Test Questions

[sclement]

I have a lot of questions as I prepare for journeyman test. All help is greatly appreciated!

What is the minimum size copper conductor needed for a 208 V three-phase machine located 140 feet from the power source? ( Machine draws 27 amperes, 3% voltage drop, use 12.9 K)

 

[luckylerado]

I have a lot of questions as I prepare for journeyman test. All help is greatly appreciated!

What is the minimum size copper conductor needed for a 208 V three-phase machine located 140 feet from the power source? ( Machine draws 27 amperes, 3% voltage drop, use 12.9 K)

I have taken several journeyman tests and I do not think that I have ever seen a voltage drop calculation on any of them. It has been a while though. As is the practice, show your method and the folks here will chime in to help.

 

[sclement]

I have taken several journeyman tests and I do not think that I have ever seen a voltage drop calculation on any of them. It has been a while though. As is the practice, show your method and the folks here will chime in to help.

I've heard the same thing, not sure if Utah has a couple voltage drop questions or not? But anyway I'm not sure exactly how to do this problem? That's kinda why I posted it here for some help with it?

 

[Unbridled]

I have a lot of questions as I prepare for journeyman test. All help is greatly appreciated!

What is the minimum size copper conductor needed for a 208 V three-phase machine located 140 feet from the power source? ( Machine draws 27 amperes, 3% voltage drop, use 12.9 K)

#10 THWN Solid= 10380 CM
Using Vd= 1.73*12.9*27*420 / 10380
You get 2.72 V
Way less than 3% of 208V which is 6.24V

12.9 = K
27= Load Current
420= Length of conductors (3X140)
1.73=Sq. Root of 3

Ampacity per 430.22 must be 125% of the FLC = 33.75 A

#10 THW Ampacity @ 75 Degrees = 35 A

 

[thomasbl7]

Test question

Test question

I've heard the same thing, not sure if Utah has a couple voltage drop questions or not? But anyway I'm not sure exactly how to do this problem? That's kinda why I posted it here for some help with it?

Have taken 3 journeymen test in the last yr for diffrent states. Haven't seen any vd questions like it neither. ..I agree with unbridles' formula. But would uses 6.24 as my voltage drop in the formula since 208 is the applied voltage. 1.732*12.9×140*27/6.24=13,534cma which is larger than a number 10awg so we must go with #8..
430.22 wouldn't apply in this case because its not stated that the motor is continous. I would take inconsideration to use what the question gives you when taken a test.They are tricky in the way the questions are asked.I'm open for corrections and learning. Thanks... formula...single phase. Cma=2×k×i×l /vd....3phase ...cma=1.732×k×i×l/vd..
2 is the lenth back and forth,k is the costant,i is applied amperage,l is the actual length.,vd is percentage of the applied voltage. We switch (2) to(1.732) in the single to 3phase. When using this formula.

 

[infinity]

My Southwire VD calculator app says #8.

 

[augie47]

agree with post #5 and #6

 

[Unbridled]

I have a lot of questions as I prepare for journeyman test. All help is greatly appreciated!

What is the minimum size copper conductor needed for a 208 V three-phase machine located 140 feet from the power source? ( Machine draws 27 amperes, 3% voltage drop, use 12.9 K)

Post #5 and 6 are correct. The way you had your question worded through me of.
I used the voltage drop of #10 CU.(2.32 V)
Your question states to use 3% Vd(6.24 V)

3% Vd is the max. per NEC. In practical use, however, I don't calculate per wire size based on the possibility of a 3% Vd, I size it per the actual Vd of the conductor I'm contemplating on using.
If you remove the wording "3%" from your question, wouldn't you calculate you conductor size just not to exceed 3%, such as #10??:thumbsup::thumbsup:

 

[jimdavis]

I have taken several journeyman tests and I do not think that I have ever seen a voltage drop calculation on any of them. It has been a while though. As is the practice, show your method and the folks here will chime in to help.

FYI- The Utah journeyman exam is actually 3 separate exams- code examination (80 questions), theory examination (50 questions), and a practical examination (7 practical exercises). There are 10 voltage drop calculation questions on the theory exam.

 

[ggunn]

3% Vd is the max. per NEC.

The NEC does not dictate a 3% max Vd.

 

[Smart $]

#10 THWN Solid= 10380 CM
Using Vd= 1.73*12.9*27*420 / 10380
You get 2.72 V
Way less than 3% of 208V which is 6.24V

12.9 = K
27= Load Current
420= Length of conductors (3X140)
1.73=Sq. Root of 3

Ampacity per 430.22 must be 125% of the FLC = 33.75 A

#10 THW Ampacity @ 75 Degrees = 35 A

You only use the line-to-load length of one conductor. Not the length of all three conductors summed.

sqrt(3) × 12.9 × 27A × 140FT ÷ (208V × 3%) = 13,535cmil

 

[sclement]

FYI- The Utah journeyman exam is actually 3 separate exams- code examination (80 questions), theory examination (50 questions), and a practical examination (7 practical exercises). There are 10 voltage drop calculation questions on the theory exam.

Thanks for the information everyone it's very helpful! The 3% voltage job factor is just a factor be used on this particular question. I don't think it applies to any real world situation

 

[sclement]

Test question

Test question

The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.

Thanks in advance!

 

[Smart $]

The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.

So how do you think it is done?

 

[sclement]

So how do you think it is done?

Well To be honest I don't know? That's kinda why I'm throwing out the problems that I'm struggling with in hopes I'll get some help with them.

 

[Smart $]

Well To be honest I don't know? That's kinda why I'm throwing out the problems that I'm struggling with in hopes I'll get some help with them.

Resistors connected in series sum arithmetically. Apply Ohm's law to the result. Current is the same through all the resistors.

That's the most I can give you without providing the actual answer.

 

[Little Bill]

The following loads are connected in series, 5 ohms, 20 ohms, 10 ohms, and 15 ohms. The voltage total is 50v. Find the voltage drop on the 10 ohm resistor.

Thanks in advance!

Is the part in red correct or did they/you mean 50 ohms?

 

[infinity]

Is the part in red correct or did they/you mean 50 ohms?

Since they're in series the current would be the same at each resistor but the voltage drop across each resistor would be different. The sum of all of the voltage drops would be 50 volts, at least that's how I read it.

 

[Little Bill]

Since they're in series the current would be the same at each resistor but the voltage across each resistor would be different. The sum of all of the voltage drops would be 50 volts, at least that's how I read it.

I got to thinking after I posted that there were no other voltages mentioned so that must be correct.:happyyes:

 

[ggunn]

Is the part in red correct or did they/you mean 50 ohms?

They probably meant 50V so it's 50V across 50 ohms to make the math easy.