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Homework Help: Power needed to light NYC?

  1. Mar 8, 2007 #1
    Power needed to light NYC? (Circuits! Please help.)

    Hi there. I have a question from my physics lab. We're supposed to write a paper about the war of currents between Edison and Tesla/Westinghouse, and part of this involves estimating "the magnitude of the design problem." The actual question is as follows.

    1. The problem statement, all variables and given/known data

    "What was the magnitude of the design problem. Estimate the power required to light the city of New York at that time [about 1900], the miles of wire required, the batteries required, the hydroelectric power required,...[.]"

    I suspect that this is mainly a matter of getting the right order of magnitude, as much of the information I would need to get a precise number is simply not findable, at least that I can see. I talked to several of the other lab groups from the class, and they couldn't find relevant information, either.

    Here is the information I could find, and some assumptions I was told it was okay to make (if they seem way off, I would vastly appreciate correction):

    - Population of NYC in 1900: 3,437,202
    - Assumption: 2 light bulbs per person.
    - Assumption: Light bulb used was the Edison-Mazda cage filament lamp, which required 110-120 V (I used an average of 115 V) and 500 W.
    - Assumption: Wires used were solid copper, cylindrical, with a diameter of 1 cm.
    - Electrical resistivity of copper: 1.68 x 10^-8 ohm-meters.
    - Land area of NYC: 303.3 mi^2.
    - One city block: 0.25 mi^2.
    - Assumption: 10 buildings per city block.
    - Assumption: 1 mile of wire per building in 1900.

    Question 1. Is there a better way to estimate the miles of wire used?

    2. My work so far

    I apologize in advance for the non-prettiness of my equations.

    3,437,202 people * 2 light bulbs/person = 6,874,404 light bulbs.

    500 W/bulb * 6,874,404 bulbs = 3.44*10^9 W required.

    6,874,404 bulbs * 115 V/bulb = 7.91*10^8 V.

    P = Vi. i = P/V = 500 W/115 V = 4.35 A/bulb. 4.35 A/bulb * 6,874,404 bulbs = 2.989*10^7 A.

    303.3 mi^2 / 0.25 mi^2 = 1.47*10^6 city blocks in New York city, which means 1.47*10^7 miles of wire using the assumptions above.

    Resistance of all this wire = R = rho(Cu) * (L/A) = 1.68*10^-8 ohm-meters * (2.37*10^10 m wire/(pi*(0.05 m)^2)) = 5.06*10^4 ohms.

    Power lost in the wire = (i^2)*R = (2.989*10^7 A)^2 * 5.06*10^4 ohms = 4.52*10^19 W.

    Question 2. Why is my power lost in the wire so much higher than the power required to light the light bulbs? I asked my prof; he said I might be using the wrong equation. He said that I might want to consider:
    P = Vi = (i^2)*R = (i^2)*(R(load)+R(line)).

    He then provided two other equations, but I don't understand what they mean or how he got them:
    P(lost) = P/(1+(P(load)/P(line)))) = P+(R(line)/(R(load)+R(line))).

    Question 3. Can someone point me to a good resource for finding out what electrical load is? Wikipedia said that the circuit connected to the output terminal is the load (which - I don't understand how that description differs from the entire circuit); my textbook, as far as I can see, didn't cover it. Should I assume that the light bulbs are the load?

    Question 4. My physics lecture prof suggested that I consider the branching of all the light bulbs in parallel, and said that I might assume that the wire branches 5 or 6 times from the power plant to the individual light bulb. How would I apply this?

    Question 5. Am I just making this a billion times more complicated than it should be? Should I just put down "42" as the answer?

    Any insight would be very much appreciated. Thank you for your time and attention.
    Last edited: Mar 8, 2007
  2. jcsd
  3. Mar 8, 2007 #2
    this seems incorrect, as wouldn't the light bulbs be wired off from the mains in a series circuit, that way if somebody's light bulb was not working, it would not affect the rest off NYC. If that is the case, then you would need to have a voltage of 120V plus the voltage lost in the copper cable, not the sum of voltage of the light bulbs.

  4. Mar 8, 2007 #3
    Sorry? Thanks for your help, unique_pavadrin, but I don't quite understand. I thought the lights had to be put in a parallel configuration in order to keep all of them from going out if one went out. If I remember correctly, if the lights are in series, the current has to travel through each light consecutively to get back to the power source, while in parallel, it can go through any of the different paths and get back to the power source, and if one light goes out, it can still travel through the other paths and come out the other end. Does that make sense? Am I mistaken?
  5. Mar 8, 2007 #4
    sorry, what i have said is incorrect and what you have said is correct, i have confused my circuit terms
  6. Mar 8, 2007 #5
    as for the problem, seeing that the lights are wired in parallel with five or six branches that would mean that the voltage required would be equal to a fifth or a sixth of the total voltage required
  7. Mar 10, 2007 #6
    Okay, I think I understand what you were saying in your first post - I wanted my equations to be for a parallel circuit, but the ones I'd been using were for a series circuit. (I asked a third (!) prof yesterday, and he kindly pointed this out to me. I don't know why the first two didn't spot this.) Thanks. But what did you mean when you said that the voltage required would be a fifth or a sixth of the total voltage required? What did you do to come to that conclusion?

    Also, I think I have a better figure for how many watts were lost in the wire (I attached the updated calculations section of my paper 'cause it's easier to see what I've done). However, do I need to use different formulae to calculate the volts and watts required to run the light bulbs in parallel? Currently, my power lost is about equal to my power used to run the light bulbs. While this is a better outcome, I'm still wondering if there's something I'd missed. Thank you.

    Attached Files:

    Last edited: Mar 10, 2007
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