Question

A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 27.2. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire.
a.) What is the current in the copper section? b.) What is the current in the silver section?

b.) What is the magnitude of E in the copper section?

c.) What is the magnitude of E in the silver section?

d.) What is the potential difference between the ends of the silver section of the wire?

Answer 1

Explanation:

Resistivity of silver wire =$1.47 e-8 ohm meter$

Resistivity of copper wire = $1.72 \mathrm{e}-8 \text { ohm meter }$

Resistance = Resistivity *length/cross sectional area

$R=\rho_{L / A}$

So,

$\begin{aligned}&I=A\left(V / P_{A g} L_{A g}+P_{C u} L C_{U}\right)\\&1=8104 \mathrm{A}\end{aligned}$

For E calculation

For copper

E = resistivity * current density

E = Resistivity * current/area

$E=4.9298 \mathrm{V} / \mathrm{m}$

For silver

$E=4.213 \mathrm{V} / \mathrm{m}$

Potential difference $=E^{*} L$

For silver

Potential difference = $4.213^{*} 1.2=5.055 \mathrm{V}$

For copper

Potential difference =$3.9438 \mathrm{V}$