression (4), the phase angle at the capacitor voltage vC
is -90o with respect to the current.
According to expression (7), the phase angle of the voltage across the
inductor is, +90o with respect to the current.
According to expression (16), the phase difference between the applied
voltage and the current is zero;  = 0.
i = 0.1 Amp  e j  t
From (16), the current at resonance (f=1006 Hz) is:
Voltage across the resistor:
vR =
Ri
Voltage across the capacitor:
vC =
1
Voltage across the inductor:
vL = j L i
j C
100  i = 10 Volts  e j  t
=
i =  j 158 Volts  e j  t
j 158 Volts  e j  t
=
This means, the instantaneous voltage drop across the two reactances cancel each other.
Accordingly, the drop across the resistor equals the applied input voltage (10 Volts).
This confirms the Kirchhoff law.
Kirchhoff’s voltage rule is valid when the phases of the currents and
voltages are taken into account.
b) A case out of resonance.
Of course Kirchhoff’ law is equally valid at frequencies away from resonance.
Consider the voltage drop around the circuit in Fig. 7 at a frequency of f= 1200 Hz.
According to expression (16), at f=1200 Hz the phase angle is,
1

 ( 2  1200  0.25 )  2  1200  10 -7 )
  arctan 
100


8





 1885  1326 
=  arctan  5.5865  = 79.85o or 1.39 radians.
 arctan 

100


According to expression (16), at f=1200 Hz the current is,
j ( t -  )
i  I o ( ) e
= I o () e-j   e j  t
VoA
=
R 2  ( L 
=
1 2
)
C
e-j   e j  t
10
100  (1885  1326)
2
2
e-j 1.39  e j  t
=
10
e-j 1.39  e j  t
567.9
=
1.76  10-2  e -j 1.39  e j  t
= 1.76  10-2  [ cos(1.39) - j sin (1.39) ]  e j  t
Voltage across the resistor:
vR = R i = 100  i = 1.76 Volts  [ cos(1.39) - j sin (1.39) ]  e j  t
= [ 0.31 - j 1.73 ]  e j  t
Voltage across the capacitor:
1
1
vC =
i =j
i
2  1200  10-7
j C
= e -j /2 1326  i
= e -j /2 1326  1.76  10-2  e -j 1.39  e j  t
= 23.3 e -j 2.96  e j  t
= 23.3 [ cos(2.96) - j sin (2.96) ]  e j  t
= [ -22.91 - j 4.2 ]  e j  t
Voltage across the inductor:
vL = j L i = j (2  1200  0.25 ) i = e j /2 1885 i
= e j /2 1885  1.76  10-2  e -j 1.39  e j  t
9
= 33.17  e j 0.18  e j  t
= 33.17  [ cos(0.18) + j sin (cos(0.18) ]  e j  t
= [ 32.63 + j 5.9 ]  e j  t
vR +
vL
+
vC = { [ 0.31 - j 1.73 ] +
[ 32.63 + j 5.9 ] +
[ -22.91 - j 4.2 ] }Volts  e j  t
= { 10.03 - j 0.03 }  e j  t
 10 Volts  e j  t
which is equal to vA  10 Volts e j t
Thus, for two different driving frequencies, f = 1006 Hz and 1200 Hz, we have
verified that the Kirchhoff’s voltage rule is valid for the circuit in Fig. 7 .
EXPERIMENTAL CONSIDERATIONS
A two-channel oscilloscope is our basic measuring tool in alternating current (AC) circuits.
The resistor Ro is a known resistor (whose value is on the order of 100 ).
The ground of the oscilloscope will be connected to G.
Channel-1 will monitor the input voltage at A.
Channel-2 will monitor the voltage at B.
Measure the frequency of the input voltage using the oscilloscope. (Do not trust the reading
from the signal generator knob).
In Fig. 7 the voltage at B is the voltage across the resistor VR = VR (t). This is the trace you see in
your oscilloscope.
Since VR(t) = R0 I (t), then the current I= I (t) can be obtained.
I (t ) 
VR ( t )
R
(18)
In Fig. 7 the voltage at A is the driving voltage VA = VA(t). It is the trace you see in your
oscilloscope.
10
vA  VoAe j t
VA(t) = Real { v A }
= Real { VoA e j t }
=VoA cos ( t )
C= 10 nF

VoA is a constant real-value.
(It is the amplitude of your sinusoidal
input voltage).
A
L
B
i  I o e j ( t - )
I(t) = Real { i }
= Real { I o e j ( t - ) }
= Io cos ( t-)
Ro
Ground

