Lab Report 4- Circuits 2 PDF

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University of the East Caloocan CampusCollege of EngineeringNEE 2202 – 1EE(CIRCUITS 2 LABORATORY)Inductive CircuitsEXPERIMENT NO. 4Submitted by: Submitted to: Ocampo, Jan Darrel B. Engr. Sinforoso D. Cimatu Jr.Student number: Date of Performance: 20181166370 February 03,Group Number: Date of Submiss...

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University of the East Caloocan Campus College of Engineering

NEE 2202 – 1EE (CIRCUITS 2 LABORATORY)

Inductive Circuits EXPERIMENT NO. 4

Submitted by: Ocampo, Jan Darrel B. Sinforoso D. Cimatu Jr.

Student number: of Performance: 20181166370 February 03,2020

Submitted to: Engr.

Date

Group Number: Date of Submission: TWO February 10, 2020

DATA COMPUTATION: RUN 1: Computed Values  Given: Trial 1 > VS = 7.1 V, VL = 1.1 V, VR = 6.5 V, I = 0.98 mA Trial 2 > VS = 7.1 V, VL = 2.1 V, VR = 6.1 V, I = 1.97 mA Trial 3 > VS = 7.1 V, VL = 3.2 V, VR = 5.3 V, I = 3.07 mA Trial 4 > VS = 7.1 V, VL = 4.2 V, VR = 4.3 V, I = 3.87 mA

TRIAL S

RESISTANCE (kΩ) Formula: R = VR / I

POWER FACTOR Formula: P.F = VR / VS

1 2 3

R = 6.5 V / 0.98 mA = 6.63 kΩ R = 6.1 V / 1.97 mA = 3.10 kΩ R = 5.3 V / 3.07 mA = 1.73 kΩ

P.F = 6.5 V / 7.1 V = 0.92 P.F = 6.1 V / 7.1 V = 0.86 P.F = 5.3 V / 7.1 V = 0.75

4

R = 4.3 V / 3.87 mA = 1.11 kΩ

P.F = 4.3 V / 7.1 V = 0.61

TRIALS

TOTAL POWER (mW) Formula: PT = (I)*(VS)* (P.F)

1

PT = (0.98 mA) * (7.1 V) * (0.92) = 6.40 mW

2

PT = (1.97 mA) * (7.1 V) * (0.86) = 12.03 mW

3

PT = (3.07 mA) * (7.1 V) * (0.75) = 16.35 mW

4

PT = (3.87 mA) * (7.1 V) * (0.61) = 16.76 mW

RUN 2: Computed Values  Given: Trial 1 > VS = 6.96 V, IT = 8.70 mA, IR = 1.2 mA, IL = 7.8 mA Trial 2 > VS = 6.96 V, IT = 9.30 mA, IR = 2.6 mA, IL = 7.8 mA Trial 3 > VS = 6.96 V, IT = 10.2 mA, IR = 4.4 mA, IL = 7.8 mA Trial 4 > VS = 6.96 V, IT = 11.4 mA, IR = 6.4 mA, IL = 7.8 mA

TRIAL S

RESISTANCE (kΩ) Formula: R = VS / IR

POWER FACTOR Formula: P.F = IR / IT

1 2 3

R = 6.96 V / 1.2 mA = 5.80 kΩ R = 6.96 V / 2.6 mA = 2.68 kΩ R = 6.96 V / 4.4 mA = 1.58 kΩ

P.F = 1.2 mA / 8.70 mA = 0.14 P.F = 2.6 mA / 9.30 mA = 0.28 P.F = 4.4 mA / 10.2 mA = 0.43

4

R = 6.96 V / 6.4 mA = 1.09 kΩ

P.F = 6.4 mA / 11.4 mA = 0.56

TRIALS

TOTAL POWER (mW) Formula: PT = (IT)*(VS)* (P.F)

1

PT = (8.70 mA) * (6.96 V) * (0.14) = 8.48 mW

2

PT = (9.30 mA) * (6.96 V) * (0.28) = 18.12 mW

3

PT = (10.2 mA) * (6.96 V) * (0.43) = 30.53 mW

4

PT = (11.4 mA) * (6.96 V) * (0.56) = 44.43 mW

Answers: Exercises: Exercise 1:

