- #1

- 3

- 0

**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.

## Homework Statement

"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.

*Is there a better way to estimate the miles of wire used?*

__Question 1.__**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.

*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:*

__Question 2.__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))).

*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 3.__*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 4.__*Am I just making this a billion times more complicated than it should be? Should I just put down "42" as the answer?*

__Question 5.__Any insight would be very much appreciated. Thank you for your time and attention.

Last edited: