TASK 01:
Solving the circuit using Kirchhoff’s Voltage Law;
By solving equations 1 and 2 simultaneously:
I1 = 3.44mA
I2 = 0.764mA
I3 = 02.68mA
Now calculating the current and voltage across resistors;
Since R1 is present in loop 1 of I1, so current across R1 is;
I1 = 3.44mA
V1 = (3.44m)(3000)
V1 = 10.32V
Since R2 is present in between loop 1 and 2 of I1 and I2, so current across R2 is;
= 3.44mA-0.764mA
= 2.676 mA
Voltage V2 can be calculated as follows;
V2 = 2.676 m x 1000
V2 = 2.676 V
Now current across R3 = I2
I2 = 0.764 mA
V3 = 0.764 m x 3500
V3 = 2.67V
Therefore;
VR1 = 10.32 V
VR2= 2.676 V
VR3= 2.66 V
Now,
Total power dissipation PT = PR1 + PR2 + PR3
PT = I12R1 + I22R2 + I32 R3
PT = (3.44×10-3)(3000) + (2.68×10-3)2(1000) + (0.76×10-3)2(3500K)
PT = 0.045W
Value of Rx such that Pt = 2.5 kW
For this, we find the equivalent circuit.
Solve 16 Ω and 12 Ω resistances in parallel combination:
 1/16 + 1/12 = 6.8517 Ω
Now solve 40 Ω and Rx resistances in parallel combination:
 (Rx + 40) / 40Rx
Therefore:
Req = 6.8517 + (40Rx / 40 + Rx)
Req = (274.068 +46.8517Rx) / 40 + Rx
Now;
Pt = V2 / Req
2500 = 2502 / ((274.068 +46.8517Rx) / 40 + Rx)
 Rx = 33.2 Ω
Now apply KVL:
Loop 1:
-260V + 12(I1-I2) + 33.22(I1-I3) = 250
12I1-12I2+33.22I1-33.22I3 = 250
45.22I1-12I2-33.22I3=250 -------------- (1)
Loop 2:
12(I2-I1)+16I2 = 0
-12I1+28I2 = 0 ---------- (2)
Loop 3:
33.22(I3-I2)+40I3 = 0
33.22I1+73.22I3 = 0 ------------ (3)
Using Thevenin rules:
VTh= -VB + VA =VA - VB
Superposition Principle:
Short circuit source V2 and calculate the parameters as shown:
Apply KVL:
Loop 1:
12I1 + 6 (I1 – I2) = 84V
18I1 – 6I2 = 84V -------- (1)
Loop 2:
6(I2 – I1) + 3I2 = 0
-6I1 + 9I2 = 0 ------------- (2)
Solving equation 1 and 2:
I1 = 6 A
I2 = 4 A
Now Short circuit source one and calculate the parameters accordingly:
Apply KVL:
Loop 1:
12I1 + 6 (I1 – I2) = 0
18I1 – 6I2 = 0 -------- (1)
Loop 2:
6(I2 – I1) + 3I2 = 21
-6I1 + 9I2 = 21 ------------- (2)
Solving equation 1 and 2:
I1 = -1 A
I2 = -3 A
I1T = 6-1 = 5 A
I2T = 4-3 = 1 A
Software simulation:
As we can see, both answer, the practical values and theoretical values, are same
TASK 02:
Given:
L = 80mH,
R = 200 ohm,
Now calculate voltage parameters:
VR = IR = (40x10-3) (200) = 8V
VL = I XL = (40x10-3) (25.1) = 1.0052V
Calculate the impedance:
VL = I XL = ( 40x10-3) ( 25.1) = 1.0052V
Xc = 1/ 2πfC
Xc = 1/ 2x3.14xfC = 144.69 ohm
C=22 uF,
f = 50Hz,
I = 40mA
VL = (40x10-3) ( 144.69) = 5.7876
Z = R + j(25.13 – 144.69) = 200 +j(25.13 – 144.69)
Z = 200-119.56j = 233< -30.87o ohm
Now;
Z= R +j(XL – XC)
|Z| = √(200)2 + (25.13 − 144.69)2
XL = 2x 3.14x fL = 25.13 ohm
Therefore;
Supply voltage: VS = IZ = 40x10-3 (233) = 9.32V
And Phase angle:
XL < Xc so circuit is capacitive current leads voltage by ΦD
ΦD = ΦL – ΦV = 30.87- 0 = 30.870
Phasor Diagram:
Calculate the current in the Coil
XL= 2 π f L
= 2 π ( 5x103) (0.3) = 9425 ohm
XC = 1/ 2 π f C = 1 / 2 π ( 5x103) (0.06x10-6) = 530.52 ohm
Now calculate the equivalent impedance:
Zeq = (400+j9424.78)||(-j30.5)
Zeq = 11.8362+j556.818
Zeq = 556.95< -88.780Ω
I =V/Z
I = 40/556.95<88.780
I = 0.0718<88.780A
Current in coil:
IL = (-530.5x0.0718<88.78)/(4000 + 8894.28j)
IL = 3.906x10-3 < -67.05A
Current in capacitor:
IC = I-IL = 0.0753 < 900 A
Total Current:
I = I RL + IC = ( 0.0039,67.9) + ( 0.0754<90)
I = 0.07539 < 88.990
Now:
ZRL = 4000 + 9425j
IRL =V/ ZRL = (40<00) / (10238< -67.90)
= 0.0039< -67.90A
Phasor Diagram:
Now calculate Power Consumed by the system:
 P = I2 R =I2 Z
P = (0.0178<88.78)2 (556.95<-88.780)
P = 0.160<88.780
