Hanoi University of Science and Technology

School of Electrical and Electronic Engineering

EXPERIMENTAL REPORT
LINEAR CIRCUIT I
( Lab #1, #2, #3, #6, #8)

Instructor : MSc. Nguyen Van Thuc

Name : Thân Ngọc Sơn
Student ID : 20195813
Class code : 720554

Ha Noi, 2022
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 Lab #1: Kirchhoff’s Current and Voltage Laws

I. OBJECTIVES
- To learn and apply Kirchhoff’s Current Law.
- To learn and apply Kirchhoff’s Voltage Law.
- To obtain further practice in electrical measurements.
- Compare experimental results with those using hand calculations.
II. BACKGROUND AND THEORY
1. Kirchhoff’s Current Law (KCL)
Kirchhoff's Current Law (KCL) states that the algebraic sum of currents leaving any
node or the algebraic sum of currents entering any node is zero. KCL can be stated as the
sum of the currents entering a node must equal the sum of the currents leaving a node.
Kirchhoff's Current Law can also be expressed as follows:
- The algebraic sum of the currents entering a junction (node) equal zero.
- The algebraic sum of the currents leaving a junction (node) equal zero.
- The algebraic sum of currents entering a node equal to the algebraic sum of currents
leaving a node.
2. Kirchhoff’s Voltage Law (KVL)
Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltages around a
closed path is zero. As you make a summation of voltages, it is suggested that you
proceed around the closed path in a clockwise direction. If you encounter a positive (+)
sign as you first enter the circuit element, then add the value of that. On the contrary, if
you first encounter a negative sign as you enter the circuit element, then subtract the
value of that voltage.
We can state Kirchhoff's Voltage Law in three ways which all really say the same
thing:
- The algebraic sum of the voltage drops around any closed path of an electric circuit
equal zero.
- The algebraic sum of the voltage rises around any closed path of an electric circuit
equal zero.
- The algebraic sum of the voltage rises equal the algebraic sum of the voltage drops
around any closed path of an electric circuit.
III. EQUIPMENT AND PARTS LIST
- Digital Multimeter (DMM)
- DC power supply
- Circuit bread board
- Resistors
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IV. PROCEDURE
1. Without substituting in numbers for R1, R2, R3, and R4 apply Kirchhoff's Current
Law at nodes 1 and 2 so as to obtain two equations in terms of the two unknown node
voltages V₁ and V2. Simplify these equations.
2. Let , , , and in the equations of Step 1.
Solve these equations by hand for V₁ and V2.
From V1 and V2, find Va, Vb, Vc, Ia, Ib, and Ic.
3. Measure the resistors. Use these values to find V1, V2, Va, Vb, Vc, Ia, Ib, and Ic as
in Step 2.
4. Construct the circuit shown in Figure below.

5. Use the multimeter to measure the indicated three currents and five voltages.
6. Compare the results of Step 2 with those obtained from the theoretical calculations
of Steps 2 and 3 above.
7. Using the measured values of the three currents, check KCL at node 2.
8. Use your measured values of the source voltage, Va, Vb, and Vc to check
Kirchhoff's Voltage Law for the outer loop of the circuit.
V. CALCULATIONS AND COMPARISONS
1. Experiment results

Ia = 36.20 mA Va = 3.602 V
Ib = 27.02 mA Vb = 7.123 V
Ic = 9.069 mA Vc = 1.333 V

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 2. Calculation and comparisions

* Calculate and compare the current values

- Testing KCL
| Ia + Ib – Ic | = | 36.20 -27.02 – 9.069 | = 0.111 ~ 0 (satisfied)

- Compare chart

Experimental results (mA) Theoretical results (mA) Differences (%)

36.2 35.6 0.6
27.02 26.67 0.35
9.069 8.89 0.179

=> The difference between experimental and theoretical results is smaller than 2%,
which is acceptable value.

* Calculate and compare the voltage values

- Testing KVL
| 12 – Va – Vb – Vc | = | 12 – 3.602 – 7.123 – 1.333 | = 0.058 ~ 0 (satisfied)
- Compare chart

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 Experimental results (V) Theoretical results (V) Differences (%)
3.602 3.5557 0.0463
7.123 7.1114 0.0116
1.333 1.3331 0.0001

=> The difference between experimental and theoretical results is smaller than 2%,
which is acceptable value.

