Power in AC Circuit Lab Report

February 21, 2018 | Author: JEJUNG | Category: Capacitor, Ac Power, Electrical Network, Amplitude, Trigonometric Functions
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Power in AC Circuit Lab Report electric circuits lab...

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Power in AC Circuit Lab Report

Aims: 1. To differentiate between true power and apparent power in AC circuit. 2. To measure power in AC circuit using volt-ampere method and oscilloscope. Apparatus: •

Function Generator.



Dual channel oscilloscope.



Digital multimeter.



Capacitors 47nF and 100nF.



Resistor 100Ω.

Method: A: Determining Power Using the Volt-Ampere Method. 1. The resistance and the capacitance are measured using RCL meter or DMM and the values are recorded in Table 8-1 and Table 8-2. 2. The circuit in figure 8-1 is connected and the frequency of the signal generator is set to 10 kHz.

3. S1 is closed. The output voltage of the signal generator is increased to 3Vrms and this is maintained through out the course of the measurement. The voltage across the resistor (VR), the voltage across the capacitor (VC) and the total current (IT) are measured and the values is recoded in Table 8-1. 4. The corresponding phase angle (θ), power factor, apparent power (PA), and the true power (PT) are calculated using the formulas given and the values that was calculated is filled into Table 8-1. 5. S1 is opened; the 47nF capacitor is replaced with 100nF. 6. S1 is closed. The output voltage of the signal generator is increased to 3Vrms and this is maintained through out the course of the measurement. The voltage across the resistor (VR), the voltage across the capacitor (VC) and the total current (IT) are measured and the values is recorded in Table 8-1. 7. With the resistance, capacitances, Vrms, and th e frequency, the theoretical values are calculated and it is filled in Table 8-2 and the value for relative or percentage errors are calculated.

Nominal

Meas.

Values R=0.1kΩ

Values R=99.5Ω

C=47nF C=46.6nF R=0.1kΩ R=99.5Ω

VR

VC

0.78

2.89

0.654

6 2.5

C=0.1µF C=0.11µF

IT 7.86mA 16.21m

Phase

Power

PA

Angle Factor 74.9˚ 0.26 0.0236 77.4

0.218

PT 6.14

X10-3 0.0486 1.06 X10-2

A Table 8-1: Measured Values.

Nominal

Meas.

VR

VC

IT

Phase

Power

PA

PT

Values R=2.2kΩ

Values R=0.1kΩ

C=47nF

C=47.1nF

869.5

2.871

8.695

Angle Factor 72.37 0.3028 7.56m

25.0m

Table 8-2: Theoretical Values.

B: Determining Power Factor with Oscilloscope. 1. The dual-trace oscilloscope is connected to series RC circuit. The output voltage should be the lowest. 2. S1 is closed. The output from the generator is increased to 3Vrms. Channel 1 is the voltage reference channel. The oscilloscope is turned on. The controls on the oscilloscope are adjusted so that a single wave, about 6 div peak-to-peak fills the width of the screen. The horizontal and vertical controls are used to center the waveform on the screen. 3. Switch to channel 2, this is the current channel. The controls on the oscilloscope are adjusted so that a single wave, about 4 div peak-to-peak fills the width of the screen. The vertical controls are used to center the waveform vertically. The horizontal controls are not to be used. 4. The oscilloscope is turned to dual-channel where the channels 1 and 2 should appear together. The location where the curves cross the horizontal axis are noted whereby these are the zero points of the two sin waves. With a centimeter scale, the horizontal distance (d) between the two positive and the two negative peaks of the sine wave is measured accurately. The value for distance (D) from 0˚ to 360˚ for the voltage sine wave is measured and all the values are recorded in Table 8-2. 5. PSpice is used to simulate this circuit.

