EMT 1150 LAB | Professor Aparicio Carranza
Electrical Circuits Laboratory Reports 4 | 5 | 6
Report Written by Galib Rahman 10-31-2016
1 | P a g e
Measurements in Series
Objective
Procedure
Materials…………………………………………………………………………….……….Page 3
Data……………………………………………………………………………………..........Page 4
Conclusion…………………………………………………………………………..……...Page 6
Measurements in Parallel
Objective
Materials
Procedure
Data…………………………………………………………………………………………..Page 9
Calculations……………………………………………………………………………….Page 12
Conclusion ………………………………………………………………………………..Page 13
Measurements in Series-Parallel
Objective
Materials
Procedure
Data………………………………………………………………………………………..Page 15
Multisim………………………………………………………………………………….Page 18
Conclusion………………………………………………………………………………Page 19
Signed Documentation……………………………………………………………………Page 20
TABLE OF CONTENTS
2 | P a g e
Measurements in Series Circuit
EMT 1150 LAB | Section D385 | Professor Aparicio Carranza
Report Written By Galib F. Rahman
3 | P a g e
Objective The objective of this laboratory experiment is to analyze a series circuit in respect to its
components and characteristics. We will use a multi-meter to measure the voltage (V), resistance (R),
and current (I) of said series circuit and use the collected data to determine how a series circuit
functions.
Procedure First, we measured the resistance of one resistor alongside a switch on a breadboard – and two
resistors in series with a switch as well. Then we measured the voltage (using the multimeter) after
connecting a power supply to the series circuits we designed onto the breadboard according to the
provided schematics using approximately 10 Volts. Afterwards we set the multimeter to measure
current (DC milliamps) and recorded the current of each requested point. Finally, we measured the
resistance in a series circuit (by setting the multimeter to read ohms) without external voltages by
removing the power supply and recorded the data corresponding to the specified points.
Materials The materials utilized in this experiment are the following:
Number of Resistors Nominal Resistor Value Number of Materials Item 2 47Ω 1 Multimeter 1 330 Ω 1 Power Supply 1 470 Ω 1 220 Ω
4 | P a g e
Data
Measure Resistors in Series
Run No. 001
5 | P a g e
Run No. 002
Column 1 Column 2
Polarity Magnitude Units of Measure Polarity Magnitude Units of Measure
A to B 0
Volts
B to A 0
Volts
B to C .8 C to B -.8
C to D 3.68 D to C -3.68
D to E 5.55 E to D -5.55
E to F 0 F to E 0
F to A -10.03 A to F 10
Total 0 Total 0
Column 1 Column 2
Polarity Magnitude Units of Measurement Polarity Magnitude Units of Measurement
A to B 0
Volts
0 B to A
Volts
B to C .45 -.45 C to B
C to D 9.58 -9.58 D to C
D to E 0 0 E to D
E to F 0 0 F to E
F to A -10.03 +10.03 A to E
Total 0 Total 0
6 | P a g e
Run No. 003 | Current Measurement in a Series Circuit Current at Point C 10.1 mA | Current at Point F 0 mA | Current at Point F 0 mA
Run No. 004 | Current Measurement in a Series Circuit
Resistance Point Measured Resistance RAB Open Load RBC 47Ω RCD 219 Ω RDE 329 Ω RAE 593 Ω
Conclusion 1. For Circuit No. 1: Does the measurement for RAH, have different values when RFG open vs RFG closed.
Explain.
The measured resistance for RAH will change depending on the switch’s state (open or closed). This
is because an open circuit will theoretically have a resistance of infinity while a closed circuit
composing of only resistors would measure a resistance close to the sum of the nominal resistances
on said resistors.
2. For Circuit No. 2: Compare RBC, RCD & RAE
For RBC the measured resistance was of 217 Ω while RCD had a measured resistance of 328 Ω. RAE has
a measured resistance of 544 Ω.
3. For Circuit No. 3: Compare RAE to (RAB+RBC+ RCD+ RDE)
When comparing the sum of the resistors to RAE they were approximately equivalent in magnitude.
This proves that in a circuit containing resistors in series arrangement, the total resistance of the
circuit, given that the only components in the circuit are said resistors, would be approximately the
sum of the resistance of each individual resistor present.
4. For Run No. 002: Compare the sum of the voltages measured from going around the loop
clockwise to those measured counter-clockwise.
When measuring the voltages of the circuit clock wise the received readings were positive while
measuring the same circuit counter-clockwise the readings were of the same magnitude but of
different signs (negative). The greatest drop in readings that were clockwise were from
measurements C to D in comparison to D to E. For both readings, however, the total readings came
up to 0 Volts.
5. Compare the currents measured in a series circuit. What do you conclude about the current in a
series circuit?
In a series circuit, the current remains constant since the measured values were constant as well.