VI. CONCLUSIONS
The measured value is quite accurate and not different so much with the nominal
value.

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 Lab #2: Nodal Analysis
I. OBJECTIVES
- To learn and apply Nodal analysis.
- To obtain further practice in electrical measurements.
- Compare experimental results with those using hand calculations.

II. BACKGROUND AND THEORY

Analysis of electrical networks involves determination of node voltages and loop or
branch currents. Nodal analysis refers to the technique of writing equations where the
unknown quantities are the node voltages of the circuit. Kirchhoff's current law is used to
define the equations at each node in the circuit, using currents obtained by Ohm's law.

III. EQUIPMENT AND PARTS LIST

- Digital Multimeter (DMM)
- DC power supply
- Circuit bread board
- Resistors

IV. PROCEDURE
1. Experiment 1
- Consider the circuit shown in Figure below.

- Measure (with the multimeter) and record the resistance of each resistor you use.
Use the multimeter to adjust the power supply.
- Measure and record all node voltages and branch currents.
2. Experiment 2
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 - Consider the bridge circuit shown in Figure below. One useful property of the
bridge circuit is the so-called balance condition that occurs when the relationship among
the bridge resistors ("legs") is such that (R1/R2) = (R3/R4). Note that in the balance state
the node voltages V2 and V3 are equal, meaning that the current in R5 must be zero. Also
note that the balance condition depends only upon the resistor ratios, not the applied
voltage V.

- The bridge circuit can be used to determine the value of an unknown resistor if
several known resistors are available. For example, if the R3 leg was an unknown resistor
we could use various combinations of known resistors for the R1, R2, and R4 legs until
the balance condition was achieved. We would then know that R3 = R4*(R1/R2).
- Using the resistance and voltages given in the Figure, determine all node voltages
and branch currents in the circuit. Using the voltage and current values, calculate the
power dissipated by each resistor.
- Determine a new value for R4, leaving the other resistors unchanged, that will
balance the bridge circuit. Assume an adjustable resistor is used so that any resistance
value can be obtained.
- Construct the circuit shown in the Figure.
- Measure and record the resistance of each resistor you use.
- Use the multimeter to adjust the power supply.
- Measure and record all node voltages and branch currents

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V. EXPERIMENT RESULTS
* Calculated value:

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 1. Experiment 1
a. Nominal value

Vs 12V
R1 100 Ω
R2 200 Ω
R3 100 Ω

b. Measured value

R1 101.013 Ω Vs 12.04V I1 60.84 mA

R2 199.56 Ω V1 12.04V I2 60.88 mA
R3 99.62 Ω V2 6.032 V

2. Experiment 2
a. Nominal value

Vs 12V
R1 100 Ω
R2 50 Ω
R3 200 Ω
R4 100 Ω
R5 200 Ω

V1 = Vs = 12 V
The power dissipated by each resistor:

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b. Measured value

V1 12.15 V R1 101.12 Ω I1 80.07 mA

V2 3.993 V R2 50.55 Ω I2 80.88 mA
V3 4.026 V R3 199.24 Ω I3 39.97 mA
R4 99.87 Ω I4 40.13 mA
R5 201.03 Ω I5 0 mA

VI. CONCLUSIONS
- The experiment confirmed the nodal analysis approach to solve this type of circuit,
and also the bridge condition when a specific condition of bridge resistors is met.
- The measurement and the theoretical values of current are quite similar, the
differences are < 2%. Measured I5 is not 0 because R1/R2 is different from R3/R4. The
differences may be caused by the changes of resistances and the voltage source.

