Published on

this report still not completed and got many errors

Published in: Education
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide


  1. 1. FACULTY OF ELETRICAL ENGINEERING UNIVERSITY TEKNOLOGI MARA ELECTRICAL ENGINEERING LABORATORY 1 (EEE230) EXPERIMENT 7 SERIES RESISTOR-CAPASITOR CIRCUITOBJECTIVE 1. To understand the relationship between voltage, current and phase angle. 2. To calculate the phase angle.LIST OF REQUIREMENTSEQUIPMENT 1. Signal generator 2. Oscilloscope 3. MultimeterCOMPONENT 1. Decade capacitor box 2. Decade resistor box1
  2. 2. THEORYIn the DC analysis of resistor circuits we examined how to calculate the total circuit resistance ofseries components. In this section we will use this approach to analyse circuits containing seriesresistors and capacitors. To do this we use the capacitative reactance as the effectiveresistance of the capacitor and then proceed in a similar manner to before.Analysing Series RC CircuitsYou will recall that a series circuit provides only one route for the current to flow between twopoints in a circuit, so for example the diagram below shows a resistor in series with a capacitorbetween the points A and B. Since the resistor and capacitor are in series,the impedance can be calculated using the equation 1.1 Z = R +XC (Equation 1.1) Values of XC depends on the frequency of the AC current flowing through the capacitor:We can calculated using equation 1.2 XC = -j/(2f C) equation 1.2) ( From Equation 1.1 and 1.2 ,We can get an expression for Z in terms of f, C and R: Z = R -j/(2f C)2
  3. 3. Figure 1.1 shows a circuit which resistor and capacitor is in series with alternating sine wavevoltage source. The current flowing in the circuit given by the equation 1.3I = Im sin ωt(Equation 1.3)Figure 1.1Using Ohm’s Law the voltage across the resistor is: VR = IR(Equation 1.4) VR = Im sin ωt × R(Equation 1.5)Hence, VR = Vm sin ωt(Equation 1.6) Vm = ImR(Equation 1.7) Im = (Equation 1.8)3
  4. 4. From equations 1.3 and 1.6 as I = Im sin ωt and Vr=Vmr sin ωt , We can observe the votageacross resistor and current flow through it are in phase or we can call same phase.We can seethe shape of the VR and I waveform as in Figure 7.2.If the voltage and current equations are written in phasor notation, then VR = VR < O° where themagnitudes of VR and I are root mean squre (rms) values. The phasor diagrams for VR and I areas in figure 7.3 which shows that they are in phase.The voltage across the capacitor VC can be derived from VC = ∫ idt(Equation 1.9) VC = ∫ Im sin ωtdt(Equation 2.0) VC = cos ωt(Equation 2.1) VC = sin (ωt-90°)(Equation 2.2)Therefore, VC = Vmc sin (ωt-90°)(Equation 2.3) Vmc = = ImXC(Equation 2.4)4
  5. 5. WhereXC is known as the capacitive reactance.From Equations 1.3 and 2.3.We can see the voltage across the capacitor lags the current by90°. In phasor notation, the voltage and the current can be written as VC = VC √-90° and I = I√0°respectively. The waveforms and the phasor diagram for C and I can be seen as in Figure 7.4and Figure 7.5 .5
  6. 6. The waveforms and the complete phasor diagram for the circuit in figure 1 are illustrated infigure 7.6 and figure 7.7By combining figure 7.2 and figure 7.4,We can see the diagram as in Figure 7.6 whilecombination of figure 7.3 and figure 7.5 results in the phasor diagram are shown in figure 7.7.The angle between the voltage and the current is called phasor angle. In figure 7.7 the angleunit is θ. From the figure , the magnitude of the voltage is given by: V= (Equation 2.5) V= (Equation 2.6) = IR(Equation 2.7)6
  7. 7. Where (Equation 2.8)And the phase angle is given by θ= (Equation 2.9) θ= (Equation 3.0)For AC. circuits, the resistance is termed impedance and is given the symbol Z. the magnitudeof impedance of a circuit containing a series combination of a resistor and a capacitor is givenby . Thus, the impedance for the circuit of figure 7.1 isWith phase angle of θ = tan¯¹ (Equation 3.1)In phasor notation, this is written as Z = R- jXC (Equation 3.2)7
  8. 8. PROCEDURES 1. The circuit was connected as in figure 7.8 2. V and VR waveforms were obtained on the oscilloscope. The waveforms were drawed and labeled.The peak values of V and VR was stated.For the circuit,the current is in phase with the voltage VR.Thus,the phase difference between V and VR is equal to the phase angle between V and I waveforms.The distance between the waves(d1) and distance for one cycle(X1) was determined. 3. The resistor and the capacitor position are changed as in Figure 7.9 and the V and VC waveforms were obtained. The distance between the waves (d2) and distance for one cycle (XC) were determined. 4. All readings were recorded in table 7.1. Figure 7.8 Figure 7.98
