Voltage drop
- Thread starter Padro
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Resistance equals (RHO x l)/A, or resistivity times length divided by area. So resistance per unit length would be RHO/A.
But is that what you are asking? Table 8 in the back of the NEC has values for DC Resistance per 1000 feet for various size conductors. Table 9 has AC impedance.
But is that what you are asking? Table 8 in the back of the NEC has values for DC Resistance per 1000 feet for various size conductors. Table 9 has AC impedance.
Voltage drop
A question in the study quide for California electricians practrice exam is
etermine the voltage drop on a branch circuit givin the following:
current=5 amps
length = 300 ft.
resistance = 4 olms/1000ft. My question is: how did they come up with 4olms/1000 is this a standard or is it just a figure there using for the problem.
A question in the study quide for California electricians practrice exam is
etermine the voltage drop on a branch circuit givin the following:current=5 amps
length = 300 ft.
resistance = 4 olms/1000ft. My question is: how did they come up with 4olms/1000 is this a standard or is it just a figure there using for the problem.
Bob NH
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Padro said:A question in the study quide for California electricians practrice exam is
current=5 amps
length = 300 ft.
resistance = 4 olms/1000ft. My question is: how did they come up with 4olms/1000 is this a standard or is it just a figure there using for the problem.
The formula is:
Voltage Drop = Current x Resistance, usually written as VD = I*R,
but sometimes written E = I*R or E = IR, where E is the electromotive force, same as voltage.
Where current (I) is Amps and Resistance (R) is Ohms.
So if you have one wire, 300 ft long, and the resistance is 4 Ohms per 1000 ft, then the resistance is:
R = (4 Ohms per 1000 ft) x 300 ft/1000 ft = 1.2 Ohms
The voltage drop: VD = IR = 5 Amps x 1.2 Ohms = 6 Volts
The information given leaves things a little uncertain because it doesn't say if the distance is the round-trip distance or the one-way distance; and it doesn't say if the voltage drop is the one-way voltage drop or the round-trip voltage drop, and it doesn't say if the resistance is for one conductor or the total for the outgoing and return conductors.
A better way to write such a problem would be:
A load 300 ft away from the circuit breaker requires 5 Amps, and the resistance of each conductor in the circuit is 4 Ohms per 1000 ft. How much less is the voltage at the load than the voltage at the circuit breaker?
That is a different problem and has a different answer. I leave that one for you to work out.
The only way to reliably get the correct answer to this kind of problem is to understand the relationship between the problem and the formula, including what each term of the formula means in a physical and mathematical sense.
Otherwise, you will be plugging numbers into a formula and sometimes you will be lucky and get it right, and sometimes you will be unlucky.
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The formula I use most frequently is:
VD = 2 x R x I x D (single phase)
(2 x Resistance x Amps x Distance)
VD = 1.73 x R x I x D (three phase)
(Square root of three x Resistance x Amps x Distance)
It's easy to remember, to rid yourself of voltage drop. Generally, as I understand it, when a distance is given they tend to mean it in the most literal sense, the breaker supplying the load is ___ feet from the load.
Hence, the 2 in the equation above: there is a hot and a neutral, two conductors. If the load is three phase, the current travels along all three wires, so it is more efficient.
VD = 2 x R x I x D (single phase)
(2 x Resistance x Amps x Distance)
VD = 1.73 x R x I x D (three phase)
(Square root of three x Resistance x Amps x Distance)
It's easy to remember, to rid yourself of voltage drop. Generally, as I understand it, when a distance is given they tend to mean it in the most literal sense, the breaker supplying the load is ___ feet from the load.
Hence, the 2 in the equation above: there is a hot and a neutral, two conductors. If the load is three phase, the current travels along all three wires, so it is more efficient.
LarryFine
Master Electrician Electric Contractor Richmond VA
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The current is 5 amps, the length of the circuit (apparently one-way) is 300 ft., and they're giving you the resistance of the conductor (rather than making you find it from a wire size).Padro said:A question in the study quide for California electricians practrice exam is
current=5 amps
length = 300 ft.
resistance = 4 olms/1000ft. My question is: how did they come up with 4olms/1000 is this a standard or is it just a figure there using for the problem.
If my statements are correct, you need to figure out the voltage drop along 600' of a conductor that has 4 ohms per 1000, when carrying 5 amps. Calculate the resistance (600/1000 of 4) and use that to calculate voltage drop.
0.6 * 4 = 2.4 ohms
2.4 * 5 = 12 volts
One of the nuances omitted is whether the 5-amp current is that of the load without a voltage drop on the supply conductors, or with the voltage drop.
Since voltages were not given, I'd say ignore what I mentioned in the previous paragraph for the purposes of the question. Plus, my numbers refer to a simple 2-wire circuit.
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Padro said:A question in the study quide for California electricians practrice exam is
current=5 amps
length = 300 ft.
resistance = 4 olms/1000ft. My question is: how did they come up with 4olms/1000 is this a standard or is it just a figure there using for the problem.
This question to me seems a little strange. First of all the amount of voltage drop is determined by the size of the conductors. For a breakdown of conductor properties for these type problems see chapter 9 table 8 in the code book. Notice that in the column under copper / uncoated / ohms per 1000 ft that a different number is given depending on the size of the wire being used. The only sized wires that come close to your particular question is #16 AWG at 4.89 ohms/kft solid or 3.07 ohms/kft for #14 solid.
The correct way of determining voltage drop for this type of problem is the calculation 2 x K x I x D for single phase and 1.732 x K x I x D for 3 phase then dividing the result by the circular mils of the wire to be used. If you are unfamiliar with this equasion: K = constant for the resistance which is 12.9ohms for copper and 21.2 ohms for aluminum.
But in the question you had posed the wire size really isn't given...only a vague 4 ohms/1000ft.
So for the sake of arguement lets suppose that they are using #14 solid copper conductor the problem would then look like this supposing that this is single phase:
2 x 12.9 = 25.8
25.8 x 300 = 7740 (assuming that the 300' is the distance from your source to the end of the circuit.
7740 x 5 amps = 38700
now divide 38700 by the circular mils of #14 solid (chap 9 table 8) which is 4110 = 9.416
so the final answer for my example is a voltage drop of 9.416 volts.
While that method is the most traditional, it is not the only correct way.bstoin said:
{1;2;√3} ? K ? I ? L ? cmil ...
...is the same as...{1;2;√3} ? R ? I ? D
...in that...R = K ? cmil
...and...L = D, which is the one-way length of circuit, source to load.
Nevertheless, both equations will yield less accurate results than formulas which include adjustments for ambient and conductor temperatures, and conductor and load impedances for AC circuits.
As the OP asked about a question in a study guide, and the same is true for tests, it's always best to work with the provided parameters.
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