1

PSU-Physics PH-315 Andrés La Rosa

RLC SERIES CIRCUIT RESONANCE

(Complex impedance) _______________________________________________________________________________

PURPOSE To observe the frequency-dependence of impedance in an alternating current (AC) circuit. We will measure the resonance frequency and use its value to determine the inductance of a coil, assuming the values of the capacitance and the resistor. The oscilloscope will be used to measure the phase lags between voltage and the current across a resistor, a capacitor and an inductor. The analysis is undertaken using complex variable.

THEORETICAL CONSIDERATIONS I. Impedance IA. The RLC Circuit in series IB Relationship between the voltage and current across each element Resistance Capacitor complex impedance Inductor complex impedance

II. Kirchhoff law for the RLC circuit

III. How to calculate and measure current amplitude Io() and phase lag () ? Complex impedance of the RLC-series circuit

Calculation of Io and

IV. Resonance Resonance condition

Relative orientation of the phasors input voltage vA and current

response i at different values of Example: Analysis at resonance and out of resonance conditions

EXPERIMENTAL CONSIDERATIONS

Experimental setup

Measurements a) Finding L

b) Plot oI as a function of .

c) Plot the experimental values of | z | = Zo = o

oA

I

Vas a function of

d) Plot the experimental values of the phase vs frequency, = ()

e) Influence of the resistance (Repeat the experiments using at least two different values)

f) Measurement of the complex voltages across R, C and L.

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I. Impedance

We will use the following notation:

vA, i, z complex quantities (lower case fonts).1

VoA, Io, Zo constant real quantities (capital fonts).

j2 = -1 (we use j instead of the more common symbol i, to avoid confusion with the current.)

IA. The RLC Circuit in series

The circuit in Fig. 1 shows the three elements, C, L and R connected in series.

The connection in series implies that the current i is the same across each element.

L

R

C

vA

i

Fig. 1 RLC series circuit. Fig. 2 Phasor vA , of amplitude VoA, rotating

with angular velocity

vA t VoA

A

Im

Real

Input voltage vA

The input voltage is given by,

t j

eV oAAv (Input driving voltage) (1)

Here oAV is a positive real number;

The diving input voltage is controlled by the user (hence, it is assumed to be known.) Current i

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. (can be positive or negative.)

1 For a brief description on complex variable see the companion file Complex Variable available online at the PH-

315 webpage. http://www.pdx.edu/nanogroup/sites/www.pdx.edu.nanogroup/files/NOTES_2013__COMPLEX_NUMBERS_for_Exp_RLC_SERIES.pdf

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)(

-tj

eIoi where oI and are unknown (2)

The peculiar characteristic of the RLC circuit is that oI and depend on the

frequency ! That is, )( oo II and )(

One of the tasks in this lab is:

Given C, L and R, as well as VoA, we have to figure out the values of Io () and ().

vA

I

VoA

VA

time

i

t

Trace I lags trace VA by

VA =VoA Cos (t); I = Io Cos (t - )

Real voltages Complex analysis measured by the oscilloscope Phasors

Io

Phasor i lags phasor vA by

Phasors vA and i rotating with

angular velocity

Illustration of possible profiles of the input voltage and the current in a RLC circuit, corresponding to a particular value of the frequency.

IB. Relationship between voltage and current across each element

Resistance Voltage across the resistor: i R Rv (3)

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.

Voltage across the capacitor:

On one hand: C

q Cv , which implies

C

i

dt

d

Cv (i)

On the other hand: The voltage across the capacitor should be changing also

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harmonically at the frequency ; like t o )(

jeCvvC .

This implies,

e j dt

d j t o)(

CC v

v

j dt

dC

C vv

(ii)

Equating the two expressions (i) and (ii), one obtains,

C

i j Cv ;

Or, equivalently,

i Cj

1

Cv (4)

Alternatively, since )

1 /2j-e

j

one can write,

i eC

/2j-

)

1

Cv Voltage across the capacitor is

lagging 90o with respect to the current.

Exercise: Assuming )(

-tjeIoi , sketch the phasors Cv and i ( together in the

same diagram) for arbitrary fixed values of , , and t.

In (4), we identify the quantity,

Cj

zC

1

(5)

Which is called the capacitance complex impedance.

Thus, one re-writes expression (4) as,

i z CCv (6)

Notice in (5) that C

z changes with the frequency .

Inductor complex impedance Voltage induced across the inductor:

On one hand: dt

i dL Lv

On the other hand: Since the current is assumed to be changing according to

5

)( -tjeIoi

Then

)( -tjeI j L

dt

i dL o Lv

i j L

Thus,

i jL Lv (7)

Since )/2jej

i eL /2j ][ )

Lv Voltage across the inductor leading

90o with respect to the current.