RUN 1: Computed Values  Given: Trial 1 > VS = 7.1 V, VL = 1.1 V, VR = 6.5 V, I = 0.98 mA Trial 2 > VS = 7.1 V, VL = 2.1 V, VR = 6.1 V, I = 1.97 mA Trial 3 > VS = 7.1 V, VL = 3.2 V, VR = 5.3 V, I = 3.07 mA Trial 4 > VS = 7.1 V, VL = 4.2 V, VR = 4.3 V, I = 3.87 mA

TRIAL S

RESISTANCE (kΩ) Formula: R = VR / I

POWER FACTOR Formula: P.F = VR / VS

1 2 3

R = 6.5 V / 0.98 mA = 6.63 kΩ R = 6.1 V / 1.97 mA = 3.10 kΩ R = 5.3 V / 3.07 mA = 1.73 kΩ

P.F = 6.5 V / 7.1 V = 0.92 P.F = 6.1 V / 7.1 V = 0.86 P.F = 5.3 V / 7.1 V = 0.75

4

R = 4.3 V / 3.87 mA = 1.11 kΩ

P.F = 4.3 V / 7.1 V = 0.61

TRIALS

TOTAL POWER (mW) Formula: PT = (I)*(VS)* (P.F)

1

PT = (0.98 mA) * (7.1 V) * (0.92) = 6.40 mW

2

PT = (1.97 mA) * (7.1 V) * (0.86) = 12.03 mW

3

PT = (3.07 mA) * (7.1 V) * (0.75) = 16.35 mW

4

PT = (3.87 mA) * (7.1 V) * (0.61) = 16.76 mW

RUN 2: Computed Values  Given: Trial 1 > VS = 6.96 V, IT = 8.70 mA, IR = 1.2 mA, IL = 7.8 mA Trial 2 > VS = 6.96 V, IT = 9.30 mA, IR = 2.6 mA, IL = 7.8 mA Trial 3 > VS = 6.96 V, IT = 10.2 mA, IR = 4.4 mA, IL = 7.8 mA Trial 4 > VS = 6.96 V, IT = 11.4 mA, IR = 6.4 mA, IL = 7.8 mA

TRIAL S

RESISTANCE (kΩ) Formula: R = VS / IR

POWER FACTOR Formula: P.F = IR / IT

1 2 3

R = 6.96 V / 1.2 mA = 5.80 kΩ R = 6.96 V / 2.6 mA = 2.68 kΩ R = 6.96 V / 4.4 mA = 1.58 kΩ

P.F = 1.2 mA / 8.70 mA = 0.14 P.F = 2.6 mA / 9.30 mA = 0.28 P.F = 4.4 mA / 10.2 mA = 0.43

4

R = 6.96 V / 6.4 mA = 1.09 kΩ

P.F = 6.4 mA / 11.4 mA = 0.56

TRIALS

TOTAL POWER (mW) Formula: PT = (IT)*(VS)* (P.F)

1

PT = (8.70 mA) * (6.96 V) * (0.14) = 8.48 mW

2

PT = (9.30 mA) * (6.96 V) * (0.28) = 18.12 mW

3

PT = (10.2 mA) * (6.96 V) * (0.43) = 30.53 mW

4

PT = (11.4 mA) * (6.96 V) * (0.56) = 44.43 mW

Exercise 2:

Answers: Question and Problems: 1. Trials

VS approximate = VR2 + VL2

1

VS Approximate = (6.5 V) 2 + (1.1 V)2 = 6.59 V

2

VS Approximate = (6.1 V) 2 + (2.1 V)2 = 6.45 V

3

VS Approximate = (5.3 V) 2 + (3.2 V)2 = 6.19 V

4

VS Approximate = (4.3 V) 2 + (4.2 V)2 = 6.01 V

> Comparing the measured value for voltage supply and the computation for vector sum of voltages VR and V L, it is observed that both of there values for every trial is different but both are relatively close to one another. The computed total voltage seems to be decreasing because of the resistance in the circuit which is being varied by decreasing its amount using parallel connections in each resistor. 2. Trials

