How Long Can Copper Wire Be in a Data Center to Limit Voltage Drop to 1V?

[cmathis]

Homework Statement

Your data center has 100,000 CPUs. Each burns 100 W at 120 V. Copper has a resistivity of 1.7E-8 Ω*m. What is the maximum length of copper wire on the power supply to give a voltage drop of no more than 1V?

Homework Equations

P = V*I
R = (rho*L)/A, where L is the length of the wire, A is the cross-section area
V= I*R

The Attempt at a Solution

Each CPU burns 100W at 120V, therefore each draws 100W/120V = 0.83A current. 0.83A * 100,000 = 83 kA total draw. Is this right? That is a MONSTROUS amount of current.

The answer to the question would be:

V = I*R

1V = 83000A * (1.7E-8 * L)/A

L = (1V * A)/(83000A * 1.7E-8)

 

[rude man]

Sorry, but the wire's length will depend on the wire's diameter.

 

[cmathis]

I know. I left that variable as "A" in the answer (cross section area) because it wasn't given to me in the problem. The rest of the work looks fine though?

 

[rude man]

Homework Statement

Your data center has 100,000 CPUs. Each burns 100 W at 120 V. Copper has a resistivity of 1.7E-8 Ω*m. What is the maximum length of copper wire on the power supply to give a voltage drop of no more than 1V?

Homework Equations

P = V*I
R = (rho*L)/A, where L is the length of the wire, A is the cross-section area
V= I*R

The Attempt at a Solution

Each CPU burns 100W at 120V, therefore each draws 100W/120V = 0.83A current. 0.83A * 100,000 = 83 kA total draw. Is this right? That is a MONSTROUS amount of current.

Yes, it's right!

The answer to the question would be:

V = I*R

1V = 83000A * (1.7E-8 * L)/A
L = (1V * A)/(83000A * 1.7E-8)

Good! I got confused between A = area and A = ampers there for a minute.