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ACSUPPLY
INDUCTOR
Li C Ci
Alternating Current RLC Circuits
1 Objectives
- To understand the voltage/current phase behavior of RLC circuits under applied al-
ternating current voltages,
- To understand the current amplitude behavior of RLC circuits under applied alternat-
ing current voltages, and
- To understand the phenomenon of resonance in RLC circuits.
2 Introduction
In previous labs
1 you studied the behavior of the RC and RL circuits under alternating
applied (or AC) voltages. Here, you will study the behavior of a similar circuit containing
series connected capacitor, inductor, and resistor. This is, quite reasonably, called an RLC
Circuit; see Figure 1.
3 Theory
Once again, let’s analyze this circuit using Kirchoff’s Rules. As always, you find
V
s (t) − V R (t) − V L (t) − V C (t) = 0 ,
1 Alternating Current RC Circuits and Alternating Current RL Circuits
Figure 1: The RLC circuit.
leading to a differential equation we have not encountered in these labs before
d
2 q(t) R dq(t) 1
dt
2 L dt LC
where q(t) is the charge on the capacitor. We will not actually solve this equation, as the
derivation is beyond the mathematics level of this course; however, in Appendix A we quote
some important results. For a sinusoidally varying source voltage
V
s (t) = V s cos ωt ,
we find the current is again out of phase, but this time, whether the current lags or leads the
applied voltage depends on whether the inductive or capacitive reactances (both defined as
before) dominate the behavior of the circuit at the driving voltage. Comparing the solutions
in the Appendix with our differential equation here, matching coefficients, we have
1 R
ω 0 = √ 2 β =.
LC
L
Putting all these definitions together, we can solve for the current and voltage profiles as a
function of frequency
V
s ω cos (ωt + δ)
I(t) = q
L (^2 )
(ω 0
2 − ω
2 ) + (2βω)
cos (ωt + δ)
V R (t) = I(t)R = V s 2 βω q
(ω 0
2 − ω
2 )
2
2
sin (ωt + δ)
VC (t) = q(t)/C = Vsω 0
2 q
(ω 0
2 − ω
2 )
2
2
d
2 q(t) sin (ωt + δ)
V L (t) = −L = V s ω
2 q
dt
2
(ω
2 (^2 )
− ω
2 )
where the phase is given by
ω
2
0 − ω
2
tan δ =.
2 βω
One useful tool for the study of two equifrequency signals is the XY mode of the oscillo-
scope. In the standard mode of the oscilloscope, you can think of the display as a standard
plot of the signal, with the independent variable, t, on the horizontal axis, and the depen-
dent variable, V (t), on the vertical axis. The XY mode can be thought of as a parametric
plot, where the independent variable, t, is implicit (not displayed), while the horizontal and
vertical axes trace out two different dependent variables, V 1 (t) and V 2 (t). The XY mode is
most useful when the two signals have commensurate frequencies (their ratio is rational),
·.·.:..:_-...
············ :.:.::·- ·-_ - ·- ·· ·- ·····.··· .......... ,., ,.....
,,,...-................
/ ................
.,~~,, ,- ··········
.,;.:•·••:;... ·:.:.::::
-0.
0
1
0 0.5 1 1.5 2 2.5 3
Phase [
π
/2]
ω [ω 0 ]
Phase angle with frequency
2 β/ω 0 = 1/
= 1/
= 1
= 2
= 3
0
1
0 0.5 1 1.5 2 2.5 3
V
R
/V
s
ω [ω 0 ]
Resistor voltage amplitude schematic
2 β/ω 0 = 1/
= 1/
= 1
= 2
= 3
(a) Phases (b) Resistor voltage amplitude
0
1
2
3
4
0 0.5 1 1.5 2 2.5 3
V
C
/V
s
ω [ω 0
]
Capacitor voltage amplitude schematic
2 β/ω 0 = 1/
= 1/
= 1
= 2
= 3
0
1
2
3
4
0 0.5 1 1.5 2 2.5 3
V
L
/V
s
ω [ω 0
]
Inductor voltage amplitude schematic
2 β/ω 0 = 1/
= 1/
= 1
= 2
= 3
(c) Capacitor voltage amplitude (d) Inductor voltage amplitude
Figure 3: The phase angle as a function of angular frequency is on the upper left. The
resistor voltage/current amplitudes are displayed on the upper right. The lower left shows
the capacitor voltage amplitude, while the lower right shows the inductor voltage amplitude. √
In all cases, the frequency is normalized in units of ω 0 = 1/ LC. Because the phase and
amplitudes are also a function of 2 β = R/L, we plot families of curves for various values of
2 β/ω 0
. The phase is normalized to π/2, while the amplitudes are normalized to V s
There are a number of other resonances for this circuit. We can, for instance, look for
the maxima of the voltage amplitudes, the so called amplitude resonances of the circuit; see
Figure 3. To predict these, we extremize the amplitudes versus the frequencies (
dA
dω
(ω)
0). Clearly the current and the voltage across the resistor will be maximized at the same
frequency:
dV R (ω)
= 0 −→ ω R = ω 0
dω ω R
or the current amplitude resonance occurs at the same frequency as the natural oscillation
frequency of the circuit. Interestingly, the amplitude resonance for the capacitor and inductor
voltages are not the same as for the current! For the capacitor,
q dV C (ω)
= 0 −→ ω C = ω 0
2 − 1 / 2 (2β)
2 ,
dω ωC
the resonant voltage amplitude across the capacitor occurs at a lower frequency than the
phase resonance! For the inductor,
s
ω
4 dV L (ω) 0 = 0 −→ ω L
dω ωL
ω
2 − 1 / 2 (2β)
2 0
the resonant voltage amplitude occurs at a frequency higher than the phase resonance. Of
course, these last two resonance conditions will only occur if the radical is real.
