German – Jordanian University (GJU)
Electrical Circuits 1 Laboratory
Section 3
Experiment 2
Simple Resistive Circuits
Post lab Report
Mahmood Hisham Shubbak
17 / 10 / 2008
1
Objectives:
 To learn how to use the Unitr@in kit and TINA software.
 To apply Ohm's law on simple resistive circuits.
Introduction and Theory:
Simple resistive circuits are kind of electric circuits which can be simplified into a circuit of
one loop.
In this experiment we will make three simple circuits consists of voltage source and
resistors combined in two ways:
1. In series:
- The same current will go through the resistors.
- The voltage drop will divide between the resistors in a direct proportion to their
resistances.
- We can replace all the resistors with an equivalent resistor whose resistance is
given by:
eq
∑
i
2. In parallel:
- The current will divide between the resistors in an inverse proportion.
- The voltage drop across the resistors will be the same.
- We can replace all the resistors with an equivalent resistor whose resistance is
given by:
∑
eq
i
 In all our calculations we will use the Ohm's law: V = I.R
This experiment consists of three parts:
The first part:
Procedure:
1. Connect the circuit shown in the figure 1 twice: First on TINA1 simulation
software and then on the bread board.
1
Here I used Multisim simulation software instead of TINA.
2
R1
1
2
1.2kΩ
R2
1.2kΩ
V1
5V
R3
2.2kΩ
0
-
Figure 1 -
2. Measure the theoretical and experimental values of V and I for R1, R2 and R3.
Results:
The theoretical and experimental results of this part are shown in this table:
% Difference2
Theoretical Results
Experimental Results
Resistors
Voltage
V (V)
Current I
(mA)
Voltage
V (V)
Current I
(mA)
Voltage V
Current I
R1 = 1.2 kΩ
R2 = 1.2 kΩ
R1 = 2.2 kΩ
3.04
1.96
1.96
2.53
1.64
0.89
3.36
1.64
1.64
2.78
1.61
0.82
10.526%
16.327%
16.327%
9.881%
1.829%
7.865%
This figure shows the flow of current through the circuits and the instantaneous
voltage and current in some important points.
V: 5.00 V
I: 2.53 mA
1
V: 1.96 V
I: 2.53 mA
R1
1.2kΩ
V: 1.96 V
I: 893 uA
2
V: 1.96 V
I: 1.64 mA
R2
1.2kΩ
V1
5V
V: 0 V
I: 1.64 mA
R3
2.2kΩ
V: 0 V
I: 893 uA
0
V: 0 V
I: 2.53 mA
2
–
3
The second part:
Procedure:
1. Connect the circuit shown in the figure 2 also twice: First on TINA3 simulation
software and then on the bread board.
R1
1
2
1kΩ
V1
10 V
R2
2.2kΩ
0
R3
2.2kΩ
R5
470Ω
R4
1kΩ
4
-
Figure 2 -
2. Measure the theoretical and experimental values of V and I for R2, R3 and R4.
3. Simplify the circuit using an equivalent resistor.
Results:
The theoretical and experimental results of this part are shown in this table:
Theoretical Results
Experimental Results
% Difference
Resistors
Voltage V
(V)
Current I
(mA)
Voltage V
(V)
Current I
(mA)
Voltage V
Current I
R2 = 2.2 kΩ
R3 = 2.2 kΩ
R4 = 1 kΩ
5.4
4.715
0.685
2.46
2.14
0.685
5.1
4.42
0.71
2.63
2.32
0.761
5.556%
6.257%
3.650%
6.911%
8.411%
11.095%
This figure below shows the flow of current through the circuits and the
instantaneous voltage and current in some important points.
1
V: 10.0 V
I: 4.60 mA
R1
V: 5.40 V
I: 4.60 mA
1kΩ
2
V: 5.40 V
I: 2.46 mA
V1
10 V
V: 0 V
I: 4.60 mA
0
R2
2.2kΩ
V: 0 V
I: 3.14 mA
V: 0 V
I: 685 uA
V: 0 V
I: 1.46 mA
V: 685 mV
I: 1.46 mA
V: 0 V
I: 2.46 mA
R5
470Ω
4
V: 685 mV
I: 685 uA
V: 5.40 V
I: 2.14 mA
R3
2.2kΩ
R4
1kΩ
V: 685 mV
I: 2.14 mA
V: 685 mV
I: 1.46 mA
3
Here I used Multisim simulation software instead of TINA.
4
We can simplify this circuit into a circuit of one loop using an equivalent resistor as
shown in this figure below.
V: 10.0 V
I: 4.60 mA
R
1
2.174kΩ
V1
10 V
0
The third part:
Procedure:
1. Connect the circuit shown in the figure 3 on the bread board.
R1
1
2
1kΩ
R2
1kΩ
3
R3
0Ω
V1
0V
5,10, 15 V
?? Ω
0
-
Figure 3 -
2. Start with a DC voltage of 5V and increase it till 15V in 5 volt increments.
3. Measure V and I through R3 in each time.
4. Draw a table of results, then plot the voltage current relationship and find the
slope.
Results:
The results of this part are shown in this table:
Vsource (V)
5
10
15
V3 (V)
1.793
3.434
5.621
I3 (mA)
2.416
4.867
7.52
R3 (kΩ) (calculated)
0.742136
0.705568
0.747473
When we plot the voltage current relationship we find that it is a linear relationship
as shown below.
5
7
6
y = 0.7356x
Voltage (V)
5
4
3
2
1
0
0
2
4
6
8
10
Current (mA)
The slope of this line is equal to 0.735 and according to Ohm's law (V = I.R)4 it
represents the resistance of R3.
When we measure R3 it was equal to 0.739kΩ, this means that we have
%difference = 0.54%
Discussion
We can calculate the values of (V) and (I) for every resistor in part 1 and 2 using the
Kirchhoff's laws.
In this experiment we had % difference in the values of V, I and R, this might be because
one of these reasons:
1. The resistance of wires used to make the circuits.
2. The resistance of the used resistors was slightly less than its theoretical value.
3. Systematic errors.
Conclusion
We can calculate the values of (V) and (I) for any simple circuit using the Kirchhoff's
laws.
1. Kirchhoff's current law: the total current into the node = the total out from.
2. Kirchhoff's voltage law: the voltage drop through a closed loop = 0
4
(V is the y-axis, I is the x-axis and R is constant).
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