7 | P a g e
6. What is the relationship between REA and RTotal in question 18 and the resistances measured?
REA and RTotal were approximately equivalent to the sum of all the present resistance of the resistors
placed in the circuit.
7. What procedure can be used to find any open circuit in a series circuit using only the voltmeter?
One can measure the voltages across each component, and if the reading was 0 given that a voltage
source was present in the system, the component giving a reading of 0 volts may indicate an open
circuit.
8. Using the original circuit place a wire across C-D. This process is called a short. How does a short
affect the circuit? Explain using your measured data to support your answer.
By replacing a wire across C-D, the electrons in the circuit would flow through the wire instead of
flowing through the resistor. This occurs, due to the nature of electrons- which choose the path of
least resistance. Therefore, the resistor present in the local of C-D would not affect the electron
flow. Thus, the resistance of the system is decreased due to the introduction of a wire in the place of
C-D.
8 | P a g e
Measurements in Parallel Circuit
EMT 1150 LAB | Section D385 | Professor Aparicio Carranza
Report Written By Galib F. Rahman
9 | P a g e
Objective In this laboratory exercise, we analyzed the characteristics of a circuit containing components in
parallel formation. We observed how the components and overall system react when introduced to a
voltage source & its influence on the current of said system.
Materials The materials utilized for this experimental exercise are the following:
Number of Materials
Material Number of Resistors
Nominal Resistor Value Number of Resistors
Nominal Resistor Value
1 Wire Kit 1 47 Ω 1 220 Ω 1 Meter 1 470 Ω 1 330 Ω 1 Switch 1 4.7 KΩ
Procedure Wire multiple circuits corresponding to each provided schematic onto the breadboard utilizing
aforementioned materials.
Data
Measurement Resistance in Parallel
RBC 217 Ω
RCD 328 Ω
RAE 132 Ω
RAF 42 Ω
RBE 47 Ω
RCD 469 Ω
10 | P a g e
Measurement Resistance in Parallel (cont.)
Run 002
RAF 42 Ω
RBE 47 Ω
RCD 469 Ω
5. Measured Voltage Provided by Power Supply: 5.02 Volts
6. Voltages Across Resistors: VR1 = 5.02 Volts VR2 = 5.02 Volts
7. Compare all the voltages: It is proven through experiment that the voltages are
equivalent among components placed parallel to one another.
8. Measure Total Current: IT = 11.8 mA
9. Current In Each Branch: IB1 = 10.70 mA IB2 = 1.1 mA
10. Compare Total Current to the Sum of the Branch Currents:
As observed with the collected data, the total current is equivalent to the sum of the
currents of each branch in this circuit. (IT = IB1 + IB2)
11. Open the Switch & Measure the Resistance of each Branch and the Total Resistance
RB1 = 467 Ω RB2 = 4.66 kΩ RTotal = 424 Ω
11 | P a g e
Run 003 After adding a resistor R3 parallel to R2 in the same circuit shown in Run 002
Run 004 After adding a resistor R4 parallel to R3 in the same circuit in Run 003
Measured Voltage Provided by Power Supply: 5.02 Volts
Voltages Across Resistors: VR1 = 5.02 Volts VR2 = 5.02 Volts VR3 = 5.02 Volts
Compare all the voltages: It is proven through experiment that the voltages are
equivalent among components placed parallel to one another.
Measure Total Current: IT = 34.7 mA
Current In Each Branch: IB1 = 10.70 mA IB2 = 1.1 mA IB3 = 22.9 mA
Compare Total Current to the Sum of the Branch Currents:
As observed with the collected data, the total current is equivalent to the sum of the
currents of each branch in this circuit. (IT = IB1 + IB2 + IB3)
Compare the Total Resistance of Run 002 to Run 003
RTotal (Run 002) = 424 Ω RTotal (Run 003) = 145.29 Ω
Thus, by adding an additional resistor parallel to the other resistors in Run 002’s schematic, the overall or
total resistance was reduced. As a result, the total current flow increased. In this exercise we observed, that by
decreasing the total resistance of a circuit with a constant DC voltage source, the total current flow increases.
Measured Voltage Provided by Power Supply: 5.02 Volts
Voltages Across Resistors: VR1 = 5.02 Volts VR2 = 5.02 Volts VR3 = 5.02 Volts VR3 = 5.02 Volts
Compare all the voltages: It is proven through experiment that the voltages are equivalent among
components placed parallel to one another.
Measure Total Current: IT = 140.81 mA
Current In Each Branch: IB1 = 10.63 mA IB2 = 1.1 mA IB3 = 22.7 mA IB4 = 106.4 mA
In comparison to previous runs, less current flows through branches 1 & 2, because a new path is available that provides the
least resistance, resistor R4 – thereby enabling more current flow to the system, overall.
