FACULTY OF ELETRICAL ENGINEERING

UNIVERSITY TEKNOLOGI MARA

ELECTRICAL ENGINEERING LABORATORY 1

(EEE230)

EXPERIMENT 7

SERIES RESISTOR-CAPASITOR CIRCUIT

OBJECTIVE

1. To understand the relationship between voltage, current and phase angle.

2. To calculate the phase angle.

LIST OF REQUIREMENTS

EQUIPMENT

1. Signal generator

2. Oscilloscope

3. Multimeter

COMPONENT

1. Decade capacitor box

2. Decade resistor box

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THEORY

In the DC analysis of resistor circuits we examined how to calculate the total circuit resistance of

series components. In this section we will use this approach to analyse circuits containing series

resistors and capacitors. To do this we use the capacitative reactance as the effective

'resistance' of the capacitor and then proceed in a similar manner to before.

 

Analysing Series RC Circuits

You will recall that a series circuit provides only one route for the current to flow between two

points in a circuit, so for example the diagram below shows a resistor in series with a capacitor

between 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/(2f 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/(2f C)

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Figure 1.1 shows a circuit which resistor and capacitor is in series with alternating sine wave

voltage source. The current flowing in the circuit given by the equation 1.3

I = Im sin ωt(Equation 1.3)

Figure 1.1

Using 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 =VmrR

(Equation 1.8)

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From equations 1.3 and 1.6 as I = Im sin ωt and Vr=Vmr sin ωt , We can observe the votage

across resistor and current flow through it are in phase or we can call same phase.We can see

the 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 the

magnitudes of VR and I are root mean squre (rms) values. The phasor diagrams for VR and I are

as in figure 7.3 which shows that they are in phase.

The voltage across the capacitor VC can be derived from

VC =1C

∫ idt(Equation 1.9)

VC =1C

∫ Im sin ωtdt(Equation 2.0)

VC = ℑωC cos ωt(Equation 2.1)

VC = ℑωC sin (ωt-90°)(Equation 2.2)

Therefore,

VC = Vmc sin (ωt-90°)(Equation 2.3)

Vmc = ℑωC = ImXC(Equation 2.4)

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Where

XC is known as the capacitive reactance.

From Equations 1.3 and 2.3.We can see the voltage across the capacitor lags the current by

90°. 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 7.4

and Figure 7.5 .

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The waveforms and the complete phasor diagram for the circuit in figure 1 are illustrated in

figure 7.6 and figure 7.7

By combining figure 7.2 and figure 7.4,We can see the diagram as in Figure 7.6 while

combination 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 angle

unit is θ. From the figure , the magnitude of the voltage is given by:

V =√V R ²+V C ² (Equation 2.5)

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V =√R ²+X c ²(Equation 2.6)

V R= IR(Equation 2.7)

Where

V c=I Xc (Equation 2.8)

And the phase angle is given by

θ= tan−1V cV R

(Equation 2.9)

θ= tan−1XCR

(Equation 3.0)

For AC. circuits, the resistance is termed impedance and is given the symbol Z. the magnitude

of impedance of a circuit containing a series combination of a resistor and a capacitor is given

by √R ²+XC ². Thus, the impedance for the circuit of figure 7.1 is

With phase angle of θ = tan¯¹ XCR

(Equation 3.1)

In phasor notation, this is written as Z = R- jXC (Equation 3.2)

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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

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Figure 7.9

RESULTS

C1 = 0.187 μF C2 = 0.085 μf

V 9.98Vp-p 10 Vp-p

VR 7.52Vp-p 4.80Vp-p

VC 6.40Vp-p 8.80Vp-p

d1 130μs 190μs

X1 1.000ms 1.000ms

d2 150.0μs 100μs

X2 1.000ms 1.000ms

Table 7.1

For each value of capacitor used, determined the phase angle using the following formulae:

θ = tan¯¹ V CV R

(Equation 3.3)

θ =( d1X1

) × 360°(Equation 3.4)

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θ = 90 - ( d2X2

) × 360°(Equation 3.5)

