8 Adik

Faculty of Engineering Department Of Electrical & Electronic Engineering EEE 3100 Electrical & Electronic Technology

Lab 8: AC Circuits (RLC)

Name: NOOR SUHAIDA GALIP Matric No: 140007 Lecturer: PN RIBHAN ZAFIRA ABDUL RAHMAN

TABLE OF CONTENT CONTENTS

PAGE

Objectives

3

Equipments and Components

3

Results and Calculation

4

Discussion

9

Exercise

10

Conclusion

12

Reference

12

LAB 8

:

AC Circuit (RC and RL)

OBJECTIVES:

1. To verify Kirchhoff’s Voltage Law in RLC series circuit 2. To verify Kirchhoff’s Current Law in RLC parallel circuit

EQUIPMENTS AND COMPONENTS:

1. Multimeter 2. Signal generator 3. Breadboard 4. Resistor: 1 kΩ, 10Ω 5. Capacitor: 0.01 µF 6. Inductor: 10 mH 7. Crocodile clips 8. Connecting wires

RESULT AND CALCULATION: 8.4.1 To verify Kirchoff’s voltage law in RLC series circuit 1. Measure the R using a multimeter. Rmeasured =

980 Ω

2. Measure voltage of every component (VR, VL and VC) with E = 8 Vp-p using an oscilloscope. VRp-p = 8 V

VCp-p = 0.8 V

3. Calculate Ip-p using this equation; Ip-p

= VRp-p / Rmeasured = 8 V / 980 Ω = 8.163 mA

4. Calculate the ZT ZT

= Ep-p / Ip-p = 8 V / 8.163 mA = 980 Ω

5. Using L and C values used and Rmeasured, calculate ZT.

VLp-p = 8 V

= 2pf = 2p(1000Hz) = 6283.19

Xc

= 1/ωc

= 1 / 6283.19 (0.01 x 10-6) = 15915.48Ω

Zc

=1/j c

= 15915.48Ω

XL

= ωL

= 15915.48 ∠ - 90°

= 6283.19 (10 x 10-3) = 62.8319W

ZL

= jωL = 62.8319 ∠ - 90°

ZT

= R + j XL - j Xc = 1000 + j62.8319 - j15915.48 = 1000W

6. Using E = 8 V ∠ 0° , calculate I, VR, VL and VC using peak-to-peak values.

I

= V / ZT = 8 Ð - 0° / 15884.16 Ð - 86.39° = 5.0366 x 10-4∠ 86.39°A

V

= IR = 5.0366 x 10-4Ð 86.39° x 1000 ∠ 0° = 0.50366 ∠ 86.39°V

VL

= Ij XL = 5.0366 x 10-4 (j62.8319) = 0.03165 ∠ 90°V

VC

= Ij XC = 5.0366 x 10-4 (-j 15915.48) = 8.01599 ∠ -90°V

7. Draw phasor diagram including I and all voltages.

8. E

= Ö(VR2 + (VL - VC)2) = Ö(82 + (8 - 8)2) = 8V

9. Vab(p-p) = E Ð 0° x (ZLC / ZT) ∠ 0°

ZLC

= ZL + ZC

= j ZL Ð j ZC = (62.8319 ∠ 90°) + (15915.48 Ð -90°) = 15852.96 ∠ -90°

Vab(p-p) = (8 ∠ 0°) x (15852.96 ∠ -90°) / 15884.16 ∠ - 86.39° = 7.9841 Ð -90°

10. Vab

= 8V

8.4.2 To verify Kirchoff’s current law in RLC parallel circuit 1. Measure the R using a multimeter. Rmeasured = 2. Using L and C values used and Rmeasured, calculate IRp-p, ILp-p, ICp-p and ISp-p

IRp-p

= 8 ∠ 0° / 970Ω

ILp-p = 8 Ð 0° / 62.8319 Ð90° ICp-p ISp-p

= 8.247 x 10-3 ∠ 0° = 0.12732 ∠ -90°

= 8 Ð 0° / 15915.48 ∠ -90° = 5.0266 Ð 90° = IRp-p + ILp-p + ICp-p

= 0.12707 ∠ 86.39°

980 Ω

3. Disconnect the generator. Connect RS as shown in Figure 8.6 to measure IS. Measured RS. Place the scope across the RS where it should have the same grounding with signal generator. The RS value is very small so that it would not interfere the circuit. Supply E = 8Vp-p and measure the peak to peak voltage across RS.