6. The 47nF capacitor is replaced with 100nF. 7. S1 is closed and steps 3 to 6 are repeated for 100nF capacitor.

Table of Results: Resistance Ω (nominal value)

Capacitance nF (nominal value)

100 100

47 100

Distance between zero points (d) cm 20 14

Width of sine wave (D) cm

Phase angle (calculated) Ө degree

Power factor % (calculated)

100 99

72˚ 68.4˚

30.9% 36.8%

Table 8-2 Calculations: For: Table 8-1 (where R=100Ω and C=45nF and VS=3V) Phase angle (Ө) = cos-1(VR / Vrms) = cos-1(0.78 / 3) = 74.9˚

P A = V S x IT = 3 x 7.86mA = 0.0236 W

Power factor = VR / VS = 0.78 / 3 = 0.26W

PT = VS x IT x cos Ө = 3 x 7.86mA x cos 74.9˚ = 6.14x10-3W

For: Table 8-2

(When C=47nF) Ө (degree) = (d /D) x 360˚ = ( 20 / 100 ) × 360 0 = 72˚

Power factor % = (cos θ) x 100% = ( cos 72 ) × 100% = 30.9%

(When C=100nF) Ө (degree) = (d /D) x 360˚ = (14 / 99 ) × 360 0 = 68.4˚

Power factor % = (cos θ) x 100% = ( cos 68.4 ) × 100% = 36.8%

Discussion and conclusion: According to the basic information and theory, power in AC circuits is consumed only by the resistive components. There is several ways to determine the power in AC circuit. Basically, apparent power PA in an AC circuit is the product of the source voltage and the line current PA = VS x IT, where V is the applied voltage and I is the current taken by the circuit. The true power dissipated by the circuit is the product of V and I and the power factor PF. The power factor is equal to the cosine of the angle between the voltage and current in the circuit in the circuit, that is, PT = VS x IT x cos θ . Besides that, other formulas for true power are PT = IT2 x R where IT is total current in the circuit in amperes, R is total resistance of the circuit in ohms, and VR is voltage measure across the total resistance of the circuit. Theoretically, we can calculate the VR by ohm’s law with VR = IT x R. The current flow in circuit can be determined by the reactant XC = 1/2 π f C, IT = VS / XC. For the phase angle between the applied circuit voltage Vs and the current IT, we can get it by determine the power factor of an AC circuit. The power factor is PT / PA and also equal to cos θ . So, we can also determine the phase angle from it by calculate cos-1 (PT / PA). As a conclusion, we can say that the different between true power and apparent power in AC circuits is about a cosine of phase angle. Power in an AC circuit may be determined by measuring the applied voltage V and the current I and the phase angle and substituting the measured values in the formula PT = VS x IT x cos θ . However the true power may be measured directly, using a wattmeter.

Figure 8-2

Figure 8-3

3 .0 V

(1 0. 00 0K ,2 .8 7 71)

- V r (4 7n F)

2 .0 V

1 .0 V

(1 0. 00 0 K, 849. 63 1m ) 0V 0 Hz

- Vr (4 7n F)

10 K Hz 2 0K H z V( C1 :2 ) V (R 1: 1, R1 :2 )

30 KH z

40 K Hz

5 0 KH z

60 KHz

7 0K Hz

80 KH z

90 KH z

1 0 0K Hz

60KHz

70KHz

80KHz

90KHz

100KHz

60KHz

70KHz

80KHz

90KHz

100KHz

Fr eq ue nc y

3.0V (10.000K,2.5402)

- Vc (100nF)

2.0V

(10.000K,1.5959)

1.0V

0V 0Hz

10KHz 20KHz V(C1:2) V(R1:1,R1:2)

- Vc (100nF)

30KHz

40KHz

50KHz Frequency

30mA

20mA

10mA

(10.000K,8.4963m)

0A 0Hz

10KHz

20KHz

; C=47nF

30KHz

40KHz

50KHz

I(V1) Frequency

30mA

20mA

(10.000K,15.959m)

; C=100nF

10mA

0A 0Hz

10KHz

20KHz

30KHz

40KHz

50KHz

I(V1) Frequency

60KHz

70KHz

80KHz

90KHz

100KHz

View more...

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