EMT 1150 LAB | Professor Aparicio Carranza Electrical Circuits · 2016-11-14 · By replacing a wire across C-D, the electrons in the circuit would flow through the wire instead of

EMT 1150 LAB | Professor Aparicio Carranza

Electrical Circuits Laboratory Reports 4 | 5 | 6

Report Written by Galib Rahman 10-31-2016

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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

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Measurements in Series Circuit

EMT 1150 LAB | Section D385 | Professor Aparicio Carranza

Report Written By Galib F. Rahman

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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

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Data

Measure Resistors in Series

Run No. 001

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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

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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.

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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.

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Measurements in Parallel Circuit



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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

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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

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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

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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


𝑅𝐵2)

𝐼𝐵2 = 11.7𝑚𝐴 × (427.27 Ω

4700 Ω)

𝐼𝐵2 = 11.7𝑚𝐴 × (. 091)

𝐼𝐵2 = 1.0647 𝑚𝐴


𝑅𝐵1)

𝐼𝐵1 = 34.43𝑚𝐴 × (145.2 Ω

470 Ω)

𝐼𝐵1 = 34.43𝑚𝐴 × (. 309)

𝐼𝐵1 = 10.639 𝑚𝐴


𝑅𝐵2)

𝐼𝐵2 = 34.43𝑚𝐴 × (145.2 Ω

4700 Ω)

𝐼𝐵2 = 34.43𝑚𝐴 × (. 0309)

𝐼𝐵2 = 1.0639 𝑚𝐴


𝑅𝐵3)

𝐼𝐵3 = 34.43𝑚𝐴 × (145.2 Ω

220 Ω)

𝐼𝐵3 = 34.43𝑚𝐴 × (. 66)

𝐼𝐵3 = 22.724 𝑚𝐴


𝑅𝐵1

)

𝐼𝐵1 = 140.81𝑚𝐴 × (35.508 Ω

470 Ω)

𝐼𝐵1 = 140.81𝑚𝐴 × (. 755)

𝐼𝐵1 = 10.638 𝑚𝐴


𝑅𝐵2

)

𝐼𝐵2 = 140.81𝑚𝐴 × (35.508 Ω

4700 Ω)

𝐼𝐵2 = 140.81𝑚𝐴 × (. 00755)

𝐼𝐵2 = 1.0638 𝑚𝐴


𝑅𝐵3

)

𝐼𝐵3 = 34.43𝑚𝐴 × (35.508 Ω

220 Ω)

𝐼𝐵3 = 34.43𝑚𝐴 × (. 1614)

𝐼𝐵3 = 22.727 𝑚𝐴


𝑅𝐵4

)

𝐼𝐵4 = 34.43𝑚𝐴 × (145.2 Ω

47 Ω)

𝐼𝐵4 = 34.43𝑚𝐴 × (3.089)

𝐼𝐵4 = 106.354 𝑚𝐴

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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.

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Measurements in Series-Parallel Circuits



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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

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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.)

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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.

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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

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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

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Signed Documentation

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Documents

EMT 1150 LAB | Professor Aparicio Carranza Electrical Circuits · 2016-11-14 · By replacing a wire across C-D, the electrons in the circuit would flow through the wire instead of