EMT 1150 Lab Report #6 - Measurements in Series- Parallel Circuits PDF

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Measurements in Series- Parallel Circuits...

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EMT 1150 D-370 New York College of Technology

Professor Chen Xu

Experiment #6: Measurement in Series - Parallel Circuits

Authors: Chadel Dowridge & Oludara Abiona April 12st, 2018

Table of Contents Objective………………………………………………………….page 3 Materials used……………………………………………………...page 4 Procedure and Results……………………………………………...page 5-17 Calculations …………………………………………………………page 18-19 Conclusion……………………………….page 21-23 Appendix…………………………………………………………..page 23-30

Objective The objective of this experiment is to determine the effect Current (I) and Resistance (R) has on circuits when you add additional branches in different ways such as series and parallel.

Materials Used 1) 2) 3) 4) 5) 6) 7) 8) 9)

Breadboard 1- Wire Kit 1-Meter 1-Switch 220Ω (Resistior) 330Ω (Resistior) 47Ω (Resisitor) 1.0kΩ (Resistior) 470Ω (Resistior)

Procedure Using the materials mentioned above, we designed many different parallel circuits in many different ways to first measure the Resistance of the circuits and then the Voltages then after that the currents of each of the parallel circuits, we also added other parallel branches to see if it would change the current in the circuit.

Results Run #1

Rah: .208kΩ Rbf: .208kΩ Rbd: .164kΩ Rce: .215kΩ Ref: .045kΩ Rdg: .206kΩ Rbg: .208kΩ Rcf: .208kΩ

Run #2

3) Total Voltage= 10.01V R1=4.69V, R2=5.32V, R3=5.32V 4) AB=4.69V BC=0V CD=5.32V DF=0V FA=-10.01V 5) AB=4.69V BE=5.32V EF=0V FA=-10.01V 6) Total Current= 21.5mA I1= 21.5mA I2= 5.4mA I3= 16.2mA 7) IT= 21.5mA and I2= 05.4mA + I3 = 16.2mA= 21.5mA

8) R1= V1/I1= 4.69V/21.5mA= 0.218kΩ R2= V2/I2= 5.32V/5.4mA= 0.985kΩ R3= V3/I3= 5.32V/16.2mA= 0.328kΩ Rbe=V2/It= 5.32V/21.5mA= 0.247kΩ Rt= 0.218kΩ+0.985kΩ+0.328kΩ+0.247kΩ=1.778kΩ 9) Nominal (color code) Values

Measured Values

Calculated values (using Nominal values

R1

.220kΩ

.217kΩ

0.218kΩ

R2

1.0kΩ

.988kΩ

0.985kΩ

R3

330kΩ

.328kΩ

0.328kΩ

RBE

N/A

.246kΩ

.247kΩ

RTotal

N/A

.463kΩ

0.468kΩ

ETotal

10V

10.01V

10.01V

ER1

N/A

4.69V

4.37V

ER2

N/A

5.32V

5.4V

ER3

N/A

5.32V

5.3V

ITotal

N/A

21.5mA

0.0213A

IR1

N/A

21.5mA

0.0213A

IR2

N/A

5.4mA

5.3mA

IR3

N/A

16.2mA

16.1mA

The Values are the same because we calculated the Values correctly and they be off like a few values but nothing is perfect.

Run #3 10) We added a 47 resistor (R4) that is parallel with (R3) 11) 2) Total Voltage= 10.01V R1=8.45V R2=1.55V R3=1.55V R4=1.55V 3) AB= 8.45V BC=0V CD=1.55V DF=0V FA=-10.01V HI=1.55V 4) AB=8.45V BE=1.55V EF=0V FA=-10.01V HI=1.55V 5)Total Current= 38.9A I1=38.8mA I2=1.6mA I3=4.7mA I4=32.6mA 6) It= 38.9mA= I2= 1.6mA + I3= 4.7mA + I4=32.6mA= 38.9mA=38.9mA. 7)R1= V1/I1= 4.69V/21.5mA= 0.218kΩ R2= V2/I2= 5.32V/5.4mA= 0.985kΩ R3= V3/I3= 5.32V/16.2mA= 0.328kΩ R4= V4/I4= 1.55V/32.6mA= 0.475kΩ Rbe=V2/It= 5.32V/21.5mA= 0.247kΩ

