Chapter 18, Problem 24P

(a)

To determine

The total resistance of two wires.

Answer to Problem 24P

The answer is $0.276\Omega$ .

Explanation of Solution

Given:

Length of wire is 10 m followed by $5.0m$ copper and then $5.0m$ aluminium. Each has a diameter $1.0mm$ . A voltage difference is $85mV$ across the composite wire. Value of $\rho$ for aluminium is $2.65\times {10}^{-8}\Omega \cdot m$ and Value of $\rho$ for copper is $1.68\times {10}^{-8}\Omega \cdot m$ .

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

The cross-sectional area:

  $A=\pi r^2$

Where, $\pi$ is constant and r is a radius.

Formula of radius:

  $r=\frac{D}{2}$

Where, $D$ is a diameter.

Calculation:

By using the formula: $R=\frac{\rho L}{\pi r^2}$ .

Put the value of $r$ in above-mentioned formula,

  $R=\frac{\rho L}{\pi {\left(\frac{D}{2}\right)}^2}$

Substitute the given values for aluminium in the formula:

  $R=\frac{\rho L}{\pi {\left(\frac{D}{2}\right)}^2}$

  $R_{Al}=\left(2.65\times {10}^{-8}\Omega \cdot m\right)\frac{\left(5.0m\right)}{\left[\pi \left(\frac{{\left(1.0\times {10}^{-3}m\right)}^2}{4}\right)\right]}$

  $R_{Al}=0.1687\Omega$

Substitute the given values for copper in the formula:

  $R=\frac{\rho L}{\pi {\left(\frac{D}{2}\right)}^2}$

  $R_{Cu}=\left(1.68\times {10}^{-8}\Omega \cdot m\right)\frac{\left(5.0m\right)}{\left[\pi \left(\frac{{\left(1.0\times {10}^{-3}m\right)}^2}{4}\right)\right]}$

  $R_{Cu}=0.1070\Omega$

Now, add the resistance of each wire,

  $R_{Total}=R_{Al}+R_{Cu}$

  $R_{Total}=0.1687\Omega +0.1070\Omega =0.276\Omega$

Conclusion:

The total resistance of two wires is $0.276\Omega$ .

(b)

To determine

The current through the wire.

Answer to Problem 24P

  $0.308A$ .

Explanation of Solution

Given:

Length of wire is 10 m followed by $5.0m$ copper and then $5.0m$ aluminium. Each has a diameter $1.0mm$ . A voltage difference is $85mV$ across the composite wire. Value of $\rho$ for aluminium is $2.65\times {10}^{-8}\Omega \cdot m$ and Value of $\rho$ for copper is $1.68\times {10}^{-8}\Omega \cdot m$ .

The total resistance of two wires is $0.276\Omega$ .

Formula used:

Formula to calculate current:

  $I=\frac{V}{R}$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

Calculation:

Put the given values in the formula, $I=\frac{V}{R}$

  $I=\frac{85\times {10}^{-3}V}{0.276\Omega }=0.308A$

Conclusion:

The current through the wire is $0.308Amp$ .

(c)

To determine

The voltages across the aluminium part and the copper part.

Answer to Problem 24P

Voltage across the aluminium part, $V_{Al}=0.052V$ and voltage across the copper part $V_{Cu}=0.033V$

Explanation of Solution

Given:

Resistance across the aluminium wire, $R_{Al}=0.1687\Omega$

Resistance across the copper wire, $R_{Cu}=0.1070\Omega$

The current through the wire is $0.308A$ .

Formula used:

Ohm’s Law:

  $I=\frac{V}{R}$ or $V=I\times R$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

Calculation:

By using the formula, $V=I\times R$

The voltages across the aluminium part, $V_{Al}=I\times R_{Al}=\left(0.308A\right)\left(0.1687\Omega \right)=0.052V$

The voltages across the copper part, $V_{Cu}=I\times R_{Cu}=\left(0.308A\right)\left(0.1070\Omega \right)=0.033V$

Conclusion:

The voltages across the aluminium part and across the copper part:

  $V_{Al}=0.052V$ and $V_{Cu}=0.033V$