G
Io = Io() is a real-value.
Fig. 7 RLC series circuit.
Suggestion: use C~ 10 nF or 100 nF for this experiment.)
Note: If a capacitor is, for example, labeled 473G, it means C = 47  103 picoFarads.
MEASUREMENTS
a) Zo The magnitude of the impedance z of the test circuit.
Zo, is determined from the ratio of amplitudes of the two signals VA(t), and I(t) :
| z |= Zo =
Amplitude of the signal VA (t ) VoA
=
Amplitude of the signal I(t )
Io
Measure and plot I o as a function of .
From the measurements above, plot also | z | = Zo as a function of .
Compare your results with the ones obtained using expression (14) above. Plot these
theoretical values for Zo as a function of .
b) The phase can be measured on the oscilloscope as a distance between the points where the
two traces cross the horizontal axis, and converted to degrees by comparing the half (or
full) wavelength as shown on the oscilloscope. See figure below.
Pay attention during the measurements to verify if  is positive or negative. That is,
whether the VR is lagging or ahead of VA.
Plot  = ().
11
Real voltages
measured by the oscilloscope
VA
Complex analysis
Phasors
vA
vR
 v
R
t
time

180o
VA =VoA Cos (t); VR = VoR Cos (t - )
Phasors vA and vR rotating with
The traces show VR lagging VA by .
angular velocity vR lags vA by .
c) During the course of measurements you take enough data to make a graph of both
impedance Zo and phase  as a function of frequency .
How do the graphs change when using a higher value of the resistance Ro?
In order to compare your results to the theory, determine first the value of the inductance L.
This is best done early on by locating the resonance frequency =o, the one that makes the
impedance Zo minimum and the phase between VA and VR equal to zero. Values of  around o
is the frequency region where a majority of your data needs to be taken.
By swapping C with R, and L with R, it is possible to measure the voltage across C and across L.
d) At resonance condition, measure the amplitude of the voltage across the three elements
(you will need to swap C and L, one at a time, with the element R). Do these voltage values
add up?
We ask you to corroborate the validity of Kirchhoff law, expression (12). Fill out the blanks in
the two boxes given below.
e) Choose a frequency * at which Io() is ~50% of the current obtained at resonance, That is,
50% of Io()
We ask you to verify experimentally and corroborate theoretically the validity of Kirchhoff
law, expression (12):
By swapping the position of the capacitor and inductor with the position of the resistor
(in the circuit shown above) measure the amplitude and phase of the individual voltages
vc, vL, vR.
Measure the complex values for vc, vL, vR.
Fill out the blanks in the two boxes given below.
For the values of R, L , C,  , that you are using, calculate [ R  jL 
verify if it coincides with the value of the input voltage.
12
1
] I o e - j ) , and
jC
Expression
Value at
Value at

_________________________________________________________
Io()

i()


Expression
Value at
Value at

___________________________________________________________
vR ()
R i()
vL ()
j L i()
vC ()
1
j C
i()
OVERCOMING some SHORTCOMINGS ENCOUNTERED in the EXPERIMENT
Problem: Variability of the input voltage amplitude
Solution: By the normalization method.
While sweeping the frequency of the driving voltage VA around the resonance frequency, it is
observed that the amplitude VoA also changes. Ideally, it would be desirable that this amplitude
remains constant.
The instability of the VoA makes somewhat inaccurate the procedure of locating the resonance
frequency by monitoring the frequency at which the current is maximum. (When you do this,
you may find that at the frequency where the current is maximum, VA(t) and I(t) are not in
phase.) Still this is still a good preliminary step, since it helps to identify the frequency range
where we have to take a closer look.
Next, sweep the frequency a bit, until you find that the phase between VA and I is zero.
Now we suggest to manually normalize the input voltage VA. By this we mean to choose a fixed
amplitude value for VA, let’s call it VA, fixed. Then:
For a given frequency, record the current amplitude and phase.
Move to another frequency (you may notice that the amplitude of VA has changed.) Turn
the knob that controls the amplitude of VA until you obtain back the predetermined fixed
value VA, fixed. Then record the current amplitude and phase.
And so on, repeat the procedure for each frequency around the resonance frequency.
13
14
Or, i 
v
1

v , which gives,
Z R  j (L  1 )
C
1
i
R 2  (L 
1 2
)
C
e j v
where
1

 (L   C )
  arctan 
R







Notice,
at very low frequencies:
at   o 
1
we have:
LC
at very high frequencies
Thus we have,
= 

 - 
v  Vo e jt
iIe
i
 + 
j (t  )
(driving voltage)
 I e jt e j
(current)
dq
I j (t  )
I j (t  ) I j (t   / 2)
j (t   / 2)
implies q 
e
j e
 e
Qe
dt
j


q  Q e j (t   / 2)  Q e j (t  )Q e-j / 2
(charge)
q lags the current i by /2
k
Using the phasors representation,
A
Im
z  Ae
jt
x
z
m
t
Real
15
Real
Im
v
v
v
q
Real
q
q
At ~ 0
At ~ 0
At   ∞
Real
Im
i
v
v
i
Real
v
i
At ~ 0
At ~ 0
16
At   ∞