Electrical Engineering: Physics Electricity Homework Problems

This document presents solutions to two physics electricity assignments. The first assignment involves calculating total current, output power, resistance, and voltage drops in a circuit. It includes detailed calculations for power loss and current flow through different resistors. The second assignment focuses on motor efficiency, calculating the efficiency of two motors connected to a generator, determining voltage drops, and calculating total energy consumption and cost. It also involves the determination of current and the number of electrons. The document utilizes formulas and step-by-step methods to solve the problems, offering a comprehensive understanding of electrical concepts. A bibliography of cited sources is also included.

Physic Electricity 1
PHYSIC ELECTRICITY
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Physic Electricity 2
Assignment # 1
1.
Vt= 120 V
Pt= P1= 1200Watts
Ƞ= 0.9
a) From P= VI
I= P
V = 1200
120 = 10 A
Total current =10 A
b) Total output power is given by the following equations
pOUT = Pt×Ƞ
pOUT = 1200× 90
100
pOUT = 108000/100
pOUT = 1080 watts
c) Resistance in each conductor
Power loss = 1200 -1080
Power loss= 120 Watts
And from Rc= Vt
Ploss
Rc= 120
120 = 1Ω
Therefore for each conductor the resistance will be ½ = 0.5 Ω
2. When Rc = 0 Ω
a)
P= V 2
R

Physic Electricity 3
R= V 2
P = 1202
1200
R= 14400
120
R= 12 Ω
b) The total resistance from the circuit diagram is given by
Rt= R1+R2||R3
Rt= R1+ R 2 R3
R 3+ R 2 = 12
But R1=R2=R3
Therefore
R+ R2
2 R = 12
2R2+R2= 24 R
R= 8 Ω
Therefore R1=R2=R3 = 8 Ω
c) V1
Current through R1 is 10 A
Therefore from V= IR
V= 10 × 8
V1= 80 Volts
d) Total resistance for R2 and R3 can be given by the following equation
1
R 23 = 1
R 2 + 1
3

Physic Electricity 4
1
R 23 = 1
8 + 1
8
R23 = 4 Ω
Therefore V23=IR23
V23= 4×10
V23= 40 Volts
But since R2=R3 therefore current flowing through R2= current flowing through R3
Thus
V2 = V 23
2 = 40
2
V2= 20 Volts
e)
V3 = V 23
2 = 40
2
V3= 20 Volts
f) Q= It
I1= 10 A
t= 0.25× 60
t=15 seconds
Q= 15×10
Q= 150 C
g) Number of electrons through R3 for 0.25 Minutes
Q= It
T=15 seconds

Physic Electricity 5
I3= 5A
Q= 5× 15
Q= 75 C
Number of electrons= Total charge
Charge of one electron
Number of electrons= 75
1.6× 10−19
Number of electrons= 4.6875×1020 electrons
Assignment #2
1.
ȠG ×ȠM1×ȠM2 = 51
100 ×ȠM1×ȠM2 = 51 …………………………………………………………………….1
ȠM1×ȠM2 = 100 ……………………………………………………………………...….2
1000feet of 8A WG = 0.641
And from 1 m = 3.28 feet
100m = 328 feet
If 1000 feet= 0.641Ω
Therefore 328 feet= 0.210248Ω
1000 feet of 10 AWG = 1.020 Ω
1m=3.28 feet
200m= 656 feet
Therefore for 656 feet = 0.66912 Ω
Therefore for M1= 0.20248 Ω
M2=0.66912+ 0.20248
M2= 0.8716Ω

Physic Electricity 6
ȠM2 = 0.8716
1.07408 = 0.8114×100
ȠM2= 81.114 %
2.
ȠM1×ȠM2 = 0.51
ȠM1×0.8114=0.51
ȠM1= 0.6285×100
ȠM1 = 62.85
3. The two motors are connected to the 4 HP generator
Difference in resistance = 0.8716- 0.20248 = 0.66912 Ω
If 0.8716 = 200 Volts
Therefore 0.66912= 0.66192× 200
0.8716
Difference in voltage= 153.53 Volts
V1= 200+153.53+
V1= 353.5 Volts
4. Current for M1
P=VI
And 4 HP = 2980 watts
I = 2980
353.5 = 8.42
I=8.42 A
5.

Physic Electricity 7
1000feet of 8A WG = 0.641
And from 1 m = 3.28 feet
100m = 328 feet
If 1000 feet= 0.641Ω
Therefore 328 feet= 0.210248Ω
1000 feet of 10 AWG = 1.020 Ω
1m=3.28 feet
200m= 656 feet
Therefore for 656 feet = 0.66912 Ω
Therefore for M1= 0.20248 Ω
M2=0.66912+ 0.20248
M2= 0.8716Ω
Therefore total resistance for the two motors are = 0.8716+ 0.20248
Total resistance = 1.07408Ω
6.
P=I2R
P= 2980 watts
I2= 2980
1.0764
I2= √2768.487
I= 52.6 A
7.
Power lost = 100−51
100 × 2980 watts
Power lost = 1.4602kW
Total energy = 1.4602×8×5×52
Total annual energy = 3037.216kWh
Total annual cost = 5 × 3037.216 dollars

Physic Electricity 8
8.
VDROP= IR
VDROP= 1.0764×52.6
VDROP= 56.61864 Volts
Voltage at the generator
= 56.186+ 200+353.5
Voltage at the generator= 609.684
9.
Q= It
T=120 seconds
I= 20A
Q= 120× 20
Q= 2400 C
Number of electrons= Total charge
Charge of one electron
Number of electrons= 2400
1.6× 10−19
Number of electrons= 1.5×1022 electrons
10.

Physic Electricity 9
Bibliography
Bird, J., 2010.
Electrical and electronics engineeringf principles. 2nd ed. Chicago: CRC.
Tharaja, B., 2012.
Textbook of electrical and electrical engineering. 2nd ed. Hull: Springer .