5. All values were recorded in table 7.2

C1 = 0.187 μF C2 = 0.085μF

Calculated 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 7.2

DISCUSSION

Before we start our experiment,we need make sure that our equipment is in good

condition.First we test our connecting wires by using multimeter.connecting wire are important ,if

not it can affect our Values in the end of experiment.We will heard a sound from multimeter if

the Connecting wires in good condition.Then we test our oscilloscope and Signal generator .The

values that we set up in signal generator must be same as in oscilloscope.So that means that

the equipment is good connection.The Decade capacitor box and Decade Resistor box was

testing by using multimeter .The values that we set on decade capacitor box or Decade

Resistor box must be same on multimeter.We start our experiment as our equipments and

components are in good condition.

In this experiment ,we use same resistor but different capacitor. The voltage peak-peak

of 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 the

distance for one cycle(X1). The distance between the waves (d2) and the distance for one

cycle(X2) are for different position of capacitor and resistor circuit.We use two capacitor in this

experiment which are 0.187μF capacitor and 0.085μF capacitor.We use two capacitor to

compare the phase angle which obtained in the end of experiment.

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From the result in the Table 7.1,The values of in circuit in circuit with has different

position of capacitor and resistor as in Figure 7.8 and Figure 7.9 are different.As in Figure

7.8 ,capacitor is in lead ,then the resistor this mean the capacitor will received the voltage.As we

know that capacitor function is stores charge and then quickly release most of this charge when

it is needed.So the values of VR are higher than Vc.As in Figure 7.9 ,The resistor received the

voltage supply first .So the capacitor only received the Voltage supply that was filtered by the

resistor.As we know Voltage total in series must be same As sum of Voltage across in each

component but in this experiment ,the values of sum of voltage across component is higher than

the Voltage source.This is because of the function of capacitor.In this experiment we also use

two different values of capacitor which are 0.187μF capacitor and 0.085μF capacitor to know

the 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 circuit

which use capacitor 0.085μF but it is between 0® and -90®.As we know we cannot measured

directly the phase angle but using mathematical way as in Equation 3.3, Equation 3.4 and

Equation 3.5.

While we are doing the experiment,there are some error that we detected.One of them

are instrunment error which affect our reading as like which is the values of voltage source is

not same as the values of voltage sources that we set up which is 10V.It should be same but

because there are a few resistance in connecting wire that affect the final values of

Voltage.Other error is when calculated phase angle using formula which is measuring

error,when we calculated using formula we get many values and need to round off the values to

minimum 2 decimal places to easy calculated.

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CONCLUSION

In the conclusion,we can conclude that the value of phase angle between V Rand V C are

depend on the value of capacitor.If the value of of capacitor is high, than the values of VR will be

smaller than the Vc and if the value of capacitor is low,So the value of VR is bigger than the

Vc.The reason this happen because high value of capacitor will hard to charging energy than

low value capacitor which is easy charging energyThis also happen in fixed resistor.It is

possible to phase angle of 0° and 90°.First ,we need to adjust the values of capacitor and the

resistor to get the value of phase angle needed.The use the Formula in Equation 3.3, Equation

3.4 and Equation 3.5 and find the value of the X1 , X2 ,D 1 , D2 ,V R ,V Cby fixing the phase value

of 0° and 90°.Then,measure the value of capacitor and value of the resistor needed.After

that,Construct Circuit as in Figure 7.8 to check and measure the value of X1 , X2 ,D 1 , D2 ,

V R ,V C whether it is the same as the value of the calculation as same as in this experiment.

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REFERENCES

Rusnani Ariffin & Mohd Aminudin Murad (2011).Laboratory Manual Electrical

Engineering Laboratory 1:University Publication Center (UPENA).

http://www.allaboutcircuits.com/vol_2/chpt_4/3.html

http://www.ceb.cam.ac.uk/pages/resistors-and-capacitors-in-series.html

http://www.ycars.org/EFRA/Module%20A/AC%20Circuit%20Analysis%204.htm

Charles K.Alexander & Mattgew N.O.Sadiku (2009).Fundamentals Of Electrical Circuit

Fourth Edition:McGRAW HILL.

http://www.kean.edu/~asetoode/home/tech1504/circuit1/srcc.htm

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