While remodeling your garage, you need to temporarily splice, end to end, an 80

Question

While remodeling your garage, you need to temporarily splice, end to end, an 80 m long copper wire that is 0.850 mm in diameter with a 49 m long aluminum wire that has the same diameter. The maximum current in the wires is 2.00 A.

(a) Find the potential drop across each wire of this system when the current is 1.75 A.
..........V (copper wire)
  ..........V (aluminum wire)

(b) Find the electric field in each wire when the current is 1.75 A.
.............mV/m (copper wire)
............... mV/m (aluminum wire)

Explanation / Answer

here,

length of copper wire, Lcu = 80m

dia of copper wire, d = 0.850mm = 0.850*10^-3 m

area of Copper wire,
Acu = pi*(d/2)^2
Acu = 3.14 * (0.850*10^-3/2)^2
Acu = 5.672*10^-7 m^2

Length of aluminium wire, Lal = 49 m

area of alluminium wire,
Aal = pi*(d/2)^2
Aal = 3.14 * (0.850*10^-3/2)^2
Aal = 5.672*10^-7 m^2

Resistance fo Copper Wire,
Rcu = Rho*Lcu/Acu ( Rho is resistivity)
Rcu = 1.59*10^-8 * 80/(5.672*10^-7)
Rcu = 2.243 ohms

Resistance fo Aluminium Wire,
Ral = Rho*Lal/Aal ( Rho is resistivity)
Ral = 2.65*10^-8 * 49/(5.672*10^-7)
Ral = 2.289 ohms

Part A:
I = 1.75 A
From ohm's Law,
V = I*R

For Copper Wire
Vcu = I*Rcu
Vcu = 1.75 * 2.243
Vcu = 3.925 V

For Alluminium Wire
Val = I*Ral
Val = 1.75 * 2.289
Val = 4.006 V

Part B:
POtential Difference = Electric field * LEngth
E = V/d

For Copper Wire
Ecu = Vcu/Lcu
Ecu = 3.925/80
Ecu = 0.049 V/m = 49 mV/m

For Aluminium Wire
Eal = Val/Lal
Eal = 4.006/49
Eal = 0.082 V/m = 82 mV/m

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