PAMANTASAN NG LUNGSOD NG VALENZUELA
COLLEGE OF ENGINEERING AND INFORMATION TECHNOLOGY
ELECTRICAL ENGINEERING DEPARTMENT
ELECTRICAL CIRCUITS WITH LAB
Laboratory Report 2
Submitted by:
Sorela, Ram Leoj R.
20-0282
BSEE 2-3
Submission Date:
January 29,2022
I. EQUIPMENT/ MATERIALS USED
A. Superposition Theorem
1. Bread board / Circuit Board
2. 2 Variable DC Power Supplies
3. 2 Analog / Digital Multimeters
4. Carbon Resistors: R1 = 360 Ω, ¼ W; R2 = 510 Ω, ¼ W; and R3 = 1 kΩ, ¼ W
5. Connecting Wires / Wire Lead
B. Thevenin’s Theorem
1. Bread board / Circuit Board
2. 2 Variable DC Power Supplies
3. 2 Analog / Digital Multimeters
4. Carbon Resistors: R1 = 1.8 kΩ, ¼ W; R2 = 2.2 kΩ, ¼ W; R3 = 6.8 kΩ, ¼ W;
and R4 = 1 kΩ, ¼ W
5. Connecting Wires / Wire Lead
II.A. RESULTS AND DISCUSSION OF SUPERPOSITION THEOREM
Figure 1 – Two-Mesh Circuit with Supply Voltages Ve1 and Ve2

Set the circuit in such a way that allows connecting to Ve1 and measure the current
and voltage contributions of each resistor
Figure 2 ––Two-Mesh Circuit with Supply Voltage Ve1 alone

Set the circuit in such a way that allows connecting to Ve2 and measure the current
and voltage contributions of each resistor.
Figure 3 ––Two-Mesh Circuit with Supply Voltage Ve2 alone
TABLES OF DATA
Table 1 – Currents and Voltages with Supply Voltage Ve1 and Ve2
𝑉𝑒1(V)
10
𝑉𝑒2 (V)
10
𝑉𝑅1 (V)
8.58
Measured Values
𝑉𝑅2 (V)
𝑉𝑅3 (V)
11.42
1.42
𝐼𝑅1 (mA)
23.82
𝐼𝑅2 (mA)
22.4
𝐼𝑅3 (mA)
1.42
Table 2 – Currents and Voltages with Supply Voltage Ve1 alone
Measured Values
𝑉𝑒1(V)
𝑉𝑅1 ’ (V)
𝑉𝑅2 ’ (V)
𝑉𝑅3 ’ (V)
𝐼𝑅1 ’ (mA)
𝐼𝑅2 ’ (mA)
𝐼𝑅3 ’ (mA)
10
5.16
4.84
4.84
14.33
9.49
4.84
Table 3 – Currents and Voltages with Supply Voltage Ve2 alone
Measured Values
𝑉𝑒1(V)
𝑉𝑅1 ’’ (V)
𝑉𝑅2 ’’ (V)
𝑉𝑅3 ’’ (V)
𝐼𝑅1 ’’ (mA)
𝐼𝑅2 ’’ (mA)
𝐼𝑅3 ’’ (mA)
10
3.42
6.58
3.42
9.49
12.91
3.42
Table 4 – Currents and Voltage across R3
Calculated Values
𝑉𝑅3 (V)
𝐼𝑅1 (mA)
𝐼𝑅2 (mA)
𝐼𝑅3 (mA)
1.4231
23.8213
22.3981
1.4231
CALCULATIONS
𝐼1 = 𝐼1′ + 𝐼1′′
𝐼2 = 𝐼2′ + 𝐼2′′
𝐼3 = 𝐼3′ + 𝐼3′′
Currents and Voltages with Supply Voltage
Ve1 alone
Currents and Voltages with Supply Voltage
Ve2 alone
𝑅𝑒𝑞 = 1k Ω ∥ 510 Ω + 360 Ω = 697.7483 Ω
𝑅𝑒𝑞 = 760 Ω ∥ 1k Ω + 510 Ω = 774.7059 Ω
10 𝑉
10 𝑉
𝐼2 ′ = 774.7059
𝐼1 ′ = 697.7483 Ω = 14.33 mA
𝐼2′ = 14.33 𝑚𝐴 ∗ 1k
1𝑘 Ω
𝐼3′ = 14.33 𝑚𝐴 ∗ 1k
510 Ω
= 9.49 mA
Ω + 510
Ω + 510
= 4.84 mA
Ω
= 12.9081 mA
𝐼3′ = 12.9081 𝑚𝐴 ∗ 360
360 Ω
𝐼1′ = 12.9081 𝑚𝐴 ∗ 360
1k Ω
Ω + 1k Ω
Ω + 1k Ω
= 3.4169 mA
= 9.4914 mA
𝑉𝑅3 = 𝐼3 𝑅3
𝑉𝑅3 = 1.4231 𝑚𝐴 ∗ 1000 Ω
𝑉𝑅3 = 1.4231 V
𝐼1 = 14.33 + 9.4913 = 23.8213 𝑚𝐴
𝐼2 = 9.49 + 12.9081 = 22.3981 𝑚𝐴
𝐼3 = 4.84 − 3.4169 = 1.4231 𝑚𝐴
Here we can see that the calculated values are the same as the measured values
