Download Experiment 10 The RLC Series Circuit, I The resonant frequency of

Experiment 10
The RLC Series Circuit, I
The resonant frequency of an RLC series circuit is found by determining the frequency at
which the current delivered to the circuit is a maximum.
Theory
Suppose that a signal generator produces a sinusoidal voltage V= Vmsin27ft. When this
voltage is delivered to an RLC series circuit, the voltage amplitude can be writtenas
V m = I m Z,
(1)
where Im is the current amplitude and Z is the impedance of the circuit. The value of the
impedance is
Z  R 2  X L  X C  ,
2
(2)
where R is the resistance in the circuit, and the inductive reactance XL and capacitive reactance
Xc are given by
X L  2 fL,
(3)
1
.
fC 2
(4)
and
XC 
L is the inductance in the circuit and C is the capacitance.
When (2) is substituted into (1), and the result solved for the current amplitude, the expression becomes
Im 
Vm
R  X L  X C 
2
2
.
(5)
Because the inductive and capacitive reactances are frequency dependent, the impedance Z varies
with frequency. The maximum current flow then occurs when the impedance is a minimum, and
the impedance is a minimum when XL = Xc . This frequency at which occurs is called the
resonant frequency and has a value of
10 - 1
f0 
1
2 LC
.
(6)
As the frequency is varied above and below the resonant frequency, the impedance increases,
thus reducing the value of the current flow
Instead of directly determining the maximum value of the current amplitude, the ratio
of the voltage across the resistor to the voltage across the RLC combination is found. This ratio
is directly proportional to the cuffent amplitude and reaches a maximum when the current
amplitude is a maximum as is shown below:
VR  I m R 
Vm R
R  X L  X C 
2
2
,
(7)
or
VR

Vm
R
R 2  X L  X C 
2
.
(8)
Note that this equation indicates that the ratio VRIV. = I at the resonant frequency. This is only
true is the total resistance of the circuit was oigy due to the resistor. However, the signal
generator and the inductor both have resistance, and a more proper expression for the ratio is
VR

Vm
R
RT2   X L  X C 
2
,
where RT = R + (resistance of the inductor + resistance of the signal generator).
Apparatus
o
o
o
o
o
o
inductor, approximately 20 mH
capacitance box, set at approximately 0. 15 F
decade resistance box, set at approximately 500 92
signal generator
oscilloscope with 2 leads
3 cycle semi-log graph paper.
10 - 2
(9)
Procedure
1) Connect the circuit as
shown in Figure 1.
2) Turn
on
the
equipment and set the
signal generator to
100 Hz. Be sure the
switches
on
the
oscilloscope are in the
calibrate positions.
3) Set the vertical mode
switch
on
the
oscilloscope so that
Figure 1. The RLC series circuit.
both channel I and
channel 2 traces are
displayed on the screen. Adjust the amplitude knob on the signal generator until an 800
mv peak-to-peak signal is displayed on the screen for channel I ' This voltage is Vm.Read
and record the peak-to-peak signal for both channel I and channel 2. The voltage across
channel 2 is VR . (Remember you can increase the accuracy with which you read the
channel 2 voltage by changing the vertical sensitivity for channel 2.)
4) Change the frequency of the signal generator to 200 Hz. Repeat step (3). Continue to
repeat step (3) for frequencies of 400 and 700 Hz and multiple of Io above these
frequencies up to I.
5) When you are close in frequency to the resonant frequency of the circuit, make more
measurements so that the exact resonant frequency can be found.
Analysis
Graph the ratios V R / V m against the corresponding. frequencies on three cycle semi-log
graph paper with the frequency graphed on the log scale. Draw a smooth curve through the
points and determine the resonant frequency from the graph. Report the experimental value, the
theoretical value, and the percentage error between the two in a table of results.
Questions
1. Draw a phasor diagram for the RLC series circuit showing the phasors that represent the
voltage amplitude across the entire circuit, the individual circuit elements, and the current
amplitude for your circuit.
10 - 3