91影视

1. Diameter of wire A,${\mathrm{D}}_{\mathrm{A}}=1\mathrm{mm}$
2. Outer and inner diameters of wire B,${\mathrm{D}}_{\mathrm{Bo}}=2.0\mathrm{mm},\mathrm{and}{\mathrm{D}}_{\mathrm{Bi}}=1.0\mathrm{mm}$

The resistivity is the opposition to the flow of current. The resistance of the wire is directly proportional to its length and inversely proportional to the area of the cross-section. The constant of proportionality is a characteristic of that material called resistivity.

We have to use the formula of resistance for wire A and wire B and take their ratio to find the required resistance ratio.

Formulae:

The resistance of the material,$R=\frac{pL}{A}$ 鈥�(颈)

Here, R is resistance, p is resistivity, L is the length, A and is the area of the cross-section of the wire.

The cross-sectional area of the wire,$A=蟺r^2$ 鈥�(颈颈)

A is the area of the cross-section of the wire, r is the area of cross-section.

For wire A, the resistance of the wire can be calculated using equation (ii) in equation (i) as follows:

$R_A=\frac{pL}{蟺r_A^2}$ 鈥�(颈颈颈)

For wire B, the resistance of the wire can be calculated using equation (ii) in equation (i) as follows:

$R_B=\frac{pL}{蟺\left(r_{B0}^2-r_{Bi}^2\right)}$ 鈥�(颈惫)

But, the values of the radius of wire A, inner radius of B and outer radius of B can be calculated as follows:

$$\begin{gathered} {\mathrm{r}}_{\mathrm{A}}=\frac{{\mathrm{D}}_{\mathrm{A}}}{2} \\ =\frac{1\mathrm{mm}}{2} \\ =0.50\mathrm{mm} \end{gathered}$$

$$\begin{gathered} {\mathrm{r}}_{\mathrm{B}}=\frac{{\mathrm{D}}_{\mathrm{Bi}}}{2} \\ =\frac{1.0\mathrm{mm}}{2} \\ =0.50\mathrm{mm} \end{gathered}$$

$$\begin{gathered} {\mathrm{r}}_{\mathrm{Bo}}=\frac{{\mathrm{D}}_{Bo}}{2} \\ =\frac{2.0\mathrm{mm}}{2} \\ =1.0\mathrm{mm} \end{gathered}$$

Dividing equation (iii) by equation (iv) and using the above values, we can get the required value of the resistance ratio as follows:

$$\begin{gathered} \frac{{\mathrm{R}}_{\mathrm{A}}}{{\mathrm{R}}_{\mathrm{B}}}=\frac{{\displaystyle \frac{\frac{\mathrm{pL}}{{\mathrm{蟺谤}}_{\mathrm{A}}^2}}{\mathrm{pL}}}}{\mathrm{蟺}\left({\mathrm{r}}_{\mathrm{Bo}}^2-{\mathrm{r}}_{\mathrm{Bi}}^2\right)} \\ =\frac{\left({\mathrm{r}}_{\mathrm{Bo}}^2-{\mathrm{r}}_{\mathrm{Bi}}^2\right)}{{\mathrm{r}}_{\mathrm{A}}^2} \\ =\frac{{\left(1.00\mathrm{mm}\right)}^2-{\left(0.50\mathrm{mm}\right)}^2}{{\left(0.50\mathrm{mm}\right)}^2} \\ =3.0 \end{gathered}$$

Therefore, the resistance ratio $R_A/R_B$measured between the ends of the two conductors is 3.0 .