in the simulation. The values of the currents in the circuit were obtained by getting the
algebraic sum of the current produced by each source.
II.B. RESULTS AND DISCUSSION OF THEVIN’S THEOREM
Figure 4 – Current and Voltage Measurements of R3 in Single-Source Circuit
Figure 5 – Vth and Rth of the Single-Source Circuit
Figure 6. – Current and Voltage Measurements of R3 in Dual-Source Circuit
Figure 7. – Vth and Rth of the Double-Source Circuit
TABLES OF DATA
Table 5 – Thevenin Equivalent Circuit of Single-Voltage Source
𝑉𝑒1(V)
10
Measured Values
𝑉𝑡ℎ (V)
5.5
Calculated Values
𝑅𝑡ℎ (Ω)
𝑉𝑡ℎ (V)
5.5
990
𝑅𝑡ℎ (Ω)
990
Table 6 – Current and Voltage of R3 in Single-Voltage Source
Measured Values
Calculated Values
𝑉𝑅3 (V)
𝐼𝑅3 (mA)
𝑉𝑅3 (V)
𝐼𝑅3 (mA)
4.8
706.03 µ
4.810
706.0334
Table 7 – Thevenin Equivalent Circuit of Dual-Voltage Source
Measured Values
𝑉𝑒1(V)
10
𝑉𝑒2 (V)
-10
𝑉𝑡ℎ (V)
-2.21
Calculated Values
𝑅𝑡ℎ (Ω)
497.49
𝑉𝑡ℎ1 (V)
𝑉𝑡ℎ2 (V)
𝑉𝑡ℎ (V)
𝑅𝑡ℎ (Ω)
2.7638
-4.9749
-2.2111
497.4874
Table 8 – Current and Voltage of R3 in Dual-Voltage Source
Measured Values
Calculated Values
𝑉𝑅3 (V)
𝐼𝑅3 (mA)
𝑉𝑅3 (V)
𝐼𝑅3 (mA)
-2.06
-302.99 µ
4.8010
706.0334
CALCULATIONS
Voltage and Current of R3 in
Single-Source Circuit
Vth and Rth of Single-Source
Circuit
𝑉𝑡ℎ =
𝑉𝑒1 𝑅2
𝐼𝑅3 =
𝑅1 +𝑅2
=
𝑅𝑃1 =
𝑅𝑡ℎ +𝑅3
5.5
(10)(2200)
= 990+6800 = 706.0334 𝑚𝐴
= 1800+2200 = 5.5 𝑉
𝑅𝑡ℎ =
𝑉𝑡ℎ
1
𝑉𝑅3 = (6800)(706.0334 𝑚𝐴)
1/𝑅1 + 1/𝑅2
= 4.8010 𝑉
1
= 990 𝛺
1/1800 + 1/2200
(2200)(1000)
= 687.5 𝛺
(2200) + (1000)
Vth and Rth of Dual-Source Circuit
𝑅𝑡ℎ =
1
1
𝑅1
1
1
+𝑅 +𝑅
2
3
1
=
= 497.4874 𝛺
1
1
1
+
+
1800 2200 1000
𝑉𝑡ℎ1 =
𝑉𝑒1 𝑅𝑃1
𝑅1 +𝑅𝑃1
𝑅𝑃2 =
(1800)(2200)
= 990 𝛺
(1800) + (2200)
Voltage and Current of R3 in
Double-Source Circuit
𝐼𝑅3 =
𝑉𝑡ℎ
𝑅𝑡ℎ +𝑅3
−2.111
= 497.4874+6800 = −302.9947 𝑚𝐴
𝑉𝑅3 = (6800)(−302.9947 𝑚𝐴)
= −2.0604 𝑉
(10)(687.5)
= 1800+687.5 = 2.7638 𝑉
𝑉𝑡ℎ2 =
=
(−10)(990)
1000+990
𝑉𝑒2 𝑅𝑃2
𝑅4 +𝑅𝑃2
= −4.9749 𝑉
𝑉𝑡ℎ = 𝑉𝑡ℎ1 + 𝑉𝑡ℎ2 = −2.2111 𝑉
Here we can see that we obtained the
load current and the load voltage by
using the Thevenin’s theorem. The data
from the simulation and the calculated
values are identical.
III. QUESTION AND ANSWER
SUPERPOSITION THEOREM
1. What consideration should be taken in performing the theorem of the effect
superposition in the circuit containing two sources?
In performing superposition theorem, you should consider the direction
of the current of each resistor when you disconnect the other source and
compare it to the flow of the currents when all the sources are connected
in the circuit. After that, you should get the algebraic sum of every
current produced by each source.