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 Lab #3: Loop Analysis

I. OBJECTIVES
- To learn and apply Loop analysis.
- To obtain further practice in electrical measurements.
- Compare experimental results with those using hand calculations.
II. BACKGROUND AND THEORY
Circuit equations involving Kirchhoffs current law (KCL) are used in nodal
analysis to determine the unknown node voltages. In loop analysis, on the other
hand, the unknown quantities arc loop currents, and the circuit equations arc
written using Kirchhoff's voltage law (KVL).
To begin the loop analysis procedure each loop of the network is assigned an
unknown current. The direction chosen for the current in the loop is arbitrary: the
calculated current will turn out to be negative if the actual current flows in the
direction opposite to the one chosen for the analysis. The loop equations arc
written by applying KVL to each loop: the sum of the voltages around each loop
must be zero. In any resistors shared by adjacent loops the net current in the
component consists of the sum or difference of the two loop currents, depending
on whether the two currents arc defined low in the same or different directions.
Once the set of current equations is solved the calculated loop currents can be
used to determine the voltage across each of the resistances using Ohm's law.
III. EQUIPMENT AND PARTS LIST
 Digital Multimeter (DMM)
 DC power supply
 Circuit bread board
 Resistors

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IV. PROCEDURE
Experiment 1:
1. Consider the circuit shown in Figure 1

2. Use loop analysis to determine the loop currents and branch currents for
the circuit of Figure I using hand calculator (pre-lab). Also calculate the
resistor voltage and the power dissipation for each resistor. Verify your
results using a symbolic mathematics computer program such as Matlab.
3. Measure and record all node voltages and branch currents.
4. Assemble the circuit of Figure 1 using the specified resistors.
5. Adjust the power supply for 12 volts. Measure the current in each branch
of the circuit, i.e., the current in R1 & R2, in R3, and in R4 & R5.
6. Now replace R5 with two 330Q resistors in parallel and re-measure the
current in each branch of the circuit. Next, replace R5 with three 33OÍ2
resistors in parallel and measure the branch currents again. Be sure to
measure and record the actual resistance of the nominal 330Q resistors
use.
7. Measure and record all node voltages and branch currents.

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Experiment 2:
1. Consider the circuit shown in Figure 2.

2. Determine all loop currents and branch currents for the circuit of Figure 2.
Note that the circuit is the same as Figure 1 except resistor Ró = 1 kQ has
been added to the circuit

3. Measure the current supplied by the voltage source? What is the power
supplied by the voltage source?

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V. EXPERIMENT RESULTS
Experiment 1:
* Calculated value

 Case 1: R1 = 25 Ω, R2 = 50 Ω, R3 = 50 Ω, R4 = 500 Ω, R5 = 200 Ω

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Nomial values:
I1 I2 V12
-92.05 mA 9.863 mA 6V
Measure value:
I1 I2 V12
-92.84 mA 10.103 mA 6V

 Case 2: R1 = 25 Ω, R2 = 50 Ω, R3 = 50 Ω, R4 = 500 Ω, 2xR5 = 200 Ω

Nomial values:
I1 I2 V12
-91.40 mA 11.420 mA 6V
Measure value:
I1 I2 V12
-92.34 mA 11.412 mA 6V

 Case 3: R1 = 25 Ω, R2 = 50 Ω, R3 = 50 Ω, R4 = 500 Ω, 3xR5 = 200 Ω

Nomial values:
I1 I2 V12
-91.17 mA 12.070 mA 6V
Measure value:
I1 I2 V12
-92.14 mA 12.065 mA 6V

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Experiment 2:
*Calculated value:

R1 = 25 Ω, R2 = 50 Ω, R3 = 50 Ω, R4 = 500 Ω, R5 = 200 Ω, R6 = 500 Ω

Nomial values:
I1 I2 I3
-92.2 mA 9.72 mA -4.98 mA
Measure values:
I1 I2 I3
-93.02 mA 9.88 mA -5.13 mA

VI. CONCLUSION
The measured value is quite accurate and not different so much with the nominal
value.
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 Lab #6: Superposition

I. Experiment 1

1. According to mesh current method, we get:

(100+200)I1-200I2 = 24
(200+150)I2-200I1= -12
500I3 = 12
 I1 = 0.0921 A

I2 = 0.0185 A
I3 = 0.024 A
Voltage across 200 Ω:
I200 = I1-I2 = 0.0736 A
V200 = 200.(I1-I2) = 14.72 V
2. Using super position:

- Turn off 24 V source:

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According to mesh current method, we get:
(100+200)I1-200I2 = 0
(200+150)I2-200I1= -12
500I3 = 12
 I1 = - 0.037 A I2 = - 0.055 A I3 = 0.024 A