Rt= 0.218kΩ+0.985kΩ+0.328kΩ+0.247kΩ+ 0.475kΩ= 2.253kΩ 9) Nominal Values

Measured Values

Calculated values (using Nominal values)

R1

.220kΩ

.217kΩ

0.218kΩ

R2

1.0kΩ

.988kΩ

0.985kΩ

R3

330kΩ

.328kΩ

0.328kΩ

R4

47Ω

47.4Ω

48Ω

Rbe

N/A

40.2Ω

.247kΩ

RTotal

N/A

0.26kΩ

0.468kΩ

ETotal

10V

10V

10V

ER1

N/A

8.45V

8.5V

ER2

N/A

1.55V

1.55V

ER3

N/A

1.55V

1.55V

ER4

N/A

1.55V

1.55V

ITotal

N/A

10.01V

38.7mA

IR1

N/A

38.8mA

38.7mA

IR2

N/A

1.6mA

1.5mA

IR3

N/A

4.7mA

4.5mA

IR4

N/A

32.6mA

32.4mA

12) I1 & I2 in this Run (Run #3) was I2= 1.6mA and I3 was 4.7mA and in the previous run, Run #2 I2=05.4mA and I3=4.7mA. 13) IT in Run #2 was 21.5mA and IT in Run #3 was 38.9mA. So the extra resistor makes a difference in Total Current. 14) Rt in Run #2 was 1.778kΩ and Rt in Run #3 was 2.253kΩ Like with the Current adding an extra resistor in a parallel circuit makes a difference.

Run #4

Nominal Values

Measure Values

Calculated values

R1

470Ω

N/A

470Ω

R2

47Ω

N/A

47Ω

R3

860Ω

N/A

860Ω

R4

220Ω

N/A

220Ω

R5

330Ω

N/A

330Ω

Rde

N/A

115.23Ω

115.23Ω

Rce

N/A

148.16Ω

148.16Ω

Rbe

N/A

618.162Ω

618.612Ω

ITotal

N/A

16.17mA

16.17mA

IR1

N/A

16.17mA

16.17mA

IR2

N/A

13.39mA

13.39mA

IR3

N/A

2.7mA

2.7mA

I4

N/A

8.034mA

8.034mA

I5

N/A

5.34mA

5.34mA

ETotal

10V

10V

10V

ER1

N/A

2.6V

2.6V

ER2

N/A

.6V

.6V

ER3

N/A

2.3V

2.3V

E4

N/A

1.7V

1.7V

E5

N/A

1.7V

1.7V

Calculations I got most my calauctions by many different formulas such as Ohm's law and KWL and CWL but mostly Ohm's law. How i got the resistance for Run #2 and Run 3 R1= V1/I1 (Voltage divided by Current) 4.69V/21.5mA= 0.218kΩ R2= V2/I2 (Voltage divided by Current) 5.32V/5.4mA= 0.985kΩ R3= V3/I3 (Voltage divided by Current)

5.32V/16.2mA= 0.328kΩ Rbe= V2/It (Voltage divided by Total Current) 5.32/21.5= 0.247kΩ

How i Calculated the Current for Run #2 and Run 3 IR1= V/R1 (Voltage divided by Resistance) 10.01V/220Ω= 21.4mA IR2=1000(R2) x 330(R3)=330000= 330000/1330= 248.1203008+200=468.120 then you would do 10V/4.68= 0.0213 The reason why this is so long is because its a special case that does IR2xIR3/IR2+IR3. (Special case) IR3= V/R3 (Voltage divided by Resistance) 16.2V/330Ω= 5.32V Run 2 and 3 was the same process to find the resistance, the Current and the Voltage but just different numbers.

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.

Appendix...