How Much Power is Lost in a Copper Wire in a Series Circuit?

[chawki]

Homework Statement

A 200m long copper wire, with cross section area of 1.5mm², is connected with a 10Ω resistor in a series circuit, and the supply voltage is 50V.

Homework Equations

Find the power loss in the copper wire, when ρ=0.0175 Ωmm²/m

The Attempt at a Solution

R=*L/A
P=V²/R
P=50²/(0.0175*(200/1.5))
P=1068.37 Watt.

 

[gneill]

Homework Statement

A 200m long copper wire, with cross section area of 1.5mm², is connected with a 10Ω resistor in a series circuit, and the supply voltage is 50V.

Homework Equations

Find the power loss in the copper wire, when ρ=0.0175 Ωmm²/m

The Attempt at a Solution

R=*L/A
P=V²/R
P=50²/(0.0175*(200/1.5))
P=1068.37 Watt.

The voltage supply is across the wire AND the resistor (they are in series), not just the wire.

 

[chawki]

yeah i think i made a mistake :/ but we don't have I

 

[SammyS]

yeah i think i made a mistake :/ but we don't have I

Then you need to find out what the current, I, is.

What is the resistance of the copper wire?

 

[chawki]

I=50/2.34=21.36Amperes

P=2.34*(21.36)²
P=1067.62 watt

somehow it looks the same answer as the one posted in post #1

 

[gneill]

I=50/2.34=21.36Amperes

P=2.34*(21.36)²
P=1067.62 watt

somehow it looks the same answer as the one posted in post #1

You still haven't accounted for the 10 Ω resistance that's in series with the wire.

Also, your wire resistance should be 2.333 Ω to three decimals, not 2.34; I know it's being picky, but early rounding of intermediate results can muck up final results when there are multiple steps. It's good to get into the habit of keeping a few extra digits of accuracy in intermediate results, and round final results.