Electrical Engineering Homework Help

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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Electrical Engineering Homework Help