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Electrical Engineering Homework Help
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About Electrical Engineering: Electrical engineering is a
dynamic subject and students of this stream are expected to
learn the latest development in the related field so that
successfully apply their academic knowledge and practical
knowhow in their profession. Electrical engineering
assignments are part of evaluating the talent and grasp of
knowledge of these students hence these assignments come
with different levels of difficulties and complexities.
Online electrical engineering assignment help services are meant to offer students
online support to solve their assignment with best accuracy, authenticity, and within tight
deadline. Professional agencies keep their subject matter expert tutors for offer students on
time service for completion of their assignments.
Sample of Electrical Engineering Homework Illustrations and Solutions:
Question-1: Determine the current flowing through the filament of a lamp having
a constant resistance of 480 and connected across 240 v mains.
Solution:
P.D. across the lamp filament, v = 240 volts
Resistance of lamp filament, R = 480
Current flowing through the filament, I =
=
240
280 = 0.5 A
Question-2: Determine the potential difference to be applied across a conductor
of resistance 12 so that a current of 15 A may flow through it.
Solution:
Resistance off conductor, R = 12
Current flowing through the conductor, I = 15 A
Potential difference required, V = IR = 15 112 = 180 volts.
Question-3: Determine the resistance of 600 meters of a cable at 200 , if the
crossectional area of the cable in 12.5 mm2 and resistivity of copper at 200 C is
1.7 10 8 m.
Solution:
Length of cable, I = 600 meters
Cross-sectional area of cable core, a = 12.5 106 2
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Resistivity of cooper, = 1.7 108 -m
Resistance of cable, R = 1
= 1.7 108
600
12.5 106 = 0.816.
Question 4: A heater element is made of nichrome wire having resistivity equal to
100 m. The diameter of the wire is 0.4 mm. Calculate the length of the wire
requirement to get a resistance of 40.
Solution:
Resistance of nichrome wire, R = 40
Resistivity of nichrome wire, = 100 108 -m
Diameter of wire, d = 0.4m = 4 104m
X-sectional area of nichrome wire, a =
4 2 =
4 4 104 2 = 4 108 2
Length of nichrome wire required, l =
=
40 4 108
100 108 = 5.03 m.
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Question-5: A price of silver wire has a resistance of 1. will be resistance of a
manganin wire of one third the length and one-third the diameter, If the specific
resistance of manganin is 30 times that of silver.
Solution:
Specific resistance of manganin, = 30 times of specific resistance of silver i.e., 30
Length of manganin wire, = one-third of length of silver wire i.e., 1
3 8
Diameter of manganin wire, dm = one-third of diameter of silver wire = 1
3 8
or area of x-section of manganin wire, =
4 ()8 =
4
8
3 1 =
36 8
2 = 1
9 A
Since resistance of any wire, R = 1
Resistance of manganin wire, 8 =
8 = 1
Resistance of manganin wire, = m
=
3088/3
8 = 90
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Question-6: Determine the resistance of the material in terms of volume v. length
1 and resistivity of the material of a conductor of volume 0.05 300 meters and
resistance 0-0306.
Solution:
Cross-sectional area of material of volume units and length l units, a =
Resistance of material R = 1
=
1
/ =
2
from expression (i) a =
Substituting v = 0.05 3, l = 300 m and R = 0.0305 in above expression we get
0.0306 = (300)8
0.05
or = 0.0305 0.05
(300) = 1.74 108
Question-7: Determine (i) length and (ii) diameter of cooper cylinder of volume
v, resistance and specific resistance in terms of v, R and .
Solution:
Volume of cylinder, v =a 1
Resistance of cylinder R = 1
Multiplying expression (i) and (ii) we get
v R = a 1 1
= 2
or 2 =
or l = /
Dividing expression (i) by expression (ii) we get
=
/ =
2
or a =
or
4 2 =
or diameter of cylinder, d = 16
3
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Question-8: One km of wire having diameter of 11.7 mm and of resistance 0.031 is drawn so that
its diameter is 5.0 mm. What does it resistance become ?
[pb Univ. Elements of Electrical Engineering Nov : 1975]
Solution:
Resistance of old wire, 1 = 0.031
Diameter of old wire, 1 = 11.7 mm = 11.7 108 m
Diameter of new wire, 8 = 5.0 mm = 5 108 m
Area of x-section of old conductor, 1 =
4 18 =
4 (5 108)2
Since resistance of a wire, R = 1
=
/
=
2 where is the volume of wire
R 1
2 if wire is of constant volume and of same material
Resistance of new wire, 2 = 1 1
3 2 = 0.031
4(11.7 10)3
22
4(5 10)3
2 = 0.929
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Question-9: Find the resistance of a cable centimeter of cooper (i) when it is drawn into a wire of
diameter 0.032 mm and (ii) when it is hammered into a flat sheet of 1.2 mm thickness, the current
flowing through the sheet from one face to another, specific resistance of cooper is 1.6 -m.
Solution:
Volume of cooper, v = 1 2 = 1 106 8
Specific resistance of cooper, = 1.6 108 m
(i)Area of x-section of cooper wire, a =
4 2 =
4 2 =
4 (0.012 108)8 = 0.80425 109 8
Length of cooper wire, I=
=
1 106
0.80425 109 = 1.243 m
Resistance of cooper wire, R = 1
= 1.6 108
1.243
0.80425 109 = 24,728
(ii) Length of current path, I = 1.2 mm = 1.2 103 m
Area of x-section, a =
=
1 106
1.2 108 =
1
1.2 1032
Resistance of cooper flat sheet, R = 1
=
1.6 108 1.2 103
1
1.2 103
= 2.304 108
Question-10: Determine the resistance of a copper tube having external diameter 8cm ; thickness
5 mm and length 5 meters. It is given that specific resistance of cooper is 1.70 cm.
Solution:
Resistivity of cooper, = 1.7 108 -m
Length of cooper tube, l = 5 meters
External diameter of tube, D = 8 cm = 0.08 meter
Internal diameter of tube, d = D 2 thickness of tube
= 0.08 2 0.005 = 0.07 m
Cross-sectional area of tube, a =
4 2 2 =
4 (2-2) =
4 (0.082 - 0.078) = 0.001178 2
Resistance of cooper tube, R = 1
=
1.7 1085
0.001178 = 72.012 .
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