Voltage across 200 Ω:

I200 = I1-I2 = 0.018 A
V200 = 200.(I1-I2) = 3.6 V
- Turn off 12V source:

According to mesh current method, we get:

(100+200)I1-200I2 = 24
(200+150)I2-200I1 = 0
500I3 = 0
 I1 = 0.129 A I2 = 0.074 A I3 = 0 A

Voltage across 200 Ω:

I200 = I1-I2 = 0.055 A
V200 = 200.(I1-I2) = 11 V
3. Data after measuring in lab

Turn off 24 V source, keep the Turn off 12 V source, keep the 24
12 V one V one
Voltage across Current across Voltage across Current across
200 Ω 200 Ω 200 Ω 200 Ω
3.809 V 0.0192 A 10.944 V 0.0554A

The results above is nearly equal to the one of theoretical solution

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II. Experiment 2

1. According to mesh current method, we get:

(100+50)I1-100I2 = 24
(100+500)I2-100I1= 12
 I1 = 0.195 A

I2 = 0.0525 A
 V50= 50 I1 = 9.75 V

V100 = 100.(I1-I2) = 14.25 V (I1-I2 = 0.1425)

V500= 500 I2 = 26.25 V
2. Suppress 12V source:

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 According to mesh current method, we get:
(100+50)I1-100I2 = 24
(100+500)I2-100I1= 0
 I1 = 0.18 A I2 = 0.03 A

 V50= 50 I1 = 9 V

V100 = 100.(I1-I2) = 15 V (I1-I2 = 0.15)

V500= 500 I2 = 15 V
3. Suppress 24V source

According to mesh current method, we get:

(100+50)I1-100I2 = 0
(100+500)I2-100I1= 12
 I1 = 0.015 A

I2 = 0.0225A
 V50= 50 I1 = 0.75 V

V100 = 100.(I1-I2) = -0.75 V (I1-I2 = -0.0075)

V500= 500 I2 = 11.25 V

4. Voltage sum of part 2 and 3:

V50= 9+ 0.75 = 9.75 V = V50 at part 1

V100 = 15-0.75 = 14.25 V = V100 at part 1
V500= 15+11.25 = 26.25 V = V500 at part 1

III. CONCLUSION

The measured value is quite accurate and not different so much with the nominal value.
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 Lab #8: Second Order Circuit

I. OBJECTIVES
- To study the step response of second order circuits.
- To understand the difference between overdamped, critically damped, and
underdamped responses.
- To determine theoretically and experimentally the damped natural
frequency in the underdamped case.
II. BACKGROUND AND THEORY
The characteristic equation is derived as

(1)
A convenient way to examine the characteristic equation is to compare a
given second-order characteristic equation with a standard form expressed as

(2)
 is called the damping factor:  is called the undamped natural
resonant frequency. If
 <1 the system response is underdamped

 =1 the system response is critically

 >1 the system response is overdamped

With (1) and (2), solving the equation we have:

III. EQUIPMENT AND PARTS LIST

- Digital Multimeter (DMM)
- DC power supply
- Circuit breadboard
- Resistors
- Pulse generator
- Oscilloscope

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 - Capacitor
- Inductor
IV. EXPERIMENT RESULTS
1. Circuit for experiment

Actual experimental data

Rtotal L C
Depend on the case 50 mH 20uF

2. Experiment results
a. Critically damped
Calculation:
- Apply (3) we have:

- In this case (Critically Damped), we have  = 1

- Apply (4):

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 Actual test result:

b. Overdamped
Calculation:
- Choose R = 500 Ω
- Apply (4) we have: => Satisfy the condition of
Overdamped.
Actual test results:

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c. Underdamped
Calculation:
- Choose R = 50 Ω
- Apply (4) we have: => Satisfy the condition of
Underdamped.
Actual test result:

V. CONCLUSIONS
This experiment shows the oscillation in the second order circuits as we
have mention in the background and theory. As we change the resistor, the
nature of the circuit is changing between overdamped, critically damped, and
underdamped response. Type of response of the circuit will result in the
different in the wave form of capacitor voltage.

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