FIGURE 8.6

RSmeasured = 10.3Ω

VRSp = 0.4 V

VRSp-p = 0.8 V

4. Using Rmeasured and VRS obtained in step 3, calculate ISp-p. V = IR ISp-p = VRS / RS = 0.8 / 10.3 = 0.08mA 5. Disconnect the signal generator. Using RSmeasured value, calculate the peak to peak current for IL and IC. To get ILp-p and ICp-p as shown in Figue 8.7 and 8.8 and measured the respective VRS.

FIGURE 8.7

ILp-p = VRS / RS ICp-p

= 0.3 x 2 / 10.3

= VRS / RS

= 0.06A

= 0.3 x 2 / 10.3

FIGURE 8.8

6. Calculate the IRp-p if VR = E IRp-p

= VR / RS = 8V / 10.3 = 0.008A

= 0.06A

7. Using E = 8 V ∠0° and all measured current obtained, draw a phasor diagram and get the phase angle between voltage and current.

8. XL

= 8V / 0.07

= 114.27Ω

9. XL

= 8V / 0.8

= 10Ω

CONCLUSION:

Analysis of series-parallel AC circuits is much the same as series-parallel DC circuits or in other word, we can verify Kirchhoff’s Voltage Law in RLC series circuit and Kirchhoff’s Current Law in RLC parallel circuit. It is important to remember that before series-parallel reduction (simplification) can begin, you must determine the impedance (Z) of every resistor, inductor, and capacitor. That way, all component values will be expressed in common terms (Z) instead of an incompatible mix of resistance (R), inductance (L), and capacitance (C).

REFERENCE:

1. Electrical & Electronic Technology (Lab Manual). 2. Principles and Applications of Electrical Engineering, Second Edition.

EXERCISE 1. Calculate the XL, XC and ZT for circuit shown in Figure 8.4 at frequency of 10kHz. Solution: Series circuit:

f = 10 kHz L = 10 x 10-3 H C = 0.01 x 10-6 C E=8V R = 1000 Ω

Since ω = 2f XL = ωL

= (2) (10 x 103) (10 x 10-3) = 628.3185 Ω

XC = 1 / ωC

= 1 / (2) (10 x 103) (0.01 x 10-6) = 1591.549431 Ω

ZT

= [R2 + (XL  XC) 2] ½ = [10002 + (628.3185  1591.549431) 2] ½ = 1388.457355

θ

= tan-1 (XL  XC / R) = tan-1 (628.3185  1591.549431) / 1000 = -43.927°

ZT

= 1388.457355 < -43.927° Ω

2. For case in question 1, is ZT value is less than every component that is parallel at 10 kHz? Solution: Parallel circuit: ZT = ZR||ZL||ZC ZT = 1 / [(1/ZL) + (1 / ZR) + (/ZC)]

Where; = 1 / jωL = 1 / j (2) (10 x 103) (10 x 10-3)

1/ZL

= -1.59155 x 10-3 j = 1/ (1/jωC) = 1/ (1/j (2) (10 x 103) (0.01 x 10-6)

1/ZC

= 6.28318537 x 10-4 j 1/ ZR = 1/1000 = 1 x 10-3

ZT

= 1 / [1 x 10-3 + 6.28318537 x 10-4 j -1.59155 x 10-3 j] = 720.2236 < 43.927° Ω.

The value for ZT for case 1 is larger than every component that is in parallel at 10 kHz.