4 Procedures
You should receive two multimeters (one of which should be a BK-5460), an oscilloscope, a
function generator, a decade resistance box, a decade capacitance box, and an inductor.
- First, select component values for testing. Select a frequency between 300 Hz and
600 Hz, and a value for C between 0. 06 μF and 0. 1 μF. Record the value of the in-
ductance, L. Measure and record the values of the inductor resistance R
0 , and C.
Calculate X = |X L
− X
C | and choose a value for R + R
0 ≈ 1. 2 X. Set, measure and
record R.
- Configure the circuit for testing shown in Figure 1. Insert the Simpson multimeter to
record the AC current.
- Using the BK Precision meter, record the frequency f , and the RMS AC voltages
across the signal generator V s , the resistor V R , the capacitor V C , and the inductor V L
Are these values consistent?
- Measure the phase shift between the current and applied voltage for your chosen fre-
quency. Connect the oscilloscope so as to measure the voltage across the resistor and
signal generator; make sure the negtive inputs share a common reference point. Make
- The “complementary” or “homogeneous” solution q c (t), when A = 0. This is the
transient solution, and decays away exponentially; we’ll not dwell on this any further.
- The “particular” solution qp(t), where we don’t drop the driving term. We previously
called this one the “steady state” solution. Typically, we take “trial solutions” of the
same form as the driving term, and see if we can come work up a function that satisfies
the equation.
Let’s try a particular solution that’s a sinusoid. Before diving in, let’s think about the form
of the physical solution that we’re looking for here. In our specific problem, the solution to
the differential equation is the charge on the capacitor, but the physics we’re interested in
is the phase difference between the current and the drive voltage. Thus, we want a current
that is of the same functional form as the drive voltage, plus a phase offset. Since we choose
a drive voltage that is a cosine, and current is the derivative of the charge, we want to choose
a steady state (particular) solution of the form
q p (t) = D sin (ωt + δ).
To show this is a solution, we need to find D and δ, which we do by substituting into the
differential equation, to obtain
−Dω
2 sin (ωt + δ) + 2βDω cos (ωt + δ) + Dω 0
2 sin (ωt + δ) = A cos ωt.
Next, we expand the sin and cos terms, since:
cos(a + b) = cos ωt cos δ − sin ωt sin δ sin(a + b) = sin ωt cos δ + cos ωt sin δ ,
which gives six terms on the left, half of which contain a sin ωt and the other have contain
cos ωt. The right hand side contains only cos ωt. If this is a solution, the coefficients of the
sin ωt and cos ωt terms must separately be equal, and must be collectively consistent. Thus,
we have two equations for the two unknowns D and δ
sin ωt : −Dω
2 cos δ − 2 βDω sin δ + Dω 0
2 cos δ = 0
cos ωt : −Dω
2 sin δ + 2βDω cos δ + Dω 0
2 sin δ = A.
The sin ωt equation can be solved for δ
ω
2 − ω
2
0 tan δ =.
2 βω
Now that we have δ, we can use the second equation to solve for D
A
D =.
(ω 0
2 − ω
2 ) sin δ + 2βω cos δ
How do we substitute for δ in this latter equation? Using the trigonometric relations
sin δ
tan δ = sin
2 δ + cos
2 δ = 1 ,
cos δ
you can show that
ω
2 − ω
2 2 βω
sin δ = q^
0 cos δ = q^.
(^2 2 2 ) (ω 0
2 − ω
2 ) + (2βω) (ω 0
2 − ω
2 ) + (2βω)
Substituting these values, you will obtain the coefficient D
A
D = q.
(ω 0
2 − ω
2 )
2
2
Post-Lab Exercises
- From your recorded inductance, and the measured resistance, capacitance, inductance,
and initial frequency, determine the impedance of your circuit. Make sure to estimate
your uncertainties. Determine the impedance experimentally via another method,
taking care of the uncertaintites. Do you get the same results?
- Estimate the uncertainties on the measured values of V s
, V
L
, V
C , and V R
. Are these
values consistent with each other? Explain what you mean by “consistent”.
- From your measurements in Step 4 of the procedure, determine the phase shift at each
of the three measured frequencies, including an estimate of the uncertainty. How do
these compare to the theoretical predictions?
- Are your three measurements of the phase resonance frequency in Step 5 consistent
with each other? With theoretical prediction?
- Describe qualitatively what happens to your signals when you vary the frequency
around the phase resonance.
- Is your data from Step 6 consistent with the predictions of theory? Specifically, do the
voltage and current amplitudes measured by oscilloscope and by multimeter match,
within uncertainties, and do they comport with theoretical expectations?
- Why did you have to change the resistance value for Step 7? Did you find the amplitude
resonances? If so, do their values agree with your theoretical predictions?
- Discuss briefly whether you have met the objectives of the lab exercises.
