400 customers at 500 KWh/month LOAD 1 600 customers at 300 KWh /month Power Plant LOAD 2 560,000 KWh /month LOAD 3 1000 customers at 180 KWh /month Charges: 1st 100 KWh/month/customer = Php 2.12 Next 150 KWh/month/customer = Php 1.59 Next 150 Kwh/month/customer = Php 1.05 Excess from 400 KWh/month/customer = Php 0.85 Efficiency of Transmission is at 90.322% 1. The system 's monthly Energy production? 2. The production cost per month of LOAD 1 is calculated? 3. The production cost of LOAD 2 is calculated? 4. For LOAD 3 the production cost per month is?

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
icon
Related questions
Question

Solving problem & need solution. thanks

400 customers at 500 KWh/month
LOAD 1
600 customers at 300 KWh /month
Power Plant
LOAD 2
560,000 KWh /month
LOAD 3
1000 customers at 180 KWh /month
Charges:
I
1* 100 KWh/month/customer = Php 2.12
Next 150 KWh/month/customer = Php 1.59
Next 150 Kwh/month/customer = Php 1.05
Excess from 400 KWh/month/customer = Php 0.85
Efficiency of Transmission is at 90.322%
1. The system 's monthly Energy production?
2. The production cost per month of LOAD 1 is calculated?
3. The production cost of LOAD 2 is calculated?
4. For LOAD 3 the production cost per month is?
5. The cost per KWhr produced is taken?
Transcribed Image Text:400 customers at 500 KWh/month LOAD 1 600 customers at 300 KWh /month Power Plant LOAD 2 560,000 KWh /month LOAD 3 1000 customers at 180 KWh /month Charges: I 1* 100 KWh/month/customer = Php 2.12 Next 150 KWh/month/customer = Php 1.59 Next 150 Kwh/month/customer = Php 1.05 Excess from 400 KWh/month/customer = Php 0.85 Efficiency of Transmission is at 90.322% 1. The system 's monthly Energy production? 2. The production cost per month of LOAD 1 is calculated? 3. The production cost of LOAD 2 is calculated? 4. For LOAD 3 the production cost per month is? 5. The cost per KWhr produced is taken?
Expert Solution
steps

Step by step

Solved in 5 steps with 5 images

Blurred answer
Knowledge Booster
Differential Amplifier
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,