IT approximate = IR2 + IL2

1

IT Approximate = (1.2 mA)2 + (7.8 mA)2 = 7.890 mA

2

IT Approximate = (2.6 mA)2 + (7.8 mA)2 = 8.220 mA

3

IT Approximate = (4.4 mA)2 + (7.8 mA)2 = 8.960 mA

4

IT Approximate = (6.4 mA)2 + (7.8 mA)2 = 10.09 mA

> Comparing the measured value for total current and the computation for vector sum of current IR and IL, it is observed that the values are relatively close to one another and are increasing for each trial which tells that the resistance in the circuit is being decreased every trial passed which is the effect of connecting the resistors in each trial to parallel, and since current varies indirectly with the resistance. 3. In Table 4.1 – series R-L circuit, it is observed that both the resistance and power factor are decreasing. The resistance R is considered to be directly proportional to the voltage across the resistor V R based on the equation: V R = R / I and its changes also causes changes in the power factor on a direct proportion relation based on its equation P.F = VR / VS. In Table 4.2 – parallel R-L circuit, it is observed that the resistance is decreasing while the power factor is increasing on every trial. The relationship between the resistance in the circuit and the power factor have an indirectly proportional relation due to the resistance indirectly proportional influence to the IR and its changes also causes changes in the power factor on a direct proportion relation. 4. Trials

P = IR2 R

1

P = (5.80 kΩ) (1.2 mA)2 = 8.350 Ma

2

P = (2.68 kΩ) (2.6 mA)2 = 18.12 mA

3

P = (1.58 kΩ) (4.4 mA)2 = 30.59 mA

4

P = (1.09 kΩ) (6.4 mA)2 = 44.65 mA

> The values indicate that the power dissipated in the resistance appears to be rising in value due to the increase in the current in the resistor/s and the decrease in resistance.

5. Given: R = 10 Ω, V = 220 V, RL = 5 Ω, f = 60 Hz, L = 0.05 h Required: a) I, b) P.F., c) PAVE(TOTAL), d) VL Solution: *Calculate for I XL = 2πf*L = 2π (60 Hz) *(0.05 h) = 18.85 Ω RT = 10 Ω + 5 Ω = 15 Ω ZT = R2 + XL2 = (15 Ω)2 + (18.85 Ω)2 = 24.09 Ω I = V/ZT = (220 V)/ (24.09 Ω) = 9.13 A

*Calculate for P.F. P.F. = cos θ = VR/VS = 91.3/220 = 0.415 (lagging) *Calculate for PAVE(TOTAL) PAVE(TOTAL) = VI cos θ or VI* P.F. = (220 V) *(9.13 A) (0.415) = 833.57 Watts *Calculate for VL VR = IR = (9.13 A) (10 Ω) = 91.3 V VL = VS2 - VR2 = (220 V)2 – (91.3 V)2 = 200.16 V

6. Given: P = 1000 Watts

F = 60 Hz

IL = 20 A

R = 100 Ω

V = 220 V Required: a) IR and IT, b) ZT, c) (P.F.) T, d) PT Solution: *Calculate for IR IR = V/R = (220 V)/ (100 Ω) = 2.2 A *Calculate for IT IT = IR2 + IL2 = (2.2 A)2 + (20 A)2 = 20.12 A *Calculate for ZT ZT = V/IT = (220 V)/ (20.12 A) = 10.93 Ω *Calculate for (P.F.) T (P.F.) T = cos θ = IR / IT = (2.2)/ (20.12) = 0.109 (lagging) *Calculate for PT PT = VIT cos θ = (220 V) (20.12 A) (0.109) = 482.48 Watts

DATA ANALYSIS AND INTERPRETATION: The purpose of the experiment is to become familiarized with the voltage and current relations in series and parallel R-L circuits, gain practical knowledge and understanding regarding relations between voltages in a series R-L circuit as well as relations between current in a parallel R-L circuit, and to obtain the power and power factor of inductive circuits.

