Calculating Voltage of 120V Load 750ft from Source

In summary, to solve this problem, we need to first determine the resistance of the wire using the given information on length and ohms per 1000 feet. Once we have this value, we can treat the load as a simple resistor and use Ohm's Law to calculate the voltage drop across it. With this information, we can then find the voltage delivered to the load by subtracting the voltage drop from the original 120 volts. It is also important to note that in this problem, there are two wires running from the source to the load, not just one.
  • #1
enzo456
3
0

Homework Statement



What Voltage is delivered to a 120-volt/5000-watt load which is fed with #10 AWG wire (1.24 ohms/1000') and located 750 feet from the 120-volt source?


Homework Equations



Ohm's Law E= I * R
E=(K*I*2*L)/Acmil


The Attempt at a Solution



I am in an apprenticeship program for Electrical. I was wondering if I could get some help on this problem. This is what I understand...I need to find Resistivity(K) since they do not tell me the type of wire being used. the Area in circular mils for #10 wire is 10380 circular mils. I have the length of 750 feet. I also need to find the current throughout the circuit. Since I am given the Power and Voltage of the load I calculated the Current from that:

I=P/E
I=5000w/120V
I=41.7 Amps

Now in order to find K resistivity I know that the resistance is 1.24 ohms/1000'. I understand the K is resistance per mil-foot, but this is where I come up with a problem. Any help would be appreciated...the apprenticeship program I am in is not that intense, so to work out these problems is not necessary. I am trying to learn this material because I intend to go for my bachelors degree afterwards. Thanks again
 
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  • #2
Welcome to PF, enzo.

Don't worry about resistivity. What can you figure out about the wire from this information:

1. The wire is 124 ohms per 1000 ft.
2. The load is 750 feet from the source.

Extra hint: how many wires would run from the source to the load?
 
  • #3
i can get .00124 ohms per foot and get .93 ohms per 750 feet. if E=I*R and multiply the current of 41.7 amps and multiply that by the resistance of .93 ohms per 750 feet than I get a voltage drop of 38.8 Volts. That value doesn't seem right to me, but is that the correct way? I believe 1 wire would run from the source to the load in this problem, is that correct?
 
  • #4
You're getting there. Yes, 750 feet of wire would be 0.93 ohms.

It might help to treat the load as a simple resistor. We know that it draws 41.7A at 120V. The actual current will be different than the 41.7A number, since the voltage at the load is different than 120V.

I believe 1 wire would run from the source to the load in this problem, is that correct?

That's what many people will assume, but if you think about it just 1 wire is not enough to make a complete circuit ...

So I would draw a circuit diagram, using resistors for the load and for the wire(s) also. From there, it's Ohm's Law to figure it all out.
 
  • #5
so drawing it out it makes sense that there are 2 wires and the resistance for the wires 1.86 but with the load if the current is different how would I calculate the resostance for the load to treat it as a simple resistor?
 
  • #6
Good question. Since we know that it draws 41.7A at 120V, we can use Ohm's Law to figure out R for the load.
 
  • #7
Yes I agree with red, find the load Resistance, you should then be able to get the voltage drop across the load R from the total resistance at 120V
 

FAQ: Calculating Voltage of 120V Load 750ft from Source

How do you calculate the voltage of a load that is 120V and 750ft from the source?

The formula for calculating voltage drop due to distance is V = IR, where V is the voltage drop, I is the current, and R is the resistance. In this case, since we are given the voltage and distance, we need to rearrange the formula to solve for current. So, I = V/R. We also need to consider the resistance of the wire, which can be found using the wire gauge and length. Once we have the current, we can use the voltage drop formula again to calculate the voltage drop at 750ft. Subtracting this voltage drop from the original voltage will give us the voltage of the load at 750ft from the source.

What is the importance of calculating voltage drop for a load at a certain distance from the source?

Calculating voltage drop is important because it helps ensure that the load receives the required voltage to function properly. If the voltage drop is too high, the load may not function at its full capacity or may not function at all. This could lead to equipment failure or safety hazards. By calculating the voltage drop, we can determine the appropriate wire gauge or adjust the distance to ensure that the load receives the necessary voltage.

How does the wire gauge affect the voltage drop at a certain distance from the source?

The wire gauge, or thickness, affects the resistance of the wire. A thicker wire has less resistance compared to a thinner wire. This means that a thicker wire will have a lower voltage drop at a given distance compared to a thinner wire. Therefore, choosing the appropriate wire gauge is crucial in minimizing voltage drop and ensuring that the load receives the required voltage.

Can voltage drop be completely eliminated?

No, voltage drop cannot be completely eliminated. However, it can be minimized by using the appropriate wire gauge and keeping the distance between the source and load as short as possible. Other factors such as temperature and type of wire also play a role in voltage drop, but these can be controlled to some extent. Ultimately, some voltage drop is expected and acceptable in most electrical systems.

How do you account for voltage drop in electrical system design?

Voltage drop is accounted for in electrical system design by considering the maximum acceptable voltage drop for the system and calculating the wire gauge and distance accordingly. This ensures that the load receives the required voltage and the system functions efficiently. Additionally, voltage drop calculations are important in determining the appropriate circuit protection and ensuring safety in the system.

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