Question

When a voltage difference is applied to a piece of copper wire, a 5.0 mA current flows. If the copper wire is replaced with a silver wire with twice the diameter of the copper wire, how much current will flow through the silver wire? The lengths of both wires are the same, and the voltage difference remains unchanged. (The resistivity of copper is 1.68 × 10^-8 Ω·m, and the resistivity of silver is 1.59 × 10^-8 Ω ·m)

a. 1.3 mA

b. 19 mA

c. 11 mA

d. 5.3 mA

e. 21 mA

Answer 1

Concepts and reason

The concept used in this problem is specific resistance of the material and Ohm’s law

Initially, determine the resistance of the silver wire by using the formula of specific resistance. Later, use Ohm’s law to calculate the current flowing the silver wire.

Fundamentals

Specific Resistance:

The expression for the specific resistance is given by,

$R = \rho \frac{L}{A}$

Here, $R$ is the specific resistance of the material, $\rho$ is the resistivity of the material, $L$ is the length of wire and $A$ is the cross section of wire.

Ohm’s Law:

According to Ohm’s Law, “the voltage is directly proportional to the current”

$V = IR$

Here, $V$ is the voltage, $I$ is current and $R$ is the resistance.

The expression for the specific resistance of the metal wire is given by,

${R_{\rm{M}}} = {\rho _{\rm{M}}}\frac{L}{{{A_{\rm{M}}}}}$ …… (1)

Here, ${R_{\rm{M}}}$ is the resistance of the metal wire, ${\rho _{\rm{M}}}$ is the resistivity of metal wire and ${A_{\rm{M}}}$ is the area of metal wire.

The area of the metal wire is,

${A_{\rm{M}}} = \frac{\pi }{4}{\left( {{d_{\rm{M}}}} \right)^2}$

Here, ${d_{\rm{M}}}$ is the diameter of metal wire.

Substitute $\frac{\pi }{4}{\left( {{d_{\rm{M}}}} \right)^2}$ for ${A_{\rm{M}}}$ in the equation (1).

$\begin{array}{c}\\{R_{\rm{M}}} = {\rho _{\rm{M}}}\frac{L}{{{A_{\rm{M}}}}}\\\\ = {\rho _{\rm{M}}}\frac{L}{{\frac{\pi }{4}{{\left( {{d_{\rm{M}}}} \right)}^2}}}\\\end{array}$

The above equation is modified as,

${R_{\rm{M}}} = \frac{{4{\rho _{\rm{M}}}L}}{{\pi {{\left( {{d_{\rm{M}}}} \right)}^2}}}$ …… (2)

The resistance of the silver wire is given by,

${R_{\rm{S}}} = \frac{{4{\rho _{\rm{S}}}L}}{{\pi {{\left( {{d_{\rm{S}}}} \right)}^2}}}$ …… (3)

Here, ${R_{\rm{S}}}$ is the resistance of the silver wire, ${\rho _{\rm{S}}}$ is the resistivity of silver wire and ${d_{\rm{S}}}$ is the diameter of silver wire.

Dividing equations (2) and (3).

$\frac{{{R_{\rm{S}}}}}{{{R_{\rm{M}}}}} = \left( {\frac{{4{\rho _{\rm{S}}}L}}{{\pi {{\left( {{d_{\rm{S}}}} \right)}^2}}}} \right)\left( {\frac{{\pi {{\left( {{d_{\rm{M}}}} \right)}^2}}}{{4{\rho _{\rm{M}}}L}}} \right)$

The above equation is modified as,

$\frac{{{R_{\rm{S}}}}}{{{R_{\rm{M}}}}} = {\left( {\frac{{{d_{\rm{M}}}}}{{{d_{\rm{S}}}}}} \right)^2}\left( {\frac{{{\rho _{\rm{S}}}}}{{{\rho _{\rm{M}}}}}} \right)$ …… (4)

It is given that,

${d_{\rm{S}}} = 2{d_{\rm{M}}}$

Substitute $2{d_{\rm{M}}}$ for ${d_{\rm{S}}}$ , $1.68 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for ${\rho _{\rm{M}}}$ and $1.59 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for ${\rho _{\rm{S}}}$ in equation (4).

$\begin{array}{c}\\\frac{{{R_{\rm{S}}}}}{{{R_{\rm{M}}}}} = {\left( {\frac{{{d_{\rm{M}}}}}{{2{d_{\rm{M}}}}}} \right)^2}\left( {\frac{{1.59 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}{{1.68 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}} \right)\\\\ = 0.236\\\end{array}$

The above ratio can be written as,

$\frac{{{R_{\rm{M}}}}}{{{R_{\rm{S}}}}} = 4.23$

The current flowing through the metal wire is,

${I_{\rm{M}}} = \frac{V}{{{R_{\rm{M}}}}}$ …… (5)

Here, ${I_{\rm{M}}}$ is the current flowing through a metal wire.

The current flowing through the silver wire is,

${I_{\rm{S}}} = \frac{V}{{{R_{\rm{S}}}}}$ …… (6)

Here, ${I_{\rm{S}}}$ is the current flowing through the silver wire.

Dividing the equations (5) and (6).

$\begin{array}{c}\\\frac{{{I_{\rm{S}}}}}{{{I_{\rm{M}}}}} = \left( {\frac{V}{{{R_{\rm{S}}}}}} \right)\left( {\frac{{{R_{\rm{M}}}}}{V}} \right)\\\\{I_{\rm{S}}} = {I_{\rm{M}}}\frac{{{R_{\rm{M}}}}}{{{R_{\rm{S}}}}}\\\end{array}$

Substitute $5{\rm{ mA}}$ for ${I_{\rm{M}}}$ and $4.23$ for $\frac{{{R_{\rm{M}}}}}{{{R_{\rm{S}}}}}$ .

$\begin{array}{c}\\{I_{\rm{S}}} = {I_{\rm{M}}}\frac{{{R_{\rm{M}}}}}{{{R_{\rm{S}}}}}\\\\ = \left( {5{\rm{ mA}}} \right)\left( {4.23} \right)\\\\ = 21.186{\rm{ mA}}\\\\ \approx {\rm{21 mA}}\\\end{array}$

Ans:

The current flowing through the silver wire is $21{\rm{ mA}}$ .