Experiment 4. RLC Resonance
1. Review
Points to recall when a sinusoidal emf source is applied to a circuit:
• The voltage across a resistor will be in phase with the current through it.
• The voltage across an ideal inductor will lead the current through it by 90o.
• The voltage across an ideal capacitor will lag the current through it by 90o.
Given a source of alternating emf v = V cos ω t applied to a resistor, a capacitor, and an
inductor all in series, how can we calculate the amplitude I and phase φ of the current in
the circuit?
•
•
The current at any instant is the same everywhere in the single-loop circuit.
The sum of all voltage drops around the complete circuit is always zero, i.e.
v − v R − v L − vC = 0
(1)
This equation may be solved for i using methods of differential equations, trigonometry,
complex numbers or phasors. We will use phasors here. We begin with just a resistor
and an inductor in series.
The RL Circuit
y
V
IωL
φ
ωt + π/2
ωt
Figure 1. RL Circuit
IR
Figure 2. Phasors for
the RL circuit.
As an example, consider the RL series circuit above (Fig. 1). The phasor for the current
is drawn at an angle of Tt to the x-axis and has magnitude I. The phasor for vR is at the
same angle, since we know it is in phase with the current, and has amplitude VR = IR.
The phasor for VL is at angle Tt + 90o , since it leads the current by 90o, and it has
amplitude ITL. From the rules of vector addition, the resultant voltage phasor has
amplitude V given by
1
I
x
( ) = IZ
V = I R 2 + ωL
2
(2)
tot
The angle φ is given by
φ = tan −1 (I ω L IR )= tan −1 (ω L R )
(3)
In summary we see that the amplitude of the applied voltage is IZ tot , given by Eq. 2, and
leads the current by a phase angle of φ, given by Eq. 3. Note the limiting cases of Eqs. 2
and 3 when L → 0 or R → 0 . Why do they make sense?
The LRC circuit and Resonance
y
IωL
V
φ
ωt + π/2
ωt
I
IR
x
I/ωC
Figure 3. Current and Voltage phasors for an LRC circuit
If a capacitor is added in series to the RL circuit above, we must add another phasor to
represent the voltage across the capacitor (see Fig. 3). Because the voltage across a
capacitor lags the current through it by 900, this phasor is antiparallel to the phasor for the
voltage across the inductor, which leads the current by 900. Adding the 3 voltage phasors
gives
(
V = I R 2 + ω L − 1 ωC
) = IZ
2
tot
(4)
for the amplitude of the sum of the 3 voltages, which is also the amplitude of the applied
voltage, by Kirchhoff’s voltage law. We also see from Fig. 3 that the voltage leads the
current by a phase angle φ given by
ω L − 1 ωC
φ = tan −1
(5)
R
(
)
2
Since the numerator can be positive or negative, the sum voltage may actually lead or lag
the current.
What happens to Z tot and φ when the numerator is zero, i.e. when ω L − 1 ωC = 0 ,
(
)
or ω = 1 L C ≡ ω 0 , called the resonant frequency. In this case
• The total impedance is at its minimum value, and thus the current is at its maximum
value at resonance.
• The current and applied voltage are in phase, in other words, the impedance behaves
like a pure resistance.
This phenomenon is known as resonance, and the circuit is described as a resonant or
tuned circuit. Such circuits are common in radios and TV’s, where the receiver must be
tuned to select a given frequency from a myriad of signals. In an older radio, the tuning
knob by which you select the radio station is actually varying the capacitance C in the
circuit, thereby changing the resonant frequency.
Mechanical oscillators, such as a pendulum, display resonance if they are driven
by an external periodic force analogous to the AC voltage above. If the frequency of the
external force is slowly varied towards resonance, i.e. towards the natural oscillation
frequency of the pendulum, the pendulum’s amplitude of oscillation will increase to a
maximum at resonance, just like the current in the LRC circuit.
2. Experiment
The graph of the current amplitude I vs. ω is often called a response curve for the circuit.
(See Fig. 31.19 pg 1079 in Young and Freedman ed 12.) The coil (inductor) that you
will use in constructing the circuit below has non-negligible resistance RL as well as the
desired inductance L. Any calculations involving the circuit must use R + RL as the
resistance. Note that the oscilloscope is monitoring the voltage across the resistor R as
the output. This voltage is proportional to the current in the circuit and is in phase with it.
Figure 4. The Resonant RLC circuit.
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Make up the circuit shown and set R = 600 Ω, C = 0.01 μF.
Explore the response curve of the circuit by making suitable measurements as a function
of frequency. Measure the phase difference between the output and input voltages (use
the two channels of the oscilloscope to do this). Plot log10 V R V vs. log10(frequency)
(
)
and the phase difference vs. log10(frequency). To get a good plot you will need to vary
the frequency in small steps around the resonant frequency.
Because resonant circuits are used to select signals of a certain frequency, the width of
the response curve is very important: a narrow resonance will be very selective. It can
be shown that this width is proportional to the resistance R (if R is not too big): a small
resistance means a very selective response. This turns out to be a very general principle
of physics: the less power loss in a resonant system, the narrower the response. For
example, a pendulum with very little friction will have a narrow response curve when it is
driven by a sinusoidal external force.
The bandwidth B of a tuned circuit is defined as the difference of 2 frequencies (which
straddle ω 0 ) for which the current is reduced to 1
2 of its maximum value. It can be
shown that B ≈ ω 0 Q where Q is a dimensionless quantity defined by Q = ω 0 L R .
Note that a large value of Q implies a small value of B, i.e. a narrow response curve.
Determine the resonant frequency and the bandwidth. Calculate Q from your data.
Compare these quantities with their theoretical values.
Note that whenever you measure the voltage with the scope, the “ground” lead should
always be connected to the “ground” of the signal generator. If this is not done, then
your scope produces a “short-circuit” which affects the response of your circuit.
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