Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by:
Answers
Answer: hello your question is incomplete attached below is the missing detail
answer :
Complex power = 2.5 ∠ 50° VA
apparent power = 2.5 VA
average power = 1.6 Watts
reactive power = 1.915 Var
power factor = 0.64 ( leading )
Explanation:
i) complex power
P = Vrms * Irms
= 17.67∠40° * 0.1414∠-10°
= 2.5∠50° VA
ii) Apparent power
s = Vrms * Irms
= 17.67 * 0.1414
= 2.5 VA
iii) Average power absorbed
Absorbed power ( p ) = Vrms * Irms * cos∅
= 17.67 * 0.1414 * cos ( 50 )
= 1.6 watt
iv) Reactive power
P = Vrms * Irms * sin∅
= 17.67 * 0.1414 * sin ( 50 )
= 1.915 VAR
v) power factor
P.F = cos ∅ = p /s
= 1.6 watt / 2.5 VA = 0.64.
Related Questions
The National Weather Service has issued an alert for a severe storm that will bring 100 mm of rainfall in one hour. A farmer in the area is trying to decide whether to sand bag the creek that drains the 40 acres of row crops. The soil for the drainage area is a sandy clay loam and has a porosity of 0.398, effective porosity of 0.330, suction pressure of 52.3 cm, a hydraulic conductivity of 0.25 cm/hr and an effective saturation of 90%. Assuming that ponding occurs instantaneously, estimate the total depth of direct runoff in mm from the event using the Green-Ampt infiltration model.
a. 80
b. 89
c. 76
d. 72
Answers
Problem 1. Network-Flow Programming (25pt) A given merchandise must be transported at a minimum total cost between two origins (supply) and two destinations (demand). Destination 1 and 2 demand 500 and 700 units of merchandise, respectively. At the origins, the available amounts of merchandise are 600 and 800 units. USPS charges $5 per unit from origin 1 to demand 1, and $7 per unit from origin 1 to demand 2. From origin 2 to demand 1 and 2, USPS charges the same unit cost, $10 per unit, however, after 200 units, the unit cost of transportation increases by 50% (only from origin 2 to demand 1 and 2).
a) Formulate this as a network-flow problem in terms of objective function and constraint(s) and solve using Excel Solver.
b) How many units of merchandise should be shipped on each route and what is total cost?
Answers
Solution :
Cost
Destination Destination Destination Maximum supply
Origin 1 5 7 600
Origin 2 10 10 800
15, for > 200 15, for > 200
Demand 500 700
Variables
Destination 1 2
Origin 1 [tex]$X_1$[/tex] [tex]$$X_2[/tex]
Origin 2 [tex]$X_3$[/tex] [tex]$$X_4[/tex]
Constraints : [tex]$X_1$[/tex], [tex]$$X_2[/tex], [tex]$X_3$[/tex], [tex]$$X_4[/tex] ≥ 0
Supply : [tex]$X_1$[/tex] + [tex]$$X_2[/tex] ≤ 600
[tex]$X_3$[/tex] + [tex]$$X_4[/tex] ≤ 800
Demand : [tex]$X_1$[/tex] + [tex]$$X_3[/tex] ≥ 500
[tex]$X_2$[/tex] + [tex]$$X_4[/tex] ≥ 700
Objective function :
Min z = [tex]$5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$[/tex]
[tex]$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$[/tex]
Costs :
Destination 1 Destination 2
Origin 1 5 7
Origin 2 10 10
15 15
Variables :
[tex]$X_1$[/tex] [tex]$$X_2[/tex]
300 300
200 400
[tex]$X_3$[/tex] [tex]$$X_4[/tex]
Objective function : Min z = 10600
Constraints:
Supply 600 ≤ 600
600 ≤ 800
Demand 500 ≥ 500
700 ≥ 500
Therefore, the total cost is 10,600.
The term variation describes the degree to which an object or idea differs from others of the same type or from a standard.
a. True
b. False
Answers
In a tension test of steel, the ultimate load was 13,100 lb and the elongation was 0.52 in. The original diameter of the specimen was 0.50 in. and the gage length was 2.00 in. Calculate (a) the ultimate tensile stress (b) the ductility of the material in terms of percent elongation
Answers
Answer:
a) the ultimate tensile stress is 66717.8 psi
b) the ductility of the material in terms of percent elongation is 26%
Explanation:
Given the data in the question;
ultimate load P = 13,100 lb
elongation δl = 0.52 in
diameter of specimen d = 0.50 in
gage length l = 2.00 inch
First we determine the cross-sectional area of the specimen
A = [tex]\frac{\pi }{4}[/tex] × d²
we substitute
A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²
A = 0.1963495 in²
a) the ultimate tensile stress σ[tex]_u[/tex]
tensile stress σ[tex]_u[/tex] = P / A
we substitute
tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495
tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi
Therefore, the ultimate tensile stress is 66717.8 psi
b) ductility of the material in terms of percent elongation;
percentage elongation of specimen = [change in length / original length]100
% = [ δl / l ]100
we substitute
% = [ 0.52 in / 2.00 in ]100
= [ 0.26 ]100
= 26
Therefore, the ductility of the material in terms of percent elongation is 26%