TABLE 4.1 (Run 1: Series R-L Circuit)

TRIAL

VS

VL

VR

I (mA)

R (kΩ)

PT (mW)

P. F

1

7.1 V

1.1 V

6.5 V

0.98 mA

6.63 kΩ

6.40 mW

0.92

2

7.1 V

2.1 V

6.1 V

1.97 mA

3.10 kΩ

12.03 mW

0.86

3

7.1 V

3.2 V

5.3 V

3.07 mA

1.73 kΩ

16.35 mW

0.75

4

7.1 V

4.2 V

4.3 V

3.87 mA

1.11 kΩ

16.76 mW

0.61

Table 4.1 above represents the set of data gathered during the run 1 of the experiment showing: VS (voltage supply), VL (the voltage across the inductor), V R (the voltage across the resistor), I (current), R (resistance), P T (total power), P. F (power factor). The V S, VL, VR and the current are obtained using the voltmeter and ammeter. The values for the remaining elements are computed: for resistance: R = VR / I, for total power: PT = (I)*(V S)*(P.F) and lastly for power factor: P.F = V R / VS. Observing the data for every trial, it shows a constant measured value for supply voltage which is 7.1 V which will be used as a component to acquire the values for the remaining elements. It is also observed that there is an increase on the measured value for V L and current which happened after toggling the switch for R4 and R2 to the “right” while R3 and R1 to the “left”; and a decrease on the measured value for V R after transferring the connection of terminal -A of voltmeter A from -B to +B based on the given procedure.

Analyzing the data shown in every trial, the voltage across the inductor is observed to be increasing which shows that V L is directly proportional to the current I in the circuit. Using the equation for computing for the resistance: R = V R / I, the voltage across the resistance is directly proportional to the resistance while inversely to the current of the circuit which concludes why the current is increasing as well as the V L while the VR and R are decreasing. Based on the equation: VS = √V2R + V 2L , it shows that the value for the supply voltage is dependent to the change in V R and VL therefore both elements will have the same value for supply voltage when acquiring their respective values according to the equation: VL = √V2S - V2L and VR = √V2S – V2L therefore VL and VR is inversely proportional to one another based on the derived equation from the VS. The voltage across the resistor is also directly proportional to the power factor using the equation: P.F = VR / VS but since the supply voltage is constant, there is a slight decrease of

values for power factor. The power factor is used as a component to acquire the total power in the circuit based on the equation: PT = (I)*(V S)*(P.F) where the total power is dependent to the change in current, power factor and voltage supply considering that the V S is constant, P.F is decreasing and current is increasing therefore the total power would also increase.

TABLE 4.2 (Run 2: Parallel R-L Circuit) TRIAL

VS

IT (mA)

IR (mA)

IL (mA)

R (kΩ)

PT (mW)

P. F

1

6.96 V

8.7 mA

1.2 mA

7.8 mA

5.80 kΩ

8.48 mW

0.14

2

6.96 V

9.3 mA

2.6 mA

7.8 mA

2.68 kΩ

18.12 mW

0.28

3

6.96 V

10.2 mA

4.4 mA

7.8 mA

1.58 kΩ

30.53 mW

0.43

4

6.96 V

11.4 mA

6.4 mA

7.8 mA

1.09 kΩ

44.43 mW

0.56

Table 4.2 above represents the set of data gathered during the run 2 of the experiment which shows now the IT (total current), IR (the current across the resistor) and IL (the current across the inductor) which is obtained using the ammeter. These will be used as a component to compute for the values of resistance, total power and power factor; and for analyzation. The values for the remaining elements are computed: for resistance: R = VS / IR, for total power: P T = (I)*(V S)*(P.F) and lastly for power factor: P.F = IR / IT. Observing the data for each trial, it shows a constant measured value for the supply voltage V S which is 6.96 V and IL which is 7.8 mA. It is also observed that there is an increase on the measured value for I T which happened after toggling the switch for R4 and R2 to the “right” while R3 and R1 to the “left”; and an increase on the measured value for IR after removing the connections between terminals A1 and A2 and transfer the connection in terminal T7 to terminal A1 then connect the terminal X 7 and T7 together based on the procedure. Analyzing the data shown in every trial, the total current is observed to be increasing every trial which shows that the value for the total current is dependent to the change in IR and IL therefore these two elements is directly proportional to IT based on the equation IT = √I2R + I2L. Both IR and IL will have the same value for total current when acquiring their respective values