Compare the Total Resistance of Run 002 to Run 003 to Run 004
RTotal (Run 002) = 424 Ω RTotal (Run 003) = 145.29 Ω RTotal (Run 004) = 35.508 Ω
12 | P a g e
Calculations 1. For Run 002, calculate each current using the current divider rule. Compare these currents to
those you measured.
𝐼𝑥 = 𝐼𝑇 × (𝑅𝑇
𝑅𝑥)
2. For Run 003, calculate each current using the current divider rule and compare it to the
corresponding value you measured.
3. For Run 004, calculate each current using the current divider rule and compare it to the
corresponding value you measured.
𝐼𝐵1 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵1)
𝐼𝐵1 = 11.7𝑚𝐴 × (427.27 Ω
470 Ω)
𝐼𝐵1 = 11.7𝑚𝐴 × (. 91)
𝐼𝐵1 = 10.647 𝑚𝐴
𝐼𝐵2 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵2)
𝐼𝐵2 = 11.7𝑚𝐴 × (427.27 Ω
4700 Ω)
𝐼𝐵2 = 11.7𝑚𝐴 × (. 091)
𝐼𝐵2 = 1.0647 𝑚𝐴
𝐼𝐵1 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵1)
𝐼𝐵1 = 34.43𝑚𝐴 × (145.2 Ω
470 Ω)
𝐼𝐵1 = 34.43𝑚𝐴 × (. 309)
𝐼𝐵1 = 10.639 𝑚𝐴
𝐼𝐵2 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵2)
𝐼𝐵2 = 34.43𝑚𝐴 × (145.2 Ω
4700 Ω)
𝐼𝐵2 = 34.43𝑚𝐴 × (. 0309)
𝐼𝐵2 = 1.0639 𝑚𝐴
𝐼𝐵1 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵3)
𝐼𝐵3 = 34.43𝑚𝐴 × (145.2 Ω
220 Ω)
𝐼𝐵3 = 34.43𝑚𝐴 × (. 66)
𝐼𝐵3 = 22.724 𝑚𝐴
𝐼𝐵1 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵1
)
𝐼𝐵1 = 140.81𝑚𝐴 × (35.508 Ω
470 Ω)
𝐼𝐵1 = 140.81𝑚𝐴 × (. 755)
𝐼𝐵1 = 10.638 𝑚𝐴
𝐼𝐵2 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵2
)
𝐼𝐵2 = 140.81𝑚𝐴 × (35.508 Ω
4700 Ω)
𝐼𝐵2 = 140.81𝑚𝐴 × (. 00755)
𝐼𝐵2 = 1.0638 𝑚𝐴
𝐼𝐵3 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵3
)
𝐼𝐵3 = 34.43𝑚𝐴 × (35.508 Ω
220 Ω)
𝐼𝐵3 = 34.43𝑚𝐴 × (. 1614)
𝐼𝐵3 = 22.727 𝑚𝐴
𝐼𝐵4 = 𝐼𝑇 × (𝑅𝑇
𝑅𝐵4
)
𝐼𝐵4 = 34.43𝑚𝐴 × (145.2 Ω
47 Ω)
𝐼𝐵4 = 34.43𝑚𝐴 × (3.089)
𝐼𝐵4 = 106.354 𝑚𝐴
13 | P a g e
Conclusion 1. For Run 001: Does RBE + RCD = RAF? Why not? No, the total resistance (RAF) is not equivalent to the sum of the nominal resistance of the two resistors, due to
the connected formation. For, the two resistors are in parallel formation as opposed to series because the two
resistors share two common nodes. When two resistors are in parallel, the equivalent resistance would be the
inverse of the inverse sums of the resistors. However, if these two resistors were placed in series formation the
aforementioned equation would be true.
2. For Run 002: Compare RAF to the original value of 47 Ω. Does this agree? Explain.
Is RAF larger or smaller than 47 Ω?
The resistance value of RAF decreases due to the presence of additional resistance placed parallel to the
previous resistors. Thus, the resistance would be smaller than 47 Ω.
3. For Run 003: Compare RAF to RBE and to RCD. Why doesn’t this value equal to 470 Ω?
Although RBE & RCD have the same resistance of approximately 470 Ω, the total resistance, RAF measured to be
235 Ω. This occurs, because both resistors are parallel to one another thus the total resistance of the given circuit
would be less than the lowest nominal resistance.
What can you say about the resistors of the same value connected in parallel?
Based on our observations one can determine that when two resistors of identical resistance are placed parallel
to one another the total resistance is equivalent to the nominal resistance divide by the number of resistors.
4. For Run 002: What do you conclude is the relationship between the voltage of the source and
the voltage of each branch of a parallel circuit? It has become quite apparent, that the voltage of each branch given that the branches are parallel to the given
voltage source, is equivalent to one another. 5. What do you conclude is the relationship between the current supplied by the source and the
current of each branch of a parallel circuit?