Calculating capacitance of capacitor:
Supply current is minimum at resonance:
XL = XC
2πfL= 1/2πfC
C = 1/(2πf)(2πfL)
C = 1/(2πx10x103)( 2πx10x103x0.1)
C = 2.53nF
Now since;
Z=R
Z = 1000Ω
Q factor = wRC
Q = RC x 1/√𝐿𝐶
Q = 1000x2.533x10-9 x 1/√(0.1 ∗ 2.533 ∗ 10^ − 9 )
Q = 0.1591
Here,
XL =2πfL = 2π x 10000 x 0.1
XL = 6283.18Ω
and XC =1/2πfC
Xc = 1/2π x 10000 x 2.533 x 10-9
XC = 6283.2587Ω
Zeq = 39975.6<-9.0430
Now find current
I =V/Z = 2/39975.6
I = 5x10-5 < 9.043A
Capacitor Current IC = ((1000+j6283.185)/1000 )x 5x10-5 < 9.043A
IC = 3.1811x 10-4< 900A
Inductor Current IL = ((-j6283.185)/1000 )x 5x10-5 < 9.043A
IL = 3.14159< -80.9570A
Calculate Bandwidth
I= IC = 5x10-5 < 9.043A
Bandwidth = Wo /Q
= 62832.22/0.15915
= 394798.743 rad/s
TASK 03:
3.A
Diodes shield circuits by limiting voltage. They are capable of converting AC to DC. Silicon
and germanium semiconductors are used to increase the performance of the diodes. They both
transport electricity, but in distinct ways, even though they do so in the same direction.
Different types of diodes exist, and each type has a particular application. A common diode
symbol is shown, as can be seen above. We can observe that the anode and cathode terminals
are two terminals in the preceding schematic. The arrowhead serves as the anode and represents
the normal current flow direction under the forward biassed condition. On the other end is the
cathode.
A diode functions as a PN junction device, through which current flows, when the appropriate
forward potential is offered. The mixture of the p and n-type semiconductor materials must
have a controlled proportion of donor and acceptor impurities. In its simplest form, pentavalent
and trivalent impurities are doped onto the two sides of a single silicon or germanium wafer.
The P area displays trivalent impurity doping, whereas the n region displays pentavalent
impurity doping. Another choice is to make the semiconductor device just by combining
distinct p and n-type materials.
3.B
Vs
0
0.5
1
2
4
6
8
10
15
Measured Vd
0
0.0213894
0.388996
1.33083
3.28483
5.25839
7.23883
9.22269
14.1884
Measured Id
0
9.72246e-005
0.00176816
0.00604922
0.014931
0.0239018
0.0329037
0.0419213
0.0644927
Vs
-5
-10
-15
Measured Vd
-4.2704
-9.22269
-14.1884
Measured Id
0.0194109
0.0419213
0.0644927
Vs
0
2
4
6
8
10
12
14
Measured Vout
Measured Iz
2
4
5.98825
6.37524
6.44144
6.48099
6.51013
-1.502e-009
-1.504e-009
-6.52658e-005
-0.00902647
-0.0197698
-0.0306612
-0.0416104
when
Measurement
Vout
Vin=0
Vout
Vin=10
Ic
Vin=0
Ic
Vin=10
10V
0.41V
9.59mA
0.097mA
3.C
Diode in full wave bridge rectifier:
A bridge rectifier employs a bridge circuit arrangement with four or more diodes to convert
alternating current to direct current. We talk about a "full-wave bridge rectifier" here. With the
help of four carefully arranged diodes, bridge rectifiers may change the incoming alternating
current (AC) from the power source into the more usable direct current (DC). The polarity of
the signal at the circuit's output is unaffected by the direction of the AC input signal. The output
may be seen across the load resistor when an ac signal is applied to terminals a and b. Diodes
D2 and D3 conduct when forward biassed during the first positive half cycle of the alternating
current signal. If the diodes D1 and D4 are biassed in the opposite direction, they will not
conduct together. The load resistor receives current from the supply through the two forwardbiased diodes. The voltage at the output will be positive at terminal d and negative at terminal.