  9. 9. RESULTS C1 = 0.187 μF C2 = 0.085 μfV 9.98Vp-p 10 Vp-pVR 7.52Vp-p 4.80Vp-pVC 6.40Vp-p 8.80Vp-pd1 130μs 190μsX1 1.000ms 1.000msd2 150.0μs 100μsX2 1.000ms 1.000ms Table 7.1For each value of capacitor used, determined the phase angle using the following formulae: θ = tan¯¹ (Equation 3.3) θ =( ) × 360°(Equation 3.4) θ = 90 - ( ) × 360°(Equation 3.5) 5. All values were recorded in table 7.2 C1 = 0.187 μF C2 = 0.085μFCalculated phase angle 40.40° 61.89°Measured phase angle , θ1 40.96° 61.23° θ2 46.48° 68.24° θ3 36.00° 54.00° Table 7.29
  10. 10. DISCUSSION Before we start our experiment,we need make sure that our equipment is in goodcondition.First we test our connecting wires by using multimeter.connecting wire are important ,ifnot it can affect our Values in the end of experiment.We will heard a sound from multimeter ifthe Connecting wires in good condition.Then we test our oscilloscope and Signal generator .Thevalues that we set up in signal generator must be same as in oscilloscope.So that means thatthe equipment is good connection.The Decade capacitor box and Decade Resistor box wastesting by using multimeter .The values that we set on decade capacitor box or DecadeResistor box must be same on multimeter.We start our experiment as our equipments andcomponents are in good condition. In this experiment ,we use same resistor but different capacitor. The voltage peak-peakof input voltage(V), voltage peak-peak of waveform voltage drop at resistor, (VR ), voltage peak-peak of waveform voltage drop at capacitor,( VC ), the distance between the waves (d1) and thedistance for one cycle(X1). The distance between the waves (d2) and the distance for onecycle(X2) are for different position of capacitor and resistor circuit.We use two capacitor in thisexperiment which are 0.187μF capacitor and 0.085μF capacitor.We use two capacitor tocompare the phase angle which obtained in the end of experiment. From the result in the Table 7.1,The values of in circuit in circuit with has differentposition of capacitor and resistor as in Figure 7.8 and Figure 7.9 are different.As in Figure7.8 ,capacitor is in lead ,then the resistor this mean the capacitor will received the voltage.As weknow that capacitor function is stores charge and then quickly release most of this charge whenit is needed.So the values of VR are higher than Vc.As in Figure 7.9 ,The resistor received thevoltage supply first .So the capacitor only received the Voltage supply that was filtered by theresistor.As we know Voltage total in series must be same As sum of Voltage across in eachcomponent but in this experiment ,the values of sum of voltage across component is higher thanthe Voltage source.This is because of the function of capacitor.In this experiment we also usetwo different values of capacitor which are 0.187μF capacitor and 0.085μF capacitor to knowthe phase angle between small values of capacitor or high values of capacitor.As in Table 7.2,We can see that circuit which use capacitor 0.187μF has a smaller phase angle than circuitwhich use capacitor 0.085μF but it is between 0® and -90®.As we know we cannot measured10
  11. 11. directly the phase angle but using mathematical way as in Equation 3.3, Equation 3.4 andEquation 3.5. While we are doing the experiment,there are some error that we detected.One of themare instrunment error which affect our reading as like which is the values of voltage source isnot same as the values of voltage sources that we set up which is 10V.It should be same butbecause there are a few resistance in connecting wire that affect the final values ofVoltage.Other error is when calculated phase angle using formula which is measuringerror,when we calculated using formula we get many values and need to round off the values tominimum 2 decimal places to easy calculated.11
  12. 12. CONCLUSION In the conclusion,we can conclude that the value of phase angle between and aredepend on the value of capacitor.If the value of of capacitor is high, than the values of VR will besmaller than the Vc and if the value of capacitor is low,So the value of VR is bigger than theVc.The reason this happen because high value of capacitor will hard to charging energy thanlow value capacitor which is easy charging energyThis also happen in fixed resistor.It ispossible to phase angle of 0° and 90°.First ,we need to adjust the values of capacitor and theresistor to get the value of phase angle needed.The use the Formula in Equation 3.3, Equation3.4 and Equation 3.5 and find the value of the , by fixing the phase value of 0°and 90°.Then,measure the value of capacitor and value of the resistor needed.Afterthat,Construct Circuit as in Figure 7.8 to check and measure the value of ,whether it is the same as the value of the calculation as same as in this experiment.12
  13. 13. REFERENCES Rusnani Ariffin & Mohd Aminudin Murad (2011).Laboratory Manual Electrical Engineering Laboratory 1:University Publication Center (UPENA). Charles K.Alexander & Mattgew N.O.Sadiku (2009).Fundamentals Of Electrical Circuit Fourth Edition:McGRAW HILL.
  14. 14. 14