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The LCR Circuit
Purpose
•
To investigate an LCR circuit.
Introduction and Theory
In the “RC Series Circuit” and the “RL Series Circuit” labs, the effects produced by an AC
voltage applied to an RC and to an RL series circuit were investigated. In this experiment the
behaviour of a series circuit containing an inductor, a capacitor, and a resistor (LCR) will be
studied.
Phasor Diagram
VL
V
C
V = Vmax sin ωt
I
L
φ
ωt
VR
R
VC
Figure 1a
Figure 1b
Figure 1. Phasor diagram for a LCR circuit
Consider the circuit in Figure 1a. A resistor, capacitor, and inductor are in series with a
sinusoidal voltage source of amplitude Vmax and angular frequencyω. For the LCR series
circuit, the current I through each element will be the same. The voltages across the resistor
VR, the capacitor VC, and the inductor VL will combine to balance the applied voltage V. (i.e.
They obey Kirchhoff’s loop rule.) These quantities are represented by phasors in Figure 1b.
In Figure 1b, VL has larger amplitude than VC. This results in the current lagging the applied
voltage by an angle φ. If VL happens to be smaller than VC, the current will lead the voltage by φ.
The phasor diagram contains all the information necessary to analyze the circuit, for either
case.
The Kirchhoff rule becomes
Vmax sin ωt = VR max sin(ωt − φ ) + VC max sin(ωt − φ − 90 o ) + VL max sin(ωt − φ + 90 o )
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Phase Angle
Let φ be the phase angle between the voltage and the current: φ = φV − φI . From the phasor
diagram:
tan φ =
VL max − VC max
VR max
Because VR max = RImax , VC max = X C Imax and VL max = X LImax ,
tan φ =
X L − XC
R
If the applied voltage is V = Vmax sin ωt , then the current in the circuit is I = Imax sin(ωt − φ ).
Note that above equations apply for both XL > XC and XL < XC. If XL > XC as drawn in Figure 1b,
φ is positive and the current I lags the voltage V. If XC > XL, φ will be negative, showing the
current I leads the voltage V.
Circuit Impedance
From the phasor diagram, we know
Vmax = (VL max − VC max )2 + VR max
2
Replace by VR max = RImax , VC max = X C Imax and VL max = X LImax , above equation becomes
Vmax = Imax ( X L − X C )2 + R 2 .
So the circuit impedance is Z =
Vmax
= ( X L − X C )2 + R 2 . That is, all three circuit elements
Imax
contribute to the impedance.
Resonance
The magnitude of the current in our LCR circuit can be expressed as:
Imax =
Vmax
Vmax
=
Z
( X L − X C )2 + R 2
(Note above equation is also true if max values are replaced by rms values.)
Above equation shows that the current takes a maximum value (Imax 0) when XL = XC. This
occurs at a specific frequency ω0(=2πf0). We call this frequency the natural frequency of the
circuit. Resonance occurs when the frequency of the voltage source is tuned to the natural
frequency of the circuit.
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The natural frequency
Let’s solve for the natural frequency f0:
X L = X C so ω0L =
1
. Therefore, we have ω0 =
ω0C
1
1
or f0 =
LC
2π LC
Imax
Imax 0
1
2
Imax 0
Δf
f0
f
Figure 2. The natural frequency and bandwidth
The bandwidth
The amplitude of the current shows a resonance peak at the natural frequency, as shown in
Figure 2. The relative width of the peak Δf f0 is called the “bandwidth”. Here Δf is defined to be
1
the full width of the frequencies when the current amplitude is above
of its peak value (see
2
Figure 2). A smaller bandwidth means a sharper peak.
The quality factor
The quality factor, or the Q factor of a resonance circuit is the inverse of its bandwidth:
Q=
f0
Δf
The quality factor is an important parameter of the circuit. A circuit exhibiting a narrow
bandwidth is also referred to as a "high Q" circuit. Because of its narrow bandwidth, in radio
tuning, a high Q circuit is harder to tune to its resonance, but has better sound quality. Many
other properties of the circuit besides bandwidth can be related to Q.
1 L
.
It can be shown that for a LCR resonance circuit, Q =
R C
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Apparatus
BK Precision 2120 dual-trace oscilloscope, BK Precision 4011 function generator, two Fluke
73III multimeters, decade resistance box, 0.22 μF capacitor, 25 mH inductor.
Internal resistance of the function generator:
Internal resistance of the Fluke multimeter through 400 mA ammeter:
Accuracy of the Fluke multimeter:
Accuracy of the LCR meter:
Accuracy of the decade box:
Actual value (and uncertainty) of the 0.22 µF capacitor:
Actual value (and uncertainty) of the 25 mH inductor:
Resistance of the 25 mH inductor:
Note: same as our last two labs, you must record the uncertainties for all measured values. You
do not need to calculate the uncertainties for the results, but you must present all results with
proper significant digits.
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Data
Part A: The size of the current and the phase angle
Set up the circuit in Figure 1a. Set the function generator to f =1 kHz and Vrms = 5.0 V. Set
R = 500 Ω, C = 0.22 μF and L =25 mH.
Measure Irms using the multimeter.
Measure the phase angle between the applied voltage and the current (using the voltage
across the resistor), using both the dual trace method and the Lissajous method.
Compare your measured values with the values given by theory (be careful with the total
resistance). Show the results below.
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Part B: Resonance
Remove the decade resistor box to make R as small as possible. Keep your L and C the same
as in Part A. Set the voltage to about 1.0 V rms.
Use a multimeter to measure Irms as you vary the frequency f over a range of values near
resonance. Record Irms and f. As you are measuring, using another multimeter to keep the
source voltage at a constant voltage of 1.0 V rms. Take enough data so that you can make a
graph like Figure 2, you may want different steps at difference frequencies.
In the space below, list the values of f and Irms in a data table. Create a graph like Figure 2
(either using Excel or graph paper). From the graph, find the values of f0 and Δf. Calculate Q
factor from f0 and Δf, and compare the result with the value calculated from R, L and C.
How may you change f0? Discuss 2 useful applications of being able to change f0.
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