Calculate the voltage drop of the train (Picture included)

In summary, the conversation is about a student's homework where they had to calculate the voltage drop over a train in a circuit. The student initially used the equation V=IR to calculate the voltage, but their teacher told them to calculate the total resistance of the circuit and then multiply it by the current. The student was confused because they thought the question only asked for the resistance over the train, but the teacher explained that it was the potential drop across the train. The student then understood and calculated the correct voltage drop.
  • #1
Simmer
2
0
Hello guys, so I had a homework and I couldn't understand the point of my teacher. The question goes like this:

http://desmond.imageshack.us/Himg3/scaled.php?server=3&filename=21976607.jpg&res=medium

So the question is,
"Calculate the voltage drop over the TRAIN when it is at the position indicated"

Alright so he asked for the voltage that goes to the train, therefore,
V=IR

V=1500
I=300
R=The resistor the current will go through before it hits the train which is 0.140

V=300*0.140 ----> V=42
So the Voltage that will go through the train is going to be 1500-42 ------> v=1458


My teacher said it's wrong, I should calculate the total resistance of the circle which is 0.140+0.044 and then multiply by the current, but I don't understand! the question asked about the resistance over the train only not over the whole circle, anyone can help?
 
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  • #2
Simmer said:
Hello guys, so I had a homework and I couldn't understand the point of my teacher. The question goes like this:

http://desmond.imageshack.us/Himg3/scaled.php?server=3&filename=21976607.jpg&res=medium

So the question is,
"Calculate the voltage drop over the TRAIN when it is at the position indicated"

Alright so he asked for the voltage that goes to the train, therefore,
V=IR

V=1500
I=300
R=The resistor the current will go through before it hits the train which is 0.140

V=300*0.140 ----> V=42
So the Voltage that will go through the train is going to be 1500-42 ------> v=1458


My teacher said it's wrong, I should calculate the total resistance of the circle which is 0.140+0.044 and then multiply by the current, but I don't understand! the question asked about the resistance over the train only not over the whole circle, anyone can help?
I think you mean that he asked for the voltage over the train, that is, the voltage someone on the train would measure if he had a meter and measured from the "top" of the train (overhead wire) to the "bottom" of the train (track rail). That is the potential drop across the train. It happens to be what's left of the total potential (1500V) when all the drops due to other resistances in the whole circuit path are accounted for. Your teacher is correct.
 
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  • #3
gneill said:
I think you mean that he asked for the voltage over the train, that is, the voltage someone on the train would measure if he had a meter and measured from the "top" of the train (overhead wire) to the "bottom" of the train (track rail). That is the potential drop across the train. It happens to be what's left of the total potential (1500V) when all the drops due to other resistances in the whole circuit path are accounted for. Your teacher is correct.


Hello

Thank you for your help, now I understand it

If we calculate the total resistance of the whole circle, it will be

R total=V/I ----> R total =1500/300 ----> R total =5
Now by taking out the resistor 0.140 and the resistor 0.044 from the total resistance, it will be 4.816, this is the resistance of the engine of the train, now calculating the Voltage
V=IR -----> V= 4.816 * 300 ------> Voltage drop over the train engine is going to be 1444.8 which is the same answer as my teacher gave me


Thank you very much for your help!
 

1. How is the voltage drop of a train calculated?

The voltage drop of a train is calculated by using Ohm's Law, which states that voltage is equal to the current multiplied by the resistance. In this case, the resistance is the sum of all the individual resistances in the train's electrical system, including the resistance of the tracks, wires, and other components.

2. Why is it important to calculate the voltage drop of a train?

Calculating the voltage drop of a train is important because it ensures that the train is receiving enough voltage to operate safely and efficiently. A significant drop in voltage can lead to malfunctions and potential safety hazards.

3. What factors can affect the voltage drop of a train?

The voltage drop of a train can be affected by several factors, including the length and quality of the tracks, the age and condition of the train's electrical components, and the amount of current being drawn by the train's motors and other systems.

4. How does the voltage drop of a train impact its performance?

The voltage drop of a train can greatly impact its performance. A higher voltage drop means that the train is receiving less voltage, which can result in slower speeds, reduced power, and increased energy consumption. It can also lead to overheating and damage to the train's electrical components.

5. Can the voltage drop of a train be reduced?

Yes, the voltage drop of a train can be reduced by improving the overall electrical system, such as using higher quality tracks and wires, properly maintaining and replacing old components, and distributing the load evenly among multiple power sources. Regular voltage drop calculations can also help identify and address any potential issues before they become major problems.

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