Experiment 3 Prelab
Charles Turner
Nick
9/8/08
Introduction:
This experiment introduces RC and RL circuits along with time constants in RC
and RL circuits. Voltage across capacitors and current through capacitors (the same for
inductors) will be examined.
Part A:
Purpose:
The purpose of this section is to introduce the measuring of time constants in RC
circuits and using the time constant to determine an unknown capacitor value.
Procedure:
The circuit in figure 1 will be assembled with a function generator range of 10V
peak-to-peak (+5V and -5V). A 1kohm resistor and a .1uF capacitor will be placed in the
circuit. The capacitor voltage and capacitor current will be measured with the
oscilloscope.
Square
Wave
Mode
+
FG
vs
R
iC
C
-
+
vC
-
Figure 1
To be able to measure current the voltage across the resistor must be known. The current
in the circuit is the voltage across the resistor divided by the value of the resistor. A
frequency will be chosen for the square wave such that it will remain at +5V long enough
to let the capacitor fully charge and discharge. The time constant of the capacitor will be
measured on the oscilloscope and compared to the expected value given by equation 1.
T=RC
(1)
This circuit will be used to measure the actual values of five capacitors over the range of
.001uF and 2.2uF. The known resistor value and measured time constant will be used to
determine the capacitors value. The measured capacitance value will be compared to the
value of the capacitors as shown by a capacitance meters. A resistor value of 500ohms
and a square wave frequency of 100Hz will be used.
Results:
The expected time constant from the original circuit with the 1kohm resistor and
.1uF capacitor is shown in equation 2.
T=1000*(.1*10^-6)=.0001s
(2)
In the circuit where the capacitance values are changing the time period was chosen to be
long enough such that even the 2.2uF capacitor could be fully charged. The time period
for the square wave is shown in equation 3.
T=(2.2*10^-6)*500=.0011s
Time period=1/100Hz=.01s
(3)
(4)
Equation 4 shows the time period given by a frequency of 100Hz. It is clear that it is
large enough that enough time constants can pass even for the largest value capacitor.
Conclusion:
The period for the square wave must be long enough for the capacitor to fully
charge and discharge to allow the scope to show the capacitor completely charging.
Part B:
Purpose:
The purpose of this section is to observe the voltage across a resistor in a RC
circuit. This circuit will demonstrate how the voltage across a resistor behaves in an RC
circuit.
Procedure:
The circuit in figure 2 will be constructed with a capacitor value of .1uF and
resistor of 1kohm. The function generator will have a frequency of 500Hz and a peak-topeak voltage of 5V (0V to +5V).
C
+ vC -
+
v S FG
-
+
R
vR
-
Figure 2
The voltage across the capacitor and resistor will be measured on the oscilloscope. The
measured time constant will be compared to the expected theoretical value.
Results:
The expected time constant for the circuit in figure 2 is given by equation 5.
T=1000*.1*10^-6
Conclusions:
(5)
The voltage across R will pulse because when the capacitor fully charges it will
behave as an open circuit, all of the voltage will be dropped across the capacitor. The
pulses of Vr will relate to the time constant because it takes four or five (depending on
when one considers the capacitor to be fully charged) to fully charge the capacitor.
Part C:
Purpose:
This section serves to introduce RL circuits, the RL time constant and to show
that the RL time constant is calculated differently than the RC time constant.
Procedure:
The RL circuit shown figure 3 will be constructed two inductors of about 1mH
and 27mH will be used.
L
+
+5
FG
-
O
vs
R
Figure 3
As shown the function generator will have a square wave with a peak-to-peak voltage of
5V (0V to +5V). Values for the period of the waveform and the resistor will be chosen
such that the period is long enough for either inductor to become fully charged. In
addition, a resistance value will be chosen to make the time constant occupy a fair
amount of horizontal space on the scope. The time constant for RL circuits is given by
equation 6.
T=L/R
(6)
One possible resistor is 500 ohms; this gives the time constant shown in equation 7 in the
Results section. A possible period for the square wave is 3kHz, which would allow the
inductors to fully charge which is shown in the Results section. The resistance of the
inductor will be measured and recorded as will the actual resistance of the resistor.
Results:
Equation 7 shows the expected time constant for the circuit in figure 3.
T=.027/500=.054ms
(7)
Equation 8 shows the amount of time it will take for the inductor to become fully
charged, or close enough.
T=5*.054*10^-3=.27ms
TimePeriod=1/3000Hz=.33ms
(8)
(9)
Equation 9 shows that a frequency 3kHz will allow enough time for the inductor to
become fully charged in on time period.
Conclusions:
An inductor behaves differently than a capacitor. The inductor resists
instantaneous changes in current not voltage. The inductor also behaves differently when
fully charged, it behaves as a short circuit when fully charged.
Part D:
Purpose:
The purpose of this section is to verify calculations made in this prelab.
Procedure:
The circuit in figure 4 will be constructed with a R value of 1kohm, an inductor
value of 1mH and a 2V peak-to-peak sinusoid.
L
+
FG
-
160 kHz
Sinusoid
R
Figure 4
The time constant will be measured in the lab by one of two ways. The actual resistance
of the circuit can be measured and divided into the inductance value. Alternatively, the
oscilloscope can be used to measure the time constant. The scope will be used to
measure the voltage across the inductor and the resistor. The current will be found by
measuring the voltage going into the circuit and knowing the resistor’s and inductor’s
actual resistance value.
Results:
The frequency in radians per second is shown in equation 10 and the equation for
current in the frequency domain is in equation 11. Equation 12 shows the current in the
time domain.
w=f*2pi=320000pi
I=Vp/(R+L*w*j)=.000498-.0005j
I(t)=.00071*sin(320000pi*t-45)
(10)
(11)
(12)
Once the current is found the voltage across the resistor and inductor can be found by
multiplying the current by the impedance as shown in equation 13 and 14.
Vl=(.000498-.0005j)*(.001*320000pi*j)=.503+.5j
Vr=(.000498-.0005j)*1000=.498-.5j
(13)
(14)
And in the time domain.
Vl(t)=.707sin(320000pi*t+45)
Vr(t)=.707sin(320000pi*t-45)
(15)
(16)
Conclusions:
The difference in phases of the voltage and current of an inductor was
demonstrated. The voltage across an inductor is 90 degrees ahead of the current through
an inductor. The voltage across a resistor and the current through a resistor are in phase
with each other.