Exercise: Assuming )(

-tjeIoi , draw the phasors Lv and i (together in the

same diagram) for arbitrary fixed values of , , and t.

The expression Lj zL (8)

is called the inductance complex impedance.

i z L

Lv (9)

Notice L

z changes with the frequency .

II. Kirchhoff law for the RLC circuit Consider the circuit shown in Fig. 1 Kirchhoff law states that, at any instant of time, the addition

of the three voltages Rv , Cv , and Lv should be equal to vA..

Av Rv + Lv + Cv

i R + i jL + i Cj

1

iCj

LjR ]

1 [

Av . (10)

Experiment assignment: Place the sinusoidal driving voltage in channel-1 of the oscilloscope. In channel-2 monitor, one at a time, the voltage across R, C, and L, respectively.

Plot all the waveforms in a single graph, for comparison.

One can write expression (10) more explicitly,

6

)(

]

1 [ V

-tt jj

eICj

Lj ReA oo

Cancelling the factor t je ,

]

1 [ V

j-eI

Cj Lj RA oo , (11)

where Io () and ()are still unknown.

Equivalently,

]

1[ ][ ][ V

j-j-j-eI

Cj eI Lj eI RA oooo

(12)

2/2/

]

1[ ][ ][ V

jj-jj-j-

eIC

eI L eI R oooAo

where )( oo II and )( are still unknown.

Kirchhoff law in phasor form

Expression (12) is very convenient for answering the questions that appear in the experimental section below [in particular see the requirement f ) in that section].

For a given , you have to measure both Io and , and subsequently verify that those experimental values fulfill expression (12).

Notice, the left side in expression (12) is a real quantity; therefore the sum of the voltages on the right side has to be real as well (i.e. when entering the experimental values the right side has to be a real number).

III. How to calculate and measure current amplitude Io() and phase lag () ? We have to calculate Io and in terms of R, L, C, and VoA . It is convenient to calculate first the total complex impedance of the circuit.

Complex impedance of the RLC-series circuit In expression (10),

iCj

LjR ]

1 [

Av ,

we identify the total impedance of the circuit iz / Av as,

) (

ωC

1L jR

Cj

1LjRz

(13)

Total impedance RLC series circuit.

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Exercise. Show that z can, alternatively, be written in the form,

j

eZz o

where (14)

1

arctanR

CL

and 22 )

1 (

CLRZ

o = Zo()

R 1/(C)

Fig. 3 Representation in the complex plane of the total complex

impedance z, given by expression(14).

L L– 1/(C)

Im

Real

Calculation of Io() and ()

From expression (11)

]

1 [ V

j-eI

Cj Lj RA oo

Using (13) and (14),

][V-j

eIz A oo

] [V-jj

eIeZA ooo Since VoA is a real number, the right side of this expression

has to be real as well. This requires that:

has to be equal to .

The latter also implies that Zo () = VoA / Io (). (15)

That is,

o

oA

ZI

V)( o =

22 )1

(

V

CLR

oA

and (16)

8

)

1 (

arctan)(R

LC

In summary, for the RLC-series circuit in Fig. 1 above,

t j

eV oAAv )

( - tjeoIi

Input voltage (given) Current (response from the circuit)

where )(

)1

(

V

2

2

oo I

CLR

I

oA

(17)

)

1 (

arctan)(R

CL

Kirchhoff law

2/2/

]

1[ ][ ][ V

jj-jj-j-

eIC

eI L eI R oooAo

IV. Resonance

Resonance condition Notice in (14) and (16) that,

when )/(1 CL ,

the impedance oZ is minimum, and the current oI is maximum

Hence,

LCo

1 is called the resonance frequency. (18)

Fig. 6 Resonance curve (from expression (16).

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Note: In the experimental section, C will be a known capacitance and the resonance frequency

will be determined experimentally by acquiring a plot of I vs . From these two values the value of L will be determined.

Relative orientation of the phasors vA and i at different values of

At very low frequencies: 0

-

)

)( /2 (

tjeoIi (current i leads the voltage vA)

at LC

o

1 : =

)

)( ( tjeoIi

o (current i in phase with the voltage vA)

at very high frequencies

)

)( /2 (

tjeoIi (current i lags the voltage vA)

vA

vA

vA

i i

~ 0 ~ 0 ∞

v

i

i

Real

Im Im Im

Real

Fig. 4 Phasor vA and phasor i rotate with the same frequency. Their phase difference

remains constant at fixed frequency, but changes as the frequency varies. Here t j

eV oAA v

A similar diagram is shown in Fig. 5.

