Voltage drop across connecting wires?

[ooohffff]

Homework Statement

Suppose you want to run some apparatus that is 95 m from an electric outlet. Each of the wires connecting your apparatus to the 120 V source has a resistance per unit length of 0.0065 Ω/m. If your apparatus draws 2.9 A, what will be the voltage drop across the connecting wires and what voltage will be applied to your apparatus?

Homework Equations

V=IR

The Attempt at a Solution

For voltage drop:
R_wire=.0065(95)=.6175Ω
V=IR=2.9*.6175=1.79*2(to apparatus and back)=3.58V which was wrong.

I got the answer to what voltage will be applied to your apparatus which was:
R_apparatus=V/I=120/2.9=411.379Ω
R_total=R_app+R_wire=41.379+.6175=41.996Ω
I_tot=V/R_tot=12-/42=2.85A
V=I_totR_wire=2.85*.6175=1.764V
120V-1.764V=118.23V

 

[SammyS]

Homework Statement

Suppose you want to run some apparatus that is 95 m from an electric outlet. Each of the wires connecting your apparatus to the 120 V source has a resistance per unit length of 0.0065 Ω/m. If your apparatus draws 2.9 A, what will be the voltage drop across the connecting wires and what voltage will be applied to your apparatus?

Homework Equations

V=IR

The Attempt at a Solution

For voltage drop:
R_wire=.0065(95)=.6175Ω
V=IR=2.9*.6175=1.79*2(to apparatus and back)=3.58V which was wrong.

I got the answer to what voltage will be applied to your apparatus which was:
R_apparatus=V/I=120/2.9=411.379Ω
R_total=R_app+R_wire=41.379+.6175=41.996Ω
I_tot=V/R_tot=12-/42=2.85A
V=I_totR_wire=2.85*.6175=1.764V
120V-1.764V=118.23V

They may be looking for the voltage drop across each wire individually.

 

[ooohffff]

They may be looking for the voltage drop across each wire individually.

Yup, you're right. It's V=(2.9)(.6175)=1.79V

 

[gneill]

The problem statement is ambiguous about whether the 2.9 A for the apparatus is meant to be its rated current at 120 V, or if it's the actual current drawn when in operation at the end of the 95 m cable. The latter seems to be the case if the voltage drop on the cables is correctly calculated using that current value. Knowing the source voltage and the cable voltage drop, KVL will tell you what's left for the load...