Physics 2
Lab 3 – Capacitors & Resistors
Opanin Kwadwo Akuffo (28302021)
Timothy Charles-Debrah (18622021)
Ashesi University
Lecturers: Dr. Nathan Amanquah & Kofi Adu-Larbi
FI: Edward Vedomey
28/09/2018
Instruction –II
(initial observations)
a)
Connect two resistors in series, place a scope across one of the resistors and ground
(0V). You can also monitor the input by connecting channel 1 scope at the input. See the
figure below.
b)
c)
Use a both a known DC and an AC source to provide the voltage across the resistors.
(For ac, try sine, triangular and square wave, one after the other)
Observe the DC/AC outputs on the scope and export the results to plot in MATLAB.
DC source = 2V
AC source = 2V
d)
For the AC sources, calculate the period and amplitude from the scope. Use the in-built
measure tool to find same values. How do the two compare?
Calculations from the scope:
Sine
Amplitude 1 V
Period
1.000000045 ms
Triangular
Amplitude
Period
1V
999.985298048 μs
Square
Amplitude
Period
1V
1.000002189 ms
Values from the in-built measuring tool:
Sine
Amplitude 993.577577542 mV
Period
1.000006039 ms
Triangular
Amplitude
Period
956.25775275 mV
1.000003561 ms
Square
Amplitude
Period
993.744931017 mV
999.994024242 μs
From calculations from the scope, the Amplitude and Period of all the AC sources are at
the same value of 1V and 1ms respectively. This value is the same when compared to
the approximate value of the inbuilt measuring tool, however the highest amplitude
measured on the inbuilt measuring tool is about 0.01V less than the calculated 1V, this
drop-in voltage is probably due to the resistance in the wire leading to the resistor.
Furthermore, for the triangular wave, the calculated amplitude gave a value of 1V
however the inbuilt measuring tool repeatedly kept increasing from 0 to 1V and then
back down to 0V every half cycle.
e)
Determine the current that should be passing through the resistor.
R2 color code: Brown, Black, Red, Gold
R2 = 10 x 100 = 1000 Ohms
From Ohm’s Law,
I = V/R
= 1/1000
= 1 mA
f)
Observe the input and output patterns of a sinusoidal waveform if resistor R1 is
replaced with a capacitor and connected as in the figure below:
When resistor R1 was replaced with capacitor C1, the voltage output across resistor R2
greatly reduced (1V to 12.55mV) but the period was approximately the equal.
PART II OBJECTIVE: TO VERIFY CAPACITOR CIRCUITS
Introduction:
Capacitor values are printed out as three numbers eg abc. The value is always in picofarads. abc
represents the capacitor ab x 10c. eg 472 is 47 x 102 pico farads = 4700pico which is 4.7nano
farads.
Instructions:
1. Choose capacitors labeled 102, 470 and 102 as C1, C2 and C3
(alternatively choose 103, 473 and 103 if the earlier set is unavailable.)
You will also be given one mystery capacitor for your experiments.
2. Connect two capacitors in series as below. Apply a 5V sinusoidal voltage. You may use a
digital multi-meter or the oscilloscope for these measurements
a. Write down the values of C1, and C2.
C1 = 10 x 102 pF = 1 x 10-9 F
C2 = 470 pF = 470 x 10-12 F
b. Measure V across C2.
Amplitude
3.475013275 V
c. Compute the value of voltage you expect across C1. (with the information on voltage
applied across the pair)
VT = 5V
CT =
470×10−12 × 1×10−9
470×10−12 + 1×10−9
= 319.7279 x 10-12 F
Q = CT x VT
= 5 x 319.7279 x 10-12
= 1.6 x 10-9 C
V1 =
1.6×10−9
1×10−9
= 1.6V
d. Measure the voltage across C1.
Amplitude
1.570191184 V
e. Compute the percentage difference as:
%difference = | measured – actual | / measured x 100
%difference = |1.570191184 – 1.6|/1.570191184 x 100
= 1.9 %
Connect three capacitors as shown in the next figure
a. Write down the values of C1, C2 and C3
C1 = 10 x 102 pF = 1 x 10-9 F
C2 = 470 pF = 470 x 10-12 F
C3 = 10 x 102 pF = 1 x 10-9 F
b. Compute the equivalent capacitance
CT =
(470×10−12 + 1×10−9 )×(1×10−9 )
(470×10−12 + 1×10−9 )+(1×10−9 )
= 595.14 x 10-12 F
c. Compute the value of voltage you expect across C2. (with the information on voltage
applied by W1)
Q = CT x V
= 595.14 x 10-12 x 5
= 2.9757 x 10-9 C
CT1 = 470 × 10−12 + 1 × 10−9
= 1.47 × 10−9 F
V2 = 2.9757 x 10-9 / 1.47 × 10−9
= 2.0243V
d. Measure the voltage across C2.
Amplitude
2.022494499 V
e. Compute the percentage difference as:
%difference = | measured – actual | / measured x 100
%difference = |2.022494499 – 2.0243|/2.022494499 x 100
= 0.0892 %
3. Connect all three capacitors in series.
a. Write down the values of C1, C2 and C3
C1 = 10 x 102 pF = 1 x 10-9 F
C2 = 470 pF = 470 x 10-12 F
C3 = 10 x 102 pF = 1 x 10-9 F
b. Compute the equivalent capacitance
1
𝐶𝑇
=
1
1×10−9
CT = 242.268 x
+
1
470×10−12
10-12
+
1
1×10−9
F
c. Compute the voltage value you expect across C2 and C3. (with the information on
voltage applied by W1)
Q = CT x V
= 242.268 x 10-12 x 5
= 1.21134 x 10-9 C
1
𝐶𝑇2
=
1
1×10−9
+
1
470×10−12
CT2 = 319.72789 x 10-12 F
V2 = 1.21134 x 10-9 / 319.72789 x 10-12
= 3.79 V
d. Measure V across C2 and C3
Amplitude
3.747130717 V
e. Compute the percentage difference as:
%difference = | measured – actual | / measured x 100
%difference = |3.747130717 - 3.79| / 3.747130717 x 100
= 1.14 %
4. Determine the value of the unknown capacitor. Hint: Set up the following circuit to find
an unknown capacitance.
From measurement,
V2 = 4.463256712 V
Therefore V1 = 5 - 4.463256712
= 0.536743288 V
Q = C1 x V1
= 1 x 10-9 x 0.536743288
= 536.743288x10-12 C
Therefore C? = 536.743288-12 / 4.463256712
= 120 pF