Chapter 18, Problem 21P

(a)

To determine

The resistance for current that passes through the solid in the $x$ direction.

Answer to Problem 21P

  $R=0.375\times {10}^{-3}\Omega$

Explanation of Solution

Given:

The resistivity, $\rho =3.0\times {10}^{-5}\Omega \cdot m$

A rectangular solid made of carbon has sides of lengths: $x=1.0cm$ , $y=2.0cm$ and $z=4.0cm$

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.

Calculation:

For x direction, current will flow from the right side to left side of the carbon rectangular. It means that the area (current path) is the right surface area $\left(A=yz\right)$ and also, the path length of the current is $L=x$ .

Since, the current passes through x-direction $L=x=1.0cm$ .

By using the formula of resistance:

  $R=\frac{\rho L}{A}=\frac{\rho x}{yz}\mathrm{...}(i)$

Put the given values in the equation $\left(i\right)$

  $R=\frac{3.0\times {10}^{-5}\times 1.0\times {10}^{-2}}{2.0\times {10}^{-2}\times 4.0\times {10}^{-2}}=0.375\times {10}^{-3}\Omega$

Conclusion:

The resistance for current that passes through the solid in the $x$ direction is $0.375\times {10}^{-3}\Omega$ .

(b)

To determine

The resistance for current that passes through the solid in the $y$ direction.

Answer to Problem 21P

  $R=1.5\times {10}^{-3}\Omega$

Explanation of Solution

Given:

The resistivity, $\rho =3.0\times {10}^{-5}\Omega \cdot m$

A rectangular solid made of carbon has sides of lengths: $x=1.0cm$ , $y=2.0cm$ and $z=4.0cm$

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.

Calculation:

For x direction, current will flow from the right side to left side of the carbon rectangular. It means that the area (current path) is $A=xz$ and also, the path length of the current is $L=y$ .

By using the formula of resistance:

  $R=\frac{\rho L}{A}=\frac{\rho y}{xz}\mathrm{...}(i)$

Put the given values in the equation $\left(i\right)$

  $R=\frac{3.0\times {10}^{-5}\times 2.0\times {10}^{-2}}{1.0\times {10}^{-2}\times 4.0\times {10}^{-2}}=1.5\times {10}^{-3}\Omega$

Conclusion:

The resistance for current that passes through the solid in the $y$ direction is $1.5\times {10}^{-3}\Omega$ .

(c)

To determine

The resistance for current that passes through the solid in the $z$ direction.

Answer to Problem 21P

  $R=6.0\times {10}^{-3}\Omega$

Explanation of Solution

Given:

The resistivity, $\rho =3.0\times {10}^{-5}\Omega \cdot m$

A rectangular solid made of carbon has sides of lengths: $x=1.0cm$ , $y=2.0cm$ and $z=4.0cm$

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.

Calculation:

For x direction, current will flow from the right side to left side of the carbon rectangular. It means that the area (current path) is $A=xy$ and also, the path length of the current is $L=z$ .

By using the formula of resistance:

  $R=\frac{\rho L}{A}=\frac{\rho z}{xy}\mathrm{...}(i)$

Put the given values in the equation $\left(i\right)$

  $R=\frac{3.0\times {10}^{-5}\times 4.0\times {10}^{-2}}{1.0\times {10}^{-2}\times 2.0\times {10}^{-2}}=6.0\times {10}^{-3}\Omega$

Conclusion:

The resistance for current that passes through the solid in the $z$ direction is $6.0\times {10}^{-3}\Omega$ .