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LRC Circuit and Resonance
LBS 267L
The purpose of this experiment is to study resonance in an series LRC (L=inductance, R=resistance, C=capacitance) circuit by
examining the response of the circuit as a function of the frequency of the applied voltage.
The amplitude of the AC current (Io) in a series LRC circuit is dependent on the amplitude of the applied voltage (Vo) and the
impedance (Z).
Io = Zo
Since the impedance depends on frequency, the current varies with frequency:
XL – XC 2 + R 2
where XL = inductive reactance = L, XC = capacitive reactance = 1  , R = resistance, and  = angular frequency = 2 ( = linear
frequency). The current will be maximum when the circuit is driven at its resonant frequency:
 =
One can show that, at resonance, XL = XC at resonance and thus the impedance (Z) is equal to R. So at resonance the impedance is the
lowest value possible and the current will be the largest possible.
LRC Write-up
Page 1
In this experiment the amplitude of the current vs. frequency is plotted. Since the current is a maximum at the resonant frequency and
is less for greater or lesser frequencies, the graph is expected to peak at the resonant frequency.
There is a phase angle  between the applied voltage and the current. It is given by
  tan 1
X L  X C 
 R
This angle can be determined as a function of frequency. It is zero right on the resonance since in that case XL=XC. Above resonance it
is positive since XL>XC, and below it is negative since XL<XC. Positive (negative) means that the voltage is ahead (behind) the current.
The phase angle can be measured by determining the time difference between the current and voltage crossing zero:
 measured 
t difference
or since T = 1/f, where f is the frequency, and T is the period.
 measured  2ft difference
This means that an oscilloscope trace can be used to find both the magnitude of the current and its phase relative to the input voltage.
LRC Write-up
Page 2
In the scope display above since the darker line is the voltage across the resistor, tdifference = ti - tv, and the phase is negative.
Macintosh Computer
Mac65 Interface
Science Workshop
Patch Cords (SE-9750)
LRC Write-up
Page 3
Power Amplifier (CI-6552)
(2) Voltage Sensors (CI-6503)
RLC Network (CI-6512)
Turn on the interface, and turn on the computer. Open the Science Workshop document titled “LRC Circuit” .
Check that the set-up corresponds to the circuit shown below, using the Power Amplifier as the voltage source.
Make sure that the DIN plugs of the Voltage Sensors are plugged into Analog Channels A and B of the interface box, and that
the Power Amplifier is plugged into input C.
The Voltage Sensor should be attached to Analog Channel A across the Function Generator output and the Voltage Sensor
attached to Analog Channel B should be across the resistor. The voltage measured at Analog Channel B will be used to calculate
the current, I, which is related to the voltage across the resistor by I = R .
Data Recording:
Set the Function Generator to a sine wave having a frequency of 10 Hz and an amplitude of 8 Volts. Click the "MONitor" button.
LRC Write-up
Page 4
As you move the frequency up, observe the current signal rising in amplitude and coming into phase with the applied voltage. At
resonance the amplitude is maximum, and the two waves are perfectly in phase.
Plotting the Current vs. Frequency:
A plot of current vs. frequency shows the resonance peak. In this case, it is not necessary to calculate the current: A plot of the voltage
across the resistor vs. frequency will show the same resonance peak. There is one additional complication: Instead of plotting just the
resistor voltage, it is necessary to plot the ratio of the resistor voltage to the source voltage just in case the source (function generator)
voltage varies because of feedback from the circuit.
To get the voltages across the resistor and the source, click on the "Smart Cursor"
button in the oscilloscope window.
Record the voltages, and then find the phase by measuring the time difference between the crossovers of the two waves. To turn
the “Smart Cursor” off, just click anywhere in the display.
Increase the frequency by 20 Hz. Record this frequency and repeat Step 1 to find the new voltages and phase.
Repeat Steps 1 and 2 until 210 Hz is reached, then increase the frequency in 40 Hz steps up to to 410 Hz. As the frequency is
increased, it will be necessary to adjust the sweep speed and the vertical scale to get a clear trace of the current on the screen. Be
sure to save a few scope displays to your floppy for the report.
Look at the data taken and determine approximately at what frequency resonance occurred by seeing where the current was a
maximum. Adjust the frequency to this value and make fine adjustments in the frequency until the trace of the current is in phase
with the voltage. The circuit is now being driven near the resonant frequency. Save the display for your report.
To check closer to see if the current is exactly in phase with the voltage, change the 'scope scales so that the horizontal input is
channel B and the vertical input is channel A only. When the two inputs are in phase, the screen will then show a straight line—
any phase difference will cause an oval trace. Be sure to transfer this scope display to your floppy for use in the report.
Use a voltmeter to measure the peak voltage across each of the components of the circuit individually when well off the
resonance. The sum of these peak voltages does not equal the applied peak voltage.
Graph the current (or VR/Vo) vs. the frequency (in Hz), with Kaleidagraph. Do the same for the phase vs. frequency.
LRC Write-up
Page 5
Using the resonant frequency in Hz found from the experiment, calculate the resonant angular frequency in radians/sec using
res = 2  res
Compare the measured resonant frequency to the theoretical value.
What would the experiment have shown if the 33 ohm resistor had been used instead of the 10 ohm one?
Is the plot of current vs. frequency symmetrical about the resonant frequency? Explain.
Why doesn’t the sum of the voltages add up to the applied voltage when off resonance?
At resonance, the reactances of the inductor and the capacitor cancel each other so that the impedance (Z) is equal to just the
resistance (R). Calculate the resistance of the circuit by using the amplitude of the current at resonance in the equation R = V
(where V is the amplitude of the applied voltage). Is this resistance equal to 10 ohms? Why not?
LRC Write-up
Page 6