Commercial Service Calculations Made Easy

Can You Answer This Commercial Service Exam Question?

How many 20 ampere general lighting circuits are required by code to supply a 5,000 sq.ft. office? 

Solution:

Step #1

Refer to NFPA Table 220.12 which requires a minimum of 3.5 va/sq.ft. for the general lighting load for office buildings.

5,000 sq.ft. x 3.5 va/sq.ft. = 17,500 va total lighting load

Step #2

Recognize that according to NFPA Article 100-Definitions the general lighting load in an office building is considered a “Continuous Load” because “the maximum current that supplies the load is expected to continue for 3 hours or more.”

Refer to NFPA 210.19(A)(1) which states that “where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.”

Remember we not trying to determine the size of conductor needed but rather how many circuits are required. Since the reciprocal of 125% is 80% a circuit supplying a continuous load can only be loaded to 80% of its rating.

Step #3

Determine the allowable capacity per circuit by multiplying the voltage of the circuit by the ampere rating  and the 80 % circuit capacity limitation for continuous loads.

120 volts x 20 amperes x 80% = 1920 va capacity per circui

Step #4

Divide total light load va by capacity per circuit va

=  17,500 va/1920 va

= 9.11 circuits or 10 circuits-final answer

(Since you cannot run a fractional portion of a circuit whenever this calculation results in a remainder always round up to the next whole number)

Knowing which formula to use and how to use it can make all the difference! Learn how to answer any formula related exam question with confidence. Call 888-919-3926 and start preparing to pass your electrical exam today!

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Three-Phase Transformer Exam Question Made Easy

Can You Answer This Three-Phase Transformer Exam Question?

What is the turns ratio of a three-phase 4-wire delta-wye transformer, with a 480 volt primary and a 208 volt/120 volt secondary?

Solution:

A turns ratio is a way of quantifying the relationship between the number of turn of the coil making up the primary verses the secondary phase coil of a transformer.

The side with the greater number of turns of coil will be the side with the higher voltage. For example a turns ratio of 2:1 indicates that the voltage on the primary side is twice as high as that on the secondary side of the transformer.

Since the question indicates that the primary is a delta-phase and line voltage is the same.  So primary side phase and line voltage is 480 volts.

However since the secondary is a four wire wye, which means that the secondary has three line conductors and one neutral conductor, the phase voltage (which would be read between any one line and the neutral conductor) is 120 volts and the line voltage (which would be read between any two line conductors) is 208 volts In a wye phase voltage x 1.732 = line voltage.

Since when determining turns ratio only phase voltages are used, simply divide the primary phase voltage by the secondary phase voltage to determine the turns ratio of the transformer.

480 primary phase volts/120 secondary phase volts = 4:1 turns ratio

Knowing which formula to use and how to use it can make all the difference! Learn how to answer any formula related exam question with confidence. Call 888-919-3926 and start preparing to pass your electrical exam today!

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Motor Efficiency Calculation Made Easy

Plan for success by learning how to answer motor efficiency calculation and related questions on your exam. Can you answer these questions correctly?

Question:

What is the efficiency of a 5 hp-230 volt single-phase ac motor?

Solution:

Use the single-phase efficiency formula

= 746 watts x Hp Rating/E x I

= 746 watts x 5 hp/230 volts x 28 amps

= 3730/6440

= .5792 x 100

=57.92% final answer.

Question:

What is the true power of a 5 hp-230 volt single-phase ac motor?

Solution:

To answer this question you need to know what true power is. True power is the power actually used by the system to do useful work. You also need to know that 746 watts equals 1 Hp.

True Power

= 746 x Hp Rating

= 746 watts x 5 hp

= 3730 watts final answer.

Question:

What is the apparent power of a 5 Hp-230 volt single-phase ac motor?

To answer this question you need to know what apparent power is. Apparent power is the power supplied to the system. It only has the potential to do work.

Apparent Power

= E x I

= 230 volts x 28 amps

= 6440 volt-amps final answer.

Question:

What is the efficiency of a 10 hp-230 volt three-phase ac motor?

Solution:

Use the three-phase efficiency formula

= 746 watts x Hp Rating/E x I x 1.732

= 746 watts x 10 hp/230 volts x 28 amps x 1.732

= 7460/11,154

= .6688 x 100

=66.88% final answer.

Question:

What is the true  power of a 10 hp-230 volt three-phase ac motor?

Solution:

True Power

= 746 watts x Hp Rating

= 746 watts x  10 hp

= 7460 watts final answer.

