Physics E & M Question about a powerline

  • Thread starter Thread starter Brandon Hawi
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around calculating the resistive power loss in a high voltage powerline operating at 500,000 V-rms with a current of 500 A and a resistance of 0.05Ω/km over 200 km. The initial attempt using the formula P = V²/R yielded an incorrect result, prompting clarification on the correct application of formulas. It was noted that the voltage given does not represent the potential drop across the entire length of the line. The correct approach involves calculating the potential drop using V = IR, leading to a power loss calculation of 2.5 MW using the formula P = I²R. The final answer aligns with the expected result, confirming the proper method for such calculations.
Brandon Hawi
Messages
8
Reaction score
1

Homework Statement


A high voltage powerline operates at a 500000 V-rms and carries an rms current of 500 A. If the resistance of the cable is 0.05Ω/km, what is the resistive power loss in 200 km of the powerline.

V = 500000 V-rms
Irms = 500 A
R/x = 0.05Ω/km
x = 200 km

Homework Equations



P = I2R
P = V2/R
P = IV

The Attempt at a Solution



I attempted to use V2/R at first, and that gave me 2.5 x 1010 Watts, however the given answer is 2.5 x 106 Watts. I get that answer when I use I2R, but I was wondering why V2/R could not be used?
EDIT: Fixed to V2/R
 
Last edited:
Physics news on Phys.org
Hello and welcome to PF!

Your formula P = V/R is incorrect. Did you mistype the formula?

Even when you have the correct formula, you have to be careful. The 500,000 V of the powerline probably represents the voltage of the line relative to ground at one end of the line. It does not represent the potential drop as you go from one end of the line to the other end of the line.

You can calculate the potential drop V and then use the (corrected) power formula that involves V and R.
 
TSny said:
Hello and welcome to PF!

Your formula P = V/R is incorrect. Did you mistype the formula?

Even when you have the correct formula, you have to be careful. The 500,000 V of the powerline probably represents the voltage of the line relative to ground at one end of the line. It does not represent the potential drop as you go from one end of the line to the other end of the line.

You can calculate the potential drop V and then use the (corrected) power formula that involves V and R.

Sorry, I meant to type V2/R. How would one then go about calculating the potential drop V?
 
Brandon Hawi said:
Sorry, I meant to type V2/R. How would one then go about calculating the potential drop V?
You're making Georg Ohm feel neglected today.:wink:
 
TSny said:
You're making Georg Ohm feel neglected today.:wink:

Alright then, so V = IR. So I get V = 5000 V. Then I used V2/R and got 2.5 MW as my answer. Thanks ! :smile:
 
Looks good!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How Does a Charged Bullet Respond to Earth's Magnetic Field?
Replies
1
Views
2K
Long distance electrical transmission efficiency
Replies
2
Views
2K
Which bulb will glow brighter in a series and parallel circuit?
Replies
5
Views
2K
Underground cable with electrical short
Replies
1
Views
1K
What Is the EMF of a Cell with Given Energy Dissipation and Current?
Replies
4
Views
4K
What is the formula for calculating the mass of air in a piston?
Replies
6
Views
2K
Radial & Tangential Acceleration of a Car at Indianapolis 500 | Physics Solution
Replies
7
Views
4K
Adiabatic process, calculating final T, enthelpy etc
Replies
11
Views
4K
Questions about finding dBm values
Replies
5
Views
2K
Change in watts and power dissipated
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top