The total current supplied by the source is equivalent to the sum of the currents flowing through each branch.
6. What will happen to the total resistance if more resistance is added in parallel? When adding additional resistors parallel to other resistive components the overall total resistance is reduced.
14 | P a g e
Measurements in Series-Parallel Circuits
EMT 1150 LAB | Section D385 | Professor Aparicio Carranza
Report Written By Galib F. Rahman
15 | P a g e
RAH 258 Ω
RBF 258 Ω
RBD 172 Ω
RCE 256 Ω
REF 43 Ω
RDG 224 Ω
RBG 258 Ω
RCF 258 Ω
Run 001
Objective In this laboratory exercise, we observed and determined the influence of the current and total
resistance when additional parallel branches were included into a give circuit.
Materials The materials utilized for this experimental exercise are the following:
Number of Materials
Material Number of Resistors
Nominal Resistor Value Number of Resistors
Nominal Resistor Value
1 Wire Kit 1 47 Ω 1 470 Ω 1 Meter 1 220 Ω 1 1 KΩ 1 Switch 1 320 Ω
Procedure Wire multiple circuits corresponding to each provided schematic onto the breadboard utilizing
aforementioned materials.
Data
Measurement Resistance in Series-Parallel
16 | P a g e
Run 002
VAB 4.08 Volts
VBC 0 Volts
VCD 5.97 Volts
VDF 0 Volts
VFA 10.06 Volts
Nominal Values Measured Values
R1 220 Ω 216 Ω
R2 1 KΩ 996 Ω
R3 330 Ω 330 Ω
RBE 248 Ω
RTotal 464 Ω
ETotal 10 Volts 10 Volts
ER1 4.7 Volts
ER2 5.4 Volts
ER3 5.4 Volts
ITotal 20.4 mA
IR1 20.4 mA
IR2 5.0 mA
IR3 15.5 mA
Measurement Resistance in Series-Parallel (cont.)
17 | P a g e
Nominal Values Measured Values
R1 220 Ω 216 Ω
R2 1 KΩ 996 Ω
R3 330 Ω 327 Ω
R4 47 Ω 48 Ω
RBE 40 Ω
RTotal 258 Ω
ETotal 10 Volts 10 Volts
ER1 8.5 Volts
ER2 1.55 Volts
ER3 1.55 Volts
ER4 1.55 Volts
ITotal 38.7 mA
IR1 38.7 mA
IR2 1.5 mA
IR3 4.5 mA
IR4 32.4 mA
In Run 003 we added a 47Ω resistor,R4, parallel to the resistor labeled R3 in the previous run.
As a result the total resistance of the circuit would decrease due to the added resistor’s formation.
18 | P a g e
Nominal Values Measured Values Calculated Values
R1 470 Ω 470 Ω
R2 47 Ω 47 Ω
R3 860 Ω 860 Ω
R4 220 Ω 220 Ω
R5 330 Ω 330 Ω
RDE 115.23 Ω 115.23 Ω
RCE 148.16 Ω 148.16 Ω
RBE 618. 162 Ω 618. 162 Ω
ITotal 16.17 mA 16.17 mA
IR1 16.17 mA 16.17 mA
IR2 13.39 mA 13.39 mA
IR3 2.7 mA 2.7 mA
I4 8.034 mA 8.034 mA
I5 5.34 mA 5.34 mA
ETotal 10 Volts 10 Volts 10 Volts
ER1 2.6 Volts 2.6 Volts
ER2 .6 Volts .6 Volts
ER3 2.3 Volts 2.3 Volts
E4 1.7 Volts 1.7 Volts
E5 1.7 Volts 1.7 Volts
Multisim Run 004
19 | P a g e
Conclusion Throughout these laboratory exercises, we have verified multiple principles in regards to circuits
structured with components in series, parallel, and series-parallel positions.
When components are in series they share the same current and the voltage is shared in ratio to
the resistance. The total resistance of a circuit containing strictly resistors in series formation is
equivalent to the sum of all resistances of each resistor. To calculate the voltage of each component one
can use the voltage divider rule
When components are in parallel to one another they share the same voltage and the current is
shared in ratio to the resistance. The total resistance of a circuit containing resistors placed parallel to
one another is equivalent of the inverse of the inverse sum of all resistances of each resistor. To
calculate the current flowing through each component one can use the current divider rule. In addition,
when adding additional resistors parallel to a given circuit the overall resistance is decreased.
Voltage Divider Rule
Current Divider Rule
Series
Parallel
20 | P a g e
Signed Documentation
21 | P a g e
Signed Documentation
22 | P a g e
Signed Documentation
23 | P a g e
Signed Documentation
24 | P a g e
Signed Documentation
25 | P a g e
Signed Documentation
26 | P a g e
Signed Documentation
27 | P a g e
Signed Documentation