QUESTIONS:
PART A
(10 Marks)
Preamble
Electrical engineering
is a professional engineering discipline that deals with the study and
application of electricity, electronics and electromagnetism. Electrical
engineering deals with power and energy systems. Energy is
transmitted from one place to another, and power is converted from one
form to another in these systems. The two types of supply available
is AC and DC supplies. AC supply is more widely used rather than DC supply due
to its advantages. In this context, debate on the topic:
“AC supply system is more superior to DC supply system
for domestic wiring.’’
Your debate
should address the following:
A1.1 Differentiate between AC supply and DC supply system.
A1.2 Discuss the preferred type of supply system
for domestic wiring.
A1.3 Discuss the types of AC supply system available.
A1.4 Justification of your stance with conclusion.
PART B
(15 Marks)
B1
In the circuit shown in Figure 1,
determine the current through the 2 kΩ resistor using the following methods:
B1.1 Mesh analysis
B1.2 Superposition
theorem
Figure 1
Note: For values of ‘x’ contact course leader
B2
A
building has
·
12 light points of 60 watts each used 5
hours a day.
·
4 fan points of 75 watts each running
10 hours a day.
·
Plug point for a 750 watts heater used
one hour a day.
·
One radio 80 watts used 6 hours a day
and
·
A ½ H.P pump of 80% efficiency running
2 hours a day.
Compute:
a.
the total connected
load in kilowatts
b. maximum possible current
c.
the daily
consumption of energy
d. monthly bill
The supply is at
‘Y’ volts and energy cost is according to Karnataka electricity board. The rent
for a meter is Rs. 20 per month. Assume the month of 30 days.
Note: For values
of ‘Y’ please contact course leader.
MARKING SCHEME
ANSWERS:
PART A (10 Marks)
“AC supply system is more
superior to DC supply system for domestic wiring.’’
AC:- In
alternating current, the electric charges flow changes its direction
periodically. AC is the most
commonly used and most preferred electric power for household equipment, office,
and buildings, etc. It was first tested, based on the principles of Michael
Faraday in 1832 using a Dynamo
Electric Generator.Alternating current can be identified in waveform called a
sine wave, in other words, it can be said as the curved line. These curved
lines represent electric cycles and are
measured per second. The measurement is read as Hertz or Hz. AC is used to
powerhouses and buildings etc because generating and transporting AC across
long distances is relatively easy. AC
is capable of powering electric motors which are used on refrigerators, washing
machine, etc
DC:-Unlike
alternating current, the flow of direct current does not change periodically.
The current
electricity flows in a single direction in a
steady voltage. The major use of DC is to supply
power for electrical devices and
also to charge batteries. Example: mobile phone batteries, flashlights,
flat-screen television, and electric vehicles. DC has the combination of
plus and
minus sign, a dotted line or a straight line.Everything that runs on a
battery and uses an AC adapter while plugging into a wall or uses a USB cable
for power relies on DC. Examples would be cellphones, electric vehicles,
flashlights, flat-screen TVs.
Differentiate between AC and DC:-
1. Ac
current changes its direction during flow while the DC current does not changes
its direction during flow and remains constant.
2. The
AC current has a frequency that shows how many times the direction of current
flow changes during flow while the frequency of the direct current is zero as it do not change the direction of flow.
3. The
power factor of AC is 0 to 1 While DC is Constant Zero.
4. The AC current is generated by the alternator while DC current
is generated by Photovoltaic cells, generators and batteries.
5. The
AC load can be capacitive, inductive or resistive but the load on DC is always resistive.
6. The
DC current graph has a constant line showing magnitude and direction is
constant while the AC current can be a sinusoidal wave, square
wave or triangular wave.
7. The AC converted into DC using a device
named rectifier while
the DC converted into AC named
inverter.
Preferred type of supply system for domestic wiring:- With increase in
voltage,line current is reduced for transmitting same power to long distance.
Hence changing the voltage level
is required for lossless transmission.AC transmission system
has the property of
transforming the level of voltage and current without change in frequency with
the help of transformer.Plant cost
for AC transmission is much lower than the equivalent DC transmission.
DC machine has brush, commutator arrangement which require more
maintenance due to its armature reaction effects.While
interrupting any large fault in the system,
the arc quenching medium in the circuit breaker puffs the arc while the sine wave current is naturally comes to zero. This makes the
interruption process easier. In DC
circuit breaker this facility is not available. External circuit is required to do the same.If we use dc transmission, rectification and inversion is required at
source and sink respectively which is a costly affair.Hence, AC is preferred
over DC for intermidoiate length transmission
Ac power supplies:-
AC adapter:-
An AC adapter is a power supply built into an AC mains power plug. AC
adapters are also known by various
other names such as "plug pack" or "plug-in adapter", or by
slang terms such as "wall wart".
AC adapters typically have a single AC or DC output that is conveyed over a
hardwired cable to a connector, but some adapters have multiple outputs that
may be conveyed over one or more
cables. "Universal" AC adapters have interchangeable input connectors
to accommodate different AC mains voltages.
Programmable power supply:-
A programmable power supply (PPS) is one
that allows remote control of its operation through an analog input or digital
interface such as RS232 or GPIB. Controlled properties may include voltage, current,
and in the case of AC output power supplies, frequency. They are used in a wide
variety of applications, including automated equipment testing, crystal growth monitoring, semiconductor fabrication, and x-ray generators.
Bipolar power supply:-
A bipolar
power supply operates in all four quadrants of the voltage/current Cartesian plane, meaning that it will generate positive and negative voltages
and currents as required to maintain regulation.When its output is controlled
by a low-level analog signal, it is effectively a
low-bandwidth
operational amplifier with high
output power and seamless zero-crossings. This type of power supply is commonly
used to power magnetic devices in scientific applications
High-voltage power supply:-
A high-voltage power supply is one that
outputs hundreds or thousands of volts. A special output connector is used that
prevents arcing, insulation breakdown and accidental human contact.
Federal Standard connectors are typically used for applications above 20 kV,
though other types of connectors (e.g., SHV connector) may be
used at lower voltages. Some
high-voltage
power supplies provide an analog input or digital communication interface that
can be used to control the output voltage. High-voltage power supplies are
commonly used to accelerate and manipulate electron and ion beams in equipment
such as x-ray generators, electron microscopes, and focused ion beam columns, and in a variety of other applications, including electrophoresis and electrostatics.
Conclusion:-
AC motors
are generally considered to be more powerful than DC motors because they can
generate higher torque by using a more powerful current.
PART B:
B1.1
Mesh analysis
Solution:-
At mesh--- 1
24-4000I1-4000(I1-I2)=0
24-4000I1-4000I1+4000I2=0
24-8000I1+4000I2=0
8000I1-4000I2=24------------------------------------- 1
At mesh -- 2
-4000(I2-I1)-2000I2-4000(I2-I3)=0
-4000I2+4000I1-2000I2-4000I2+4000I3=0
-8000I2+4000I1-2000I2+4000I3=0
10000I2-4000I1-4000I3=0------------------------------ 2
At mesh--- 3
-4000(I3-I2)-4000I3-24=0
-4000I3+4000I2-4000I3-24=0
8000I3-4000I2=-24 3
From equation 1,2 and 3
8000I1-4000I2+0I3=24
4000I1-10000I2+4000I3=0
0I1+4000I2-8000I3=24
I1=3mA
I2=0mA
I3= -3mA
Current through 2KΩ = 0mA
B1.2
Superposition theorem:
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