2. What method or law is required to be used in the superposition method of circuit
analysis?
You can use the Ohm’s Law and the KCL in the superposition method
of circuit analysis.
3. When you apply superposition analysis to a circuit, what should be done to the
source/s?
When analyzing a circuit using the superposition theorem, you should
disconnect the other source/s to the circuit then get the values of the
current of each resistor.
4. Do the measurements of the voltage and current agree with the principle of
superposition? Explain.
Yes, because according to the superposition theorem, the current and
voltage for every element of a linear circuit with multiple voltage and
current sources is the algebraic sum of the currents and voltages
produced by each source functioning independently.
5. What are the limitations of the Superposition theorem?
You cannot get calculate the power of the circuit using the superposition
theorem because it is a non-linear quantity
THEVENIN'S THEOREM
1. With respect to the open terminals of the electrical network, what does Thevenin’s
theorem create?
The Thevenin’s theorem creates a single voltage in series with a single
resistance connected across the load from several voltages and
resistances in the original circuit.
If Thevenin’s theorem is applied in a network, what should have to be done in the
current sources if they exist?
If there is a current source in the circuit, you need to transform it into an
open circuit
What is the safest way of determining Thevenin’s resistance in a single-source
circuit?
You can easily get the Thevenin’s Resistance in a single-source circuit
by making it a parallel circuit and then getting its total resistance.
Is there a large difference in the load current with respect to the network load if
Thevenin’s theorem is used in circuit analysis compared to other network
theorems? Why or why not?
No, because it operates in a linear circuit therefore it should have little
to no difference in the load current with respect to the network load
What are the limitations of Thevenin’s theorem in circuit analysis?
Thevenin’s theorem cannot be used in a nonlinear circuit.
-
2.
3.
4.
5.
IV. CONCLUSION
Overall, I can say that the Superposition method and Thevenin’s theorem is very useful
in making an analysis of a circuit. You can apply the superposition theorem to easily
identify the voltages of an electrical network. On the other hand, by using Thevenin’s
theorem, you can also simplify electrical networks into a single source and single
resistance. The simulation results and data are identical, demonstrating that the
procedures are practically valid.