EEE101: Fundamentals of Electrical Circuits - Semester 1

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ELECTRICAL AND
ELECTRONIC
FUNDAMENTALS
1

Table of Contents
Task 1:......................................................................................................................................................3
a............................................................................................................................................................3
b...........................................................................................................................................................3
c............................................................................................................................................................5
d...........................................................................................................................................................5
Task 2:......................................................................................................................................................6
Task 3.......................................................................................................................................................9
a............................................................................................................................................................9
b,..........................................................................................................................................................9
c..........................................................................................................................................................10
Task 4:....................................................................................................................................................11
a..........................................................................................................................................................11
b.........................................................................................................................................................11
c..........................................................................................................................................................11
d.........................................................................................................................................................12
e..........................................................................................................................................................12
f..........................................................................................................................................................13
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Task 1:
a.
The ohms law says that when the current flowing in the circuit and the resistance R is
connected in series so the voltage drop across the circuit is directly proportional to current.
VαI
To remove the sign of proportion a new constant will come in the circuit that is the resistance
R.
V = IR
To plot and determine the I-V curve the current is flowing the given circuit that is as follows:
When the current is flowing the circuit the following data is obtained by the analysis which is
as follows:
Current (I) Voltage (V) Resistance (R)
85 12 0.141
90 14 0.155
95 15 0.157
Hence it can easily be depicted that the graph that is obtained from the above reading is the
straight line.
b.
Kirchhoff Voltage Law: According to this law the total sum of the voltage in the circuit is
zero. The sum will be zero only because of the positive and the negative voltage will cancel
their voltage drop so the total voltage in the circuit will be zero (Gekelman et al, 2018).
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Voltage (V) Voltage (V1) Voltage (V2) Voltage (V3) E = V2 + V3
45 15 10 5.1 15.1
50 25 18 6.9 24.9
Kirchhoff Current Law: According to this when the current is leaving or exit from a circuit
then the overall current of the circuit is zero. In this phenomenon when the circuit current
enters in at the same point of time equal amount of current will leave the circuit that is
equivalent to the circuit.
The data obtained in the lab is as follows:
Voltage (V4) Current (I1) Current (I2) Current (I3) Current (I4)
10 10.5 2.8 7.8 10.6
20 15.6 4.5 11 15.5
Hence it proves that the current I4 is equal to the current I1 i.e. is the submission of I2 and I3.
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c.
The validation of the KCL and KVL have already done in the above question. To calculate
the division in the circuit we can assume the voltage is Vc which is as follows:
According to ohms law:
V = IR
Vc = (Rc/R1 + R2)*V
Voltage (V4) Current (I1) Current (I2) Current (I3) Current (I4)
10 10.5 2.8 7.8 10.6
20 15.6 4.5 11 15.5
To validated KVL the division in the circuit we can assume the voltage is Vv which is as
follows:
Vv = (Rv/R1 + R2)*E
Voltage (V) Voltage (V1) Voltage (V2) Voltage (V3) E = V2 + V3
45 15 10 5.1 15.1
50 25 18 6.9 24.9
d.
Thevenin Theorem: This theorem states that when the voltage across the terminal is been
replaced by a resistance of R ohm. The voltage across the circuit having the terminal A and B
having an equal magnitude of the resistance R.
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Task 2:
a. As some of the data is given in the question which are as follows:
L = 80 mH
R = 200 ohm
C = 22 μF
I = 40 mA
f = 50 Hz
1. Voltage drop in the resistance:
Voltage drop = Vr = IR
Put the values from the above:
V = 40 * 10-3 * 200
V = 8 Volt
2. Voltage drop in the capacitor:
Voltage drop = Vc = IXc
For the calculation of Xc = 1/j2πfC
Xc = -144.5j ohm
Now put the values in the above equation we get:
Vc = 40 * 10-3* (-144.5j)
Vc = -5.8 j Volt
3. Voltage drop in the inductor:
Voltage drop = IXL
For the calculation of XL = j2πfL
After putting the values we get XL = 25.2 j ohm
Not putting the value in the above equation:
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VL= 40 * 10-3* (25.2 j)
VL= 1 j Volt
4. Impedance of the circuit is as follows:
Impedance = Z = R + Xc + XL
Z = 200 – 120 j
5. Supply Voltage is as follows:
V = Voltage drop in the resistance + Voltage drop in the capacitor + Voltage drop in the
inductor
After putting the above calculated values we get Supply voltage which is as follows:
V = 8 – 4.8 j Volt
6. Current in the circuit = I = 40 * 10-3 A
7. Phase Angle is as follows:
φ=cos−1 R
|Z|
For the calculation of |Z| = 233 ohm
On putting the values in the above equation we get
φ = cos−1 200
233
φ = 30. 8 Degree
b.
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Task 3
a.
(i) Diode: This is an electronic device that controls the flow of current in a circuit and
circulates the current where ever needed in the circuit. The operation of the diode is done
over two different modes i.e. reverse and forward biased. When we apply the negative
voltage over the circuit and their no current will flow then there is no current that flows in the
circuit it will behave like an open circuit than the condition is of reverse biased and on the
other hand when the positive voltage applied over the circuit and current will flow from the
circuit and behave like an open circuit when the condition is known as forward biased.
(ii) Zener Diode: These are the diode that performs both the function of forward as well as
the reversely biased function. The operation of this diode is done over the reverse break down
voltage wherein the circuit the current will start conducting, also at this voltage the no
damage will occur. It can perform their function over a wide variety of voltage range. The
application of this Zener diode can be used in a voltage regulator. The wide range of voltage
gives an advantage to this diode that it protect the circuit from damage.
(iii) The transistor as a switch: When transistor will be used as a switch then it protects the
circuit from damage. In the operation of these two different p-n junctions are used. There are
mainly four different areas are performed by this switch i.e. reverse, forward, cut off and
region of saturation. But most of the application is found of cut off and saturation function
(Gekelman et al, 2016).
b,
The most common semiconductor that is found in the semiconductor theory are as follows:
1. FET: This is also known as Field effect transistor. There are mainly three different gates
are found in this semiconductor i.e. gate, drain and source. This semiconductor is mainly used
as a switch. The switch operation is used by the gate terminal and the connection in between
the drain and the source is performed by voltage difference in the circuit. The FET is used in
various application like Chopper, current limiter etc. The type FET are found that are
generally N channel and P channel FET.
2. MOSFET: It is also known as Metal oxide semiconductor field effect transistor. The gate if
the MOESFET are totally insulated and they provide current to the teneraransistor for the
working. The basic difference in between the JFET and MOESFET that the terminal of
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MOSFET is totally insulated with metal oxide where the insulation is not present over the
JFET gate.
3. SCR: This is also known as Silicon controlled rectifier. There is a different name also of
SCR i.e. Thyristor. This SCR is a four-layered device having different PNPN and NPNP
diode present in them. There are total three terminals is present in the SCR i.e. anode, cathode
and gate terminal. They are generally used in the area where are switch are required at a very
high voltage source. Generally, the working of SCR is performed over the three terminals like
OFF mode, ON mode and Reverse biased mode. The ON mode is also known as forwarding
blocking mode whereas OFF mode is known as conducting mode.
4. JFET: This is also known as Junction field effect transistor. This transistor is the most
common type of FET. The current flow through the conductor in the totally managed way so
that the management of current is done with the gate terminal. The direction of current in the
JFET is from source to drain terminal. When the current is reversed in the transistor the
voltage drop increases and it will behave like a cut-off area. The main application of JFET in
current limiter and analogue switch (Pensiaet al, 2018)
c.
They both BJT and FET have their different application that makes them different from each
other. In the BJT the base terminal requires a totally biased current so that the current can
easily flow from emitter to collector. On the other hand, the FET requires high voltage on
their gate terminal so that the current can easily flow from drain to source gate. There is also
one difference between both the transistor that at the output the FET provide high current
gain whereas on the other hand BJT provide high voltage gain. There is also a difference in
between their impedance that FET has high input impedance and on the other hand, BJT has
low input of impedance.
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Task 4:
a.
The electronics are divided into majorly 2 different categories i.e. first can be understood by
analogue and others can be understood by digital electronics. The analogue electronics are the
old dynamic electronics that are used at that time. But on the other hand, the digital
electronics are the electronics that are discovered later and having more advanced features as
compared to others. The main difference between the analogue and digital electronics are as
follows:
1. The analogue electronics have their signals that are continuous in nature while on the
other hand, digital electronics have discrete signals.
2. The digital signals are represented by square waveform while on the other hand the
analogue signal is represented by sin waves.
3. The recording of the analogue signal can be done on a continuous basis while on the
other hand the digital signals the analogues signal are limited and they are processed
in limited number.
4. The analogue signal is noisy in nature while on the other hand, the digital signal is
silent in nature.
b.
The truth table for the given figure is as follows:
A B C
0 0 1
0 1 0
1 0 0
1 1 0
c.
The name of the logic function of the above figure is NOR gate.
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A
B
C