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Fig.3 Notice, the phase is measured relative to

the voltage phasor

vA

i

Case: <o Case: >o < 0 > 0

vA

i

Real Real

Im Im

Fig. 5 The phasor voltage and the phasor current rotating counterclockwise. Left: The current leads the voltage. Right: The current lags the voltage.

Example: Analysis at resonance and out of resonance conditions (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.

The input voltage is Av 10 Volts t je

After substituting the values for R, C and L into Eq. (17),

)1

(

V )(

22

CLR

Io

oA , it is

found that the current in the circuit changes with frequency as illustrated in Fig. 8.

The maximum current is I0,max = VoA / Zo() = = VoA / R = 10V/100 = 0.1 Amps, which occurs

at the resonance frequency = o . Using o 2fo,

Hz1006 101.01025028.6

1

2

1

6-3-

LCfo

L= 250 mH

R= 100

C= 0.1 F

10V

i

Fig. 7 RLC series circuit.

A

Fig. 8 Resonance curve of circuit 7.

Io (mA)

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a) Case of resonance 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 = C

1

I =

F1 7-10 Hz006 2

1

0.1 A = 158 Volts

Voltage amplitude across the inductor: VL = L I = Hz006 2 1 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) i Cj

1

Cv , the phase angle of the voltage Cv across

the capacitor is -90o with respect to the current.

According to expression (7) i jL Lv , the phase angle of the voltage Lv across

the inductor is, +90o with respect to the current.

According to expression (3) i R Rv , the phase difference between

the applied voltage and the current is zero;

= 0.

From (17), the current at resonance (f=1006 Hz) is: i = 0.1 Amp t j e

Voltage across the resistor: vR = R i = 100 i = 10 Volts t j e

Voltage across the capacitor: vC = Cj

1

i = j 158 Volts t j

e

Voltage across the inductor: vL = Lj i = j 158 Volts t j e

Notice, the instantaneous voltage drop across the two reactances (capacitor and inductor) cancel each other out.

Thus , adding the three voltages vR + vC + vL one obtains 10 Volts t j e ,

which equals the input voltage Av 10 Volts t je .

This confirms the Kirchhoff law.

Kirchhoff’s voltage rule is valid when both the magnitude and phases of the currents and voltages are taken into account.

b) Case: Selecting a frequency different than the resonance frequency

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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 (17), at f=1200 Hz the phase angle is,

)0 12002

1 ) 25.012002(

arctan7-

100

1

3261 8851 arctan

100 = .58655 arctan = 79.85o or 1.39 radians.

According to expression (17), at f=1200 Hz the current is,

)

)( (

- tjeoIi =

)(

tj -j eeoI

= 22 )

1 (

V

CLR

oA -j e

tj e

= 22 )13261885(00

10

1

.391 -j e

tj e

= 9.567

10 .391 -j e

tj e

= 1.76 10-2 .391 -j e

tj e

= 1.76 10-2 [ cos(1.39) - j sin (1.39) ]

tj e

Voltage across the resistor:

vR = R i = 100 i = 1.76 Volts [ cos(1.39) - j sin (1.39) ] t j e

= [ 0.31 - j 1.73 ] t j e

Voltage across the capacitor:

vC = Cj

1

i = j

7-0 12002

1

1 i =

/2-j e

1326 i

= /2-j

e 1326 1.76 10-2 .391 -j e t j

e

= 23.3 2.96-j e

tj e

= 23.3 [ cos(2.96) - j sin (2.96) ]

tj e

13

= [ -22.91 - j 4.2 ] t j e

Voltage across the inductor:

vL = Lj i = j ) 25.012002( i = /2j e

1885 i

= /2j e

1885 1.76 10-2 .391 -j e t j e

= 33.17 .18j e

0

tj e

= 33.17 [ cos(0.18) + j sin (cos(0.18) ]

tj e

= [ 32.63 + j 5.9 ] t j e

Rv + Lv + Cv = { [ 0.31 - j 1.73 ] +

[ 32.63 + j 5.9 ] +

[ -22.91 - j 4.2 ] }Volts t j e

= { 10.03 - j 0.03 } t j e

10 Volts t j e

which is equal to Av 10 Volts t je , thus verifying Kirchhoff law

In summary, 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

Experimental setup

Build the circuit shown in Fig. 9. 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 50

Suggestion: Try and one at the time.

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.

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B

G

C= 10 nF

L

Ro

A

Ground

t A

jeVoAv VA(t) = Real { Av }

= Real { t

jeVoA }

=VoA cos (t )

VoA is a constant real-value. It is the amplitude of your sinusoidal input voltage. Try VoA = 0.2 Volts

)(

-tj

eI oi I (t) = Real { i }

= Real { )(

-tjeIo }

= Io cos (t-)

Io = Io() is a real-value.

vA

where Io =Io () and = (). Fig. 9 RLC series circuit.