Question:

What is the apparent power of a 10 Hp-230 volt three-phase ac motor?

Apparent Power

= E x I x 1.732

= 230 volts x 28 amps x 1.732

= 11,154 volt-amps final answer.

Knowing which formula to use and how to use it can make all the difference! Learn how to answer any formula related exam question with confidence. Call 888-919-3926 and start preparing to pass your electrical exam today!

 

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Can You Answer This Cooking Equipment Demand Factor Exam Calculation Question?

What is the demand load on the feeder for 1-14 kw 240 volt single-phase range installed in a dwelling unit?

Solution:

To answer this question correctly we need to know what the term demand as used in the code really means. The term demand always refers to a reduction from connected load. Which means that in this case the connected load or the nameplate rating is 14 kW and we need to reduce that rating by applying the appropriate code rules.

Refer to Table 220.55, Column C and Note #1 in the National Electrical Code.

Step One:

Subtract the cutoff rating for Column C from the nameplate rating or connected load of the range.

14 kW -12kW = 2 kW difference.

Step Two:

Use the kW difference to increase 5% per Note #1 of Table 220-55.

Increasing 5% by 2kW turns 5% into 10%.

Step Three:

Increase the kW demand value for 1 range in Column C by 10%.

8 kW increased by 10% = 8.8 kW demand-Final answer.

Learn everything you’ll need to know about how to effectively use this code table and the many others you’ll be required to refer to on the exam. We’ll help you learn what you need to know to pass your electrical exam. Call 888-919-3926 today.

 

 

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Prevent Injuries from Contact of Overhead Power Lines with Metal Ladders

Prevent Injuries from Contact of Overhead Power Lines with Metal Ladders

According to the The National Institute for Occupational Safety and Health following the seven preventative safety protocols listed below can significantly improve work place safety.

■ Identify the location of overhead power lines as a routine part of all initial worksite surveys for jobs involving the use of ladders. Always note power line heights and distances from work areas on site diagrams to provide key information for site supervisors and workers.

■ Avoid or limit proximity to power lines whenever possible. Consider ladder length and room for ladder staging (safely raising and lowering ladders).

■ Notify the local electric utility company for assistance if work needs to be done near energized, overhead power lines.

■ Ensure that workers maintain a safe working distance between power lines and equipment or structures that require periodic maintenance or access.

■ Do not store materials or equipment below or near overhead power lines.

■ Eliminate the use of metal ladders near energized overhead power lines.

■ Ensure that workers keep conductive objects at least 10 feet away from unguarded, energized lines up to 50 kilovolts. For every 10 kilovolts above 50 kilovolts, maintain an additional 4 inches of clearance.

The electrician with the license pulls the permit and controls the safety protocols that are followed. Prepare to pass your exam and pull your own permits. We’ll help you reach your goal! Call 888-919-3926 today.

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How Much Do Electricians Make?

How Much Do Electricians Make?

According to US News and World Report in 2016, the median wage for an electrician was $52,720. The highest-paid earned $90,420, while the lowest-paid electricians earned around $31,800 that year. An apprentice usually makes between 30 to 50 percent less than someone who is fully trained. 

Licensed electricians earn three times as much as their lower paid counterparts.

The highest paid electricians work in the metropolitan areas of San Rafael, California, San Francisco, and San Jose, California. The Chicago area also pays well, as does the city of Fairbanks, Alaska.

San Rafael, California $100,500
San Francisco $97,910
San Jose, California $79,990
Chicago $79,650
Fairbanks, Alaska $79,550

 

Thinking of relocating to find work? Get your license first! We can prepare you to pass any electrical exam nationwide.

Best Paying States for Electricians

The states and districts that pay electricians the highest mean salary are Alaska ($81,600), Hawaii ($74,770), Illinois ($73,160), New York ($73,010), and the District of Columbia ($71,960).

Alaska $81,600
Hawaii $74,770
Illinois $73,160
New York $73,010
District of Columbia $71,960

Unlock your earning potential! Prepare to pass the electrical exam and get your electrical license. We can help. Call 888-919-3926 today.

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Can You Answer General Light Load Calculations Exam Questions?

You’ll need to know how to answer general lighting load calculation exam questions to get a passing score on your electrical exam. Understanding how to answer dwelling service calculation questions like the one below can help you get a passing score.

What is the general lighting demand load for a 5000 square foot single family dwelling?