d.
Advantage and Disadvantage of Analogues Electronics are as follows:
1. The analogue signals can be converted into digital signals.
2. The accuracy and resolution of the analogue signal are high.
3. The analogues signals are continuous so they didn’t miss any data from their signals.
4. They are very noisy and prone signals.
5. The noise ratio of the analogue’s signals is very high.
Advantage and Disadvantage of Digital Electronics are as follows:
1. The digital signals have no tendency to be converted into analogue signals.
2. The digital signals make very less noise and prone. The noise ratio is also very low in
these signals.
3. The signals are present in the form of digital values and numerical values, on the other
hand, they don’t have analogue pad information.
4. The signal of digital electronics are discrete in nature so that the signal values can be
disturbed, many time the information gets missed before reaching their source
destination.
e.
Application of Digital Electronics are as follows:
1. They have used encryption and in the processing of the data.
2. The data is transferred in the form of digital data packets, so their packets are used in
many digital communication applications.
3. During the encryption of the data digital signals are used for decoding and encoding
of signals.
4. They are used in the microcontroller, microprocessor etc.
5. The signals that are processed are error free and they are very fast.
Application of Analogue Electronics are as follows:
1. They are used to filter the data of the signals and they are used in high power devices.
2. They are mainly used in the sensors.
3. When the different range of frequency is not used in the signal so they are filtered in
the analogue.
4. They are used in long-range communication because they are continuous in nature.
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