FACULTY OF ELECTRICAL ENGINEERING
ELECTRICAL ENGINEERING LABORATORY 1
Lab Report No. 6 Experiment No. 7
Title Series Resistor-Capacitor Circuit
Date Performed 15 / 8 / 2013
Date Due 22 / 8 / 2013
Date submitted 22 / 8 / 2013
Working days late: ____________ equates to ____________% reduction at 5% per day
NAME UiTM NO. GROUP
MOHD NORHAIDIE BIN ROSLI 2012605084 EE1113J
NUR YASMINE BINTI ABDUL SAMAT 2012218936 EE1113J
NURUL NUR BINTI NOR AZMI 2012235954 EE1113J
Report Format / 5
Introduction / Theory / 15
Results / 30
Discussion / Questions / 25
Conclusion / 20
References / 5
Total Marks / 100
Final Marks after Penalty / 100
Lecturers Name ROSZITAIDA BINTI ADZEMIN
Report submission slip (Student’s copy)
Students: 1. Mohd Norhaidie Bin Rosli
2. Nur Yasmine Binti Abdul Samat
3. Nurul Nur Binti Nor Azmi
Expt No. & Title: 7 & Series Resistor-Capacitor Circuit
(Signature & Stamp)
TABLE OF CONTENTS
List of Requirements 3
Questions / Discussion 10
SERIES RESISTOR-CAPACITOR CIRCUIT
The main purposes of this experiment are:
1. To understand the relationship between voltage, current and phase angle.
2. To calculate the phase angle.
LIST OF REQUIREMENTS
The equipments that are used in this experiment are listed as below:
1. Signal generator
The components that are used in this experiment are listed as below:
1. Decade capacitor box
2. 1 kΩ resistor
Figure 7.1 shows a resistor and a capacitor connected in series supplied with an alternating
sine wave voltage. Let the current flowing in the circuit given by the equation 7.1.
Using Ohm’s Law the voltage across the resistor is
From equations 7.1 and 7.4, as and , it can be seen that the
voltage across R and current flow through it are in phase (same phase). The shape of and
waveforms are as depicted in Figure 7.2.
If the voltage and current equations are written in phasor notation, then VR = VR < 0° where
the magnitudes of VR and I are root mean square (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 capacitor VC can be derived from
XC is known as the capacitive reactance.
From Equation 7.1 and 7.11 it can be seen that the voltage across 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. Figure 7.4 and 7.5 show the waveforms and the phasor diagram for C
The waveforms and the complete phasor diagram for the circuit in Figure 1 are illustrated in
Figure 7.6 and Figure 7.7.
Figure 7.6 is obtained by combining Figure 7.2 and Figure 7.4 while combination of Figure
7.3 and Figure 7.5 results in phasor diagram shown in Figure 7.7. The angle between the
voltage and the current is called phase angle. In Figure 7.7 this angle is denoted by θ. From
the figure, the magnitude of the voltage is given by:
The phase angle is given by (7.17)
For a.c. 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 . Thus, the impedance for the circuit of Figure 7.1 is
With phase angle of (7.20)
In phasor notation, this is written as
1. The circuit as in Figure 7.8 is connected.
2. The V and VR are obtained on the oscilloscope. The waveforms is drawn and labelled
completely. The peak values of V and VR are 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. Determine the distance between the
waves (d1) and the distance for one cycle (X1).
3. The position of resistor and capacitor is changed and the V and VC waveforms are
obtained. The distance between the waves (d2) and the distance for one cycle (X2)
4. All the readings are recorded in Table 7.1.
1) From this experiment, we have to understand the relationship between
voltages, current and phase angle. We have to understand the relationship
from a simple circuit that consist only voltage source, resistor and capacitor
respectively. The voltage supply is in alternating sine wave. For the component
that we use is decade capacitor box and 1 kΩ resistor. The value of voltage
across the resistor can be determined by using Ohm’s Law.
But from this experiment, we use equipment to measure the voltage across the
resistor and the capacitor. The value of output voltage will be show at
oscilloscope. Oscilloscope will be the equipment to measure the voltage. The
oscilloscope will be use two probe. First probe will use to measure the voltage
across resistor and the second probe will use to measure the voltage across
capacitor. We determine the two voltage value and the phase different by this
two probe us using.
Based on the result obtain, the value that we are measure is have slightly
different with the value that we calculated. There must be having a faulty in a
component, the equipment or error when doing this experiment. The value of
VC of C1 and C2 is nearly approximate. It is because the value of capacitor C1 is
bigger than C2. The value of calculated phase angle and measure phase angle
is also slightly different. The value of phase angle is depending on the value of
voltage that we measure. Phase angle have relationship with the voltage. If the
value of voltage is changing, then the phase angle also will change. It also
depends on the value of distance between the waves and distance for one
2) The factor that contributes to errors that we observe in the readings which are
parallax error and the value is not approximate because the analog function
3) For the parallax error, we must see the scale with eyes perpendicular to the
scale. We must make sure that ours is directly perpendicular to the scale. For
the analog function generator, we just have to change the analog function
generator to digital function generator.
1. For any value of R and C, can we determine the phase angle between VR and VC?
What are the values of these angles?
Yes, we can actually determine the phase angle between VR and VC by using this
In order to gain the value of VR and VC, we need to find the value of R and C first by
using these formulae:
Hence, the answer is 85.58° for C1 and 87.33° for C2.
2. What will happen to the phase angle if the value of C: a) increases and b)
decreases? What happens instead if the value of C is fixed and the value of R is
a) If the value of C increases, then the phase angle becomes smaller.
b) If the value of C decreases, then the phase angle becomes larger.
However, if the value of C fixed and the value of R is changed, for example the value
of R decrease, then the phase angle becomes larger while the increased value of R
will make the phase angle becomes smaller.
3. Based on Figure 7.1, given that the R and C values can be changed, it is possible to
make the phase angle: a) 0° and b) 90°?
To calculate current in the above circuit, we first need to give a phase angle
reference for the voltage source, which is generally assumed to be zero. The total
impedance (resistance) of this circuit is the contribution from both the capacitor and
resistor. Any resistance and any reactance, separately or in combination
(series/parallel), can be and should be represented as single impedance. As with the
purely capacitive circuit, the current wave is leading the voltage wave.
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