During the AC signal's negative half-cycle, diodes D1 and D4 will be biassed forward, while
diodes D2 and D3 will be biassed in the opposite direction. The cathode of D1 will be at a
negative voltage whereas the anode of D4 will be at a positive voltage in this setup. Remember
that the direction of the current via the load resistor will change from the positive half cycle.
Thus, the signal polarity at the output does not depend on the signal polarity at the input. The
fact that the negative half cycle of the AC signal is now being recorded as a positive voltage at
the output is also a contributing factor.
Zener Diode in voltage stabilizer:
A voltage stabiliser is an electrical component that, regardless of the input voltage, always
delivers the same voltage across its terminals. It protects the gadget from harmful voltage
fluctuations, overvoltage, and undervoltage. A resistor is placed in series with the diode and
the remainder of the circuit to regulate the current flowing through the diode. Firmly fastened
to the d.cpositive lead's point. In the event of a malfunction, the reverse-biased is meant to
continue functioning. Since a reverse bias exceeding a junction diode's breakdown voltage will
damage it, this type of diode is not employed. Even with the lowest input voltage and the
highest load current, the Zener diode current must remain small.
When the input voltage and required output value of a Zener diode are both known, it is simpler
to select a diode whose output voltage is roughly equal to the load voltage (VZ = VL).
Bipolar transistor as an amplifier:
Transistors can be used as amplifiers to boost weak signals. Applying a constant DC bias
voltage ensures that the emitter base junction remains in a forward biassed state. In this case,
the signal's polarity makes no difference to the inherent forward bias. Down below, you'll see
a picture of a transistor-based amplifier. Low input resistance means that even a tiny change in
the input signal will have a large impact on the output. Collector and emitter currents from the
input signal add up to a significant voltage drop across the load resistor RL. Consequently, the
transistor's amplification capability is shown by the fact that a little input voltage generates a
huge output voltage.
TASK 04:
Basis of
Difference
Analog Electronics
Digital Electronics
Analyzing systems based on analogue
signals is a focus of the field of electronics
known as analogue electronics.
Analyzing systems based on digital
signals is a focus of the branch of
electronics known as digital
electronics.
Type of
signal used
Analog electronics includes the use of
continuous time (analogue) signals.
Digital electronics uses discrete
time signals or two state signals.
Components
used
In analogue electronics, passive circuit
components like resistors, capacitors, and
others are widely utilised. However,
active components like transistors are also
occasionally used.
Digital electronics only employ
active elements.
Power
consumption
Analog electronics-based systems
consume more energy.
Digital electronics consume
significantly less power.
Power loss
Analog devices suffer from a little amount
of power loss.
Digital equipment does not
experience power loss.
Voltage &
current
Analog electronics uses a comparatively
high voltage and high current compared to
digital electronics.
Extremely low voltages and
currents are necessary for digital
electronics.
Noise &
distortion
Analog electronics have significant levels
of signal distortion and noise.
Digital devices have very little
signal noise and distortion.
Safety
Analog devices do pose certain electrical
safety issues, but they are not very great.
Digital electronics do not pose any
dangers to electrical safety.
The main functions of analogue
electronics are wireless transmission,
rectification, and continuous time signal
amplification.
The main applications of discrete
time signals in digital electronics
include multiplexing, encoding,
decoding, analysis, switching,
mixing, etc.
Definition
Processes
involved
Basis of
Difference
Analog Electronics
Digital Electronics
Used for
Analog electronics' main use is to collect
data from systems.
Using digital electronics facilitates
system data analysis.
Analogue technology is extensively used
in FM radios, TVs, telephones, and other
radio and audio devices.
Digital electronics are used
extensively in computers, data
processing and storage, robotics,
digital watches, and many other
digital devices.
Applications
4.B
GAIN:
Gain is the ratio of output voltage to input voltage of an amplifier,
where VIN1 and VIN2 are two inputs, subtracted.