according to the equation: IL = √I2T – I2L and IR = √I2T – I2L. Since the measured value for IL is constant and the IR is observed increasing therefore the IT would also increase. Based on the equation R = VS / IR, the resistance considered to be directly proportional to voltage supply and inversely to the current across the resistor based on the data since the V S is constant while the IR is increasing therefore the R would decrease its value. The power factor of a parallel circuit is dependent to the change in the measured value of the current across the resistor and the total current based on the equation P.F = IR / IT therefore the IR is directly proportional to the power factor whereas the increase on the IR would mark the increase in value for the power factor. The power factor is used as a component to acquire the total power in the circuit based on the equation: PT = (I)*(V S)*(P.F) where the total power is dependent to the change in voltage supply, total power and power factor considering that the V S is constant and both total current and P.F is increasing therefore the total power would increase.

FINDINGS & CONCLUSION: For this experiment, the group is able to become familiarized with the voltage and current relations in series and parallel R-L circuits, gain practical knowledge and understanding regarding relations between voltages in a series R-L circuit as well as relations between current in a parallel R-L circuit, and to obtain the power and power factor of inductive circuits. As it was shown in Table 4.1, it is observed that the voltage across the inductor and the current is increasing while the resistance decreases in every trial which means that the V L is directly proportional to the current of the circuit while the resistance is inversely proportional to the current using the ohms law: I = V L / R . The ohms law can also be the basis as to why the

current increases using the voltage across the resistance and the value of resistance as the component for attaining with the equation I = V R / R. Since the V L is directly proportional while the R is inverse to current and that R is directly proportional to the V R, therefore the voltage across the inductor is inversely proportional to the voltage across the resistance based on the vector sum VR and V L derived equation: VL = √V2S - V2L and V R = √V2S – V2L whereas both equation uses the same voltage supply either it be constant or increasing. Using the equation for power factor: P.F = VR / VS, it is concluded that the voltage across the resistor is also directly proportional to the power factor but since the supply voltage is constant, there is a slight decrease of values for power factor. That power factor is used as a component to acquire the total power in the circuit based on the equation: PT = (I)*(V S)*(P.F) where the total power is dependent to the change in current, power factor and voltage supply considering that the V S is constant, P.F is decreasing and current is increasing therefore the total power would also increase. As it was shown in Table 4.2, the total current is dependent to the change in current across the resistor and current across the inductor whereas both of these elements are directly proportional to the total current based on the equation IT = √I2R + I2L. This means that the same total current value will be used to compute for the value of I R and IL according to these derived equation IL = √I2T – I2L and IR = √I2T – I2L. The resistance in the circuit shows to be decreasing while the power factor seems to have a raise in value. In which we can specify the relation between the resistance in the circuit and the power factor have an indirectly proportional relation due to the resistance indirectly proportional influence to the IR and its changes also causes changes in the power factor on a direct proportion relation. That power factor is used as a component to acquire the total power in the circuit based on the equation: PT = (I)*(VS)*(P.F) where the total power is dependent to the change in current, power factor and voltage supply considering that the VS is constant, P.F is decreasing and current is increasing therefore the total power would also increase.

REFERENCE:  https://www.allaboutcircuits.com/textbook/alternating-current/chpt3/parallel-resistor-inductor-circuits/  https://www.electronics-tutorials.ws/accircuits/parallel-circuit.html  http://www.learningaboutelectronics.com/Articles/What-is-a-constantvoltage-source.php

 https://www.allaboutcircuits.com/textbook/alternating-current/chpt-3/seriesresistor-inductor-circuits/  https://courses.lumenlearning.com/physics/chapter/23-10-rl-circuits/...