5000 sq.ft. x 3.5 va/sq.ft.                                                       =   17,500 va

2 small appliance branch circuits @ 1500 va/circuit         =    3,000 va

1 laundry circuit @ 1500 va                                                  =    1,500 va                                                                                                                                        =  22,000 va  connected load

1st 3000 va @ 100 %                                                              =    3,000 va

Next 19,000 va @ 35%                                                           =    6,650 va                                                                                                                                      =    9,650 va demand load

Now that you know how the math works, do you know where the code articles are that support the math? We can help! We’ll show you how to find and apply the code references that will allow you to answer any calculation based question on your exam.

Call us at 888-919-3926 or click here to find a seminar in your state!

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Ohm’s Law Electrical Exam Question Calculations

Can you answer ohm’s law electrical exam questions related to total resistance in a series verses a parallel circuit? Here’s what you need to know.

Resistance in a series circuit is always additive whether you have two or two hundred resistances in series. To determine whether a resistance is in series, just open the resistance and ask yourself what happens to the circuit. If the circuit is opened and no current flows, the resistance is a series resistance.

Resistance in a parallel circuit is not additive. In fact the greater number of resistances in parallel, the lower total overall circuit resistance will be. The total resistance in a parallel circuit will always be less than the lowest resistive branch.

Take a look at this handy Ohm’s Law reference sheet:

 

Need to know how amperage, wattage and voltage operates in series, parallel and series-parallel combination circuits? We can help! Contact us today at 888-919-3926 or click here to view available courses for your area!

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How To Get An Electrical License

How To Get An Electrical License

Why is your most valuable possession as an electrician your own electrical license?

Licensed electricians have greater earning potential and are more employable. They don’t  walk away from jobs because they aren’t licensed and can’t pull the required permits. Licensed electricians also don’t have to share their hard eared profits with other electricians who are licensed because they can pull permits for themselves.

Your ability to take full advantage of opportunities that come your way depends on your ability to pull your own electrical permits. The only way to be able to pull your own permits is to pass your electrical exam.

man working at computer

At Electrical Exam Seminars, we present the material you’ll need to know in a format that is easy to understand and directly relates to the work you’re already performing in the field. We also provide personalized one-on-one private tutorial instruction that guarantees that you’ll receive the attention you need to succeed.

Our seminars provide the individualized attention, proven test preparation techniques, and the self-paced learning environment you need to successfully prepare to pass your exam on the first try. We’re confident that our program can help you reach your goal. With a 95% first time pass rate, we can afford to be!

That’s why we can offer our “Repeat until You Pass” guarantee. Our system has been fine-tuned and perfected; helping electricians to successfully prepare for their exams for over thirty years.

Our seminars begin with assessment testing so that we can determine how best to meet your specific needs. The learning atmosphere is relaxed and  self-paced in order to provide a stress-free learning experience. The best part is that once you’ve finished the seminar, you’ll be confident, informed and completely ready to pass your electrical exam.

Once you’ve got your electrical license in your hand, you’ll be capable of pulling your own permits, submitting more competitive bids and putting more money in your own pocket. If you want to gain control over your jobs and increase your earning potential, check out Electrical Exam Seminars. In just a few days, you could be adding “licensed electrician” after your name on your business cards.

Call us at 888-910-3926 to learn how we can help, or click HERE to find a seminar for your city or state’s licensing exam.

Learn More:

Read about our “Repeat Until You Pass” guarantee HERE.
View Frequently Asked Questions about the electrical exam HERE.

 

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Electrical Exam Practice Questions

Electrical exam practice questions can be easily answered if you know which formula to use and how to use it. Motor calculation questions are commonly asked on electrical exams. Can you answer these practice questions?

 

What is the efficiency of a 5 horse power, 230 Volt, single-phase alternating current motor?

A. 75.9%

B. 57.9%

C. 33.4%

D. None of the above

Answer: B. 57.9%.

 

What is the required conductor ampacity of a 5 horse power, 230 Volt, single-phase alternating current motor?

A. 28 amps

B. 32 amps

C. 35 amps

D. None of the above

Answer: C. 35 amps.

 

Upon completion of the seminar, you will have mastered the 3 keys to electrical exam success:

  • Understanding exam questions and identifying key words.
  • Performing code calculations with ease and precision.
  • Finding the code articles required to answer questions within the allotted time.

Our unique approach ensures that you will receive balanced, thorough preparation at the pace that is best suited to meet your needs. Call 888-919-3926 or click here to start preparing to pass your electrical exam today.

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