In a real circuit, the gain will be frequency dependent, but let us start with consideration of
the gain in an ideal amplifier.
BANDWIDTH:
The Bandwidth (BW) of a device is the range between the frequency limitations of the
amplifier.
The range of frequencies that make up a band is referred to as the "bandwidth."
An amplifier, often known as an amp, is a piece of electronics that strengthens signals. It uses
electricity to increase a signal's amplitude at the first input, resulting in an output signal with
constantly growing amplitude.
The range of frequencies or frequency band that an amplifier can magnify most effectively is
represented numerically as a bandwidth. Due to the fact that each frequency has a lower limit
(which could possibly be 0) and an upper limit, amplitude augmentation cannot be adequately
performed for all of the frequencies that an amplifier can magnify.
If an amplifier is designed to magnify, for instance, the frequencies between f1f1 and f2f2, the
bandwidth will equal the difference between these two frequencies. Therefore, the bandwidth
of the amplifier will be equal to the difference between f2f2 and f1f1.
BW=f2−f1
I/O RESISTANCE:
The voltage to current ratio at these terminals determines the input and output impedance. The
load impedance, RL, across the terminals influences the output impedance more than the power
supply employed by the amplifier does.
When Z is connected to power, the amplifier circuit becomes a load for the amplification of
the AC input signals. In a bipolar transistor circuit, an amplifier's input impedance can range
from a few tens of ohms to several thousand ohms (Kilo-ohms k) or even millions of ohms
(Mega-ohms M).
DISTORTION LEVEL:
In an amplifier, distortion essentially describes a shift in the waveform that is received at the
output in comparison to the applied input. The unwanted alterations created during
amplification are referred to as distortion.
In a pure signal, there is always a single frequency component where the voltage varies evenly
in both the positive and the negative directions. In the event that this variation is less than a full
3600 cycles, the signal is deemed to be distorted.
4.C
A NOR gate and a Nand gate are used in the universal gate circuit seen in the diagram. NOR
gates can be used to create any additional straightforward logic gate. The NOT-OR gate,
sometimes referred to as the NOR gate, is the result of combining the two. When all inputs are
low, the NOR gate will produce a high state at its output. In other words, the OR gate delivers
one thing, and this gate provides the exact opposite.
The Boolean equation for a NOR gate is created by multiplying the minus sign by the logical
products of the inputs. (A+B)'=Y
4.D
The logic function of combinational logic gates is identical to that of NOR gates; that is, they
both provide an output of 0 on 01 10 11, and act like NOR gates. Attached to the C code is also
a truth table for the function.
4.E
Data can be transmitted using analogue or digital electric signals. Both methods work by
translating data (such as a movie or song) into electrical signals. In contrast to digital systems,
which use numerical representations of data, analogue ones convert data into amplitudevariable electric pulses. Each "bit" in digital technology represents one of two possible
"amplitudes" in binary code (zero or one). The ability to transform analogue signals into digital
ones is built into many modern gadgets. A couple of the most prominent analogue equipment
are audio speakers and microphones. While analogue technology costs less, it can only transmit
so much data before it becomes unreliable.
Many devices' functions have been entirely transformed by digital technology. Once the data
has been transmitted, it will be decoded from its binary representation. They are flexible
because modifications may be made rapidly. The cost of digital devices tends to be higher than
that of analogue ones.
Analog communications require significantly less bandwidth than digital ones. Changes in
physical phenomena like sound, light, temperature, position, or pressure are more faithfully
represented by analogue signals. This means that the electrical tolerance of analogue
communication systems is more significant.
Equipment and media for VCR video recording are all examples of modern technology that
use analogue signals. Any clock is assumed to be an analogue clock without a digital readout.
Due to advancements in digital technology, gadgets are now more portable, lightweight,
speedy, and versatile than ever before.
Data can be created, stored, or processed using various digital technologies. Social networking
sites, online video games, multimedia presentations, and mobile phones are just a few examples
of the many forms of electronic media and methods of contact available today. The term
"digital learning" is used when discussing education whenever digital instruments are used.
1. F
Communication systems that transmit a continuous stream of speech, data, images, or video
typically use similar signals. The input signal is combined with a carrier signal during analogue
transmission using one of two primary techniques. Many applications continue to collect
signals from analogue sensors and transducers, and radio and audio equipment also extensively
use analogue circuitry before transforming those signals to digital for storage and processing.
The applications for digital audio systems are numerous. Television and film editors can benefit
from digital audio technology. Car stereos, concert sound systems, and studio recordings all
use some form of digital audio technology.