Suggestion: Try C~ 2 nF , 20 nF and 100 nF (one at the time) for this experiment.

Note: If a capacitor is, for example, labeled 473G, it means C = 47 103 picoFarads.

Pico means 10-12.

Measure the frequency of the input voltage using the oscilloscope. (Do not trust the reading from the signal generator knob.)

In Fig. 9 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.

R

tVtI

)()( R (18)

The voltage at A is the driving voltage VA = VA(t). It is the trace you see in your oscilloscope. Measurements

a) 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 i)

the impedance Zo minimum, ii) the phase between VA and VR equal to zero, and iii) the current I0 maximum. (See expressions 14 and 15 above.)

Select values of around o (this is the frequency region where a majority of your data need to be taken).

b) Plot oI as a function of .

(At the bottom of this file see the suggestion about how to overcome the unwanted variability of the input voltage amplitude as the frequency changes).

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c) Plot the experimental values of | z | = Zo as a function of .

Here Zo is the magnitude of the complex impedance z of the RLC circuit (see expression 14).

Zo, is determined experimentally from the ratio of amplitudes of the two signals VA(t), and I(t) (see expression (15):

| z |= Zo()= )(I

)(VA

t

t

signal the of Amplitude

signal the of Amplitude=

𝑉𝑜𝐴

𝐼𝑜(𝜔)

Compare your experimental results with the ones obtained using expression (14) above. Plot the

theoretical (curve) and experimental values (discrete points) of Zo as a function of (both in the same graph, for comparison).

d) 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 10 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 the experimental values of the phase vs frequency, = ().

Plot also the theoretical values for the phase predicted by expression (17).

vA

vR

180o

VA

time

Phasors vA and vR rotating with

angular velocity vR lags vA by .

vR

t

VA =VoA Cos (t); VR = VoR Cos (t - )

The traces show VR lagging VA by .

Real voltages Complex analysis measured by the oscilloscope Phasors

Fig. 10 Methodology to measure the phase using the oscilloscope.

e) During the course of measurements you take enough data to make a graph of both

impedance Zo and phase as a function of frequency . (as requested above).

How do the graphs change when using a higher or lower value of the resistance Ro? Repeat the experiments using at least two different values

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f) Measurement of the complex voltages across R, C and L. By swapping C with R, and L with R, it is possible to measure the voltage across C and across

L respectively.

f1) Measurement at resonance condition:

Measure the amplitude and phase (relative to vA) of the current i.

Express the experimental value of current i in phasor format.

Measure the amplitude and phase (relative to vA) of the individual voltages vc, vL, vR. Express their experimental values in phasor format. Verify if Kirchhoff law is fulfilled

Verify if i Cj

1

Cv .

Verify if i jL Lv

Hint: About the relative phase between the phases:

Compare vc with the input voltage; then compare i with the input voltage.

That way you figure out the relative phase difference between vc and i.

Compare vL with the input voltage; then compare i with the input voltage.

That way you figure out the relative phase difference between vL and i.

Hint: To verify, for example, the expression vc =(1/jwc) i you may proceed as follows:

Measure the amplitude of vc. Measure the phase between vc and the driving voltage) Measure the amplitude of i (which is Vr/R). Measure the phase between i and the driving voltage.

Verify that the magnitude of vc coincides with the magnitude of i /C.

Verify graphically that the vc phasor lags the current phasor by 90 degrees.

f2) Measurement at out of resonance condition

Choose a frequency * at which Io() is ~50% of the current obtained at resonance.

Measure the amplitude and phase (relative to vA) of the current i. Express the

experimental value of i in phasor format.

Measure the amplitude and phase (relative to vA) of the individual voltages vc, vL, vR. Express their experimental values in phasor format. Verify if Kirchhoff law is fulfilled.

Verify if i Cj

1

Cv .

Verify if i jL Lv

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_______________________________

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.

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19

Or, v1v

i)

1(

CLjR

Z

, which gives,

)1

( 22

v1

je

CLR

i

where

)

1 (

arctanR

CL

Notice,

at very low frequencies: +

at LC

o

1 we have: =

at very high frequencies -

Thus we have,

tjeoV

v (driving voltage)

jtjtj eee IIi )

(

(current)

dt

dqi implies

)

)

)

)

2/(2/(((

tjtjtjt eeee QII

jj

Iq

j

2/)()2/(

-jtjtj eee QQQq

(charge)

q lags the current i by /2

Using the phasors representation,

t

k

A

x

tjeA

z

At

x≠0 Real

zm

Im

20

Real

Im

v

q

Real

v

q

v

q

At ~ 0 At ∞ At ~ 0

At ~ 0

Real

Im

v i

At ∞

Real

v

i

At ~ 0

v

i