91Ó°ÊÓ

Figure 26-16 showing cross sections of wires.

The amount of resistance in an electrical circuit represents the resistance to current flow. We use the concept of resistance related to resistivity, length, and area. The given material and length of wire are the same, so the resistance is only proportional to the area of the cross-section. We can find the cross-section area of each wire from the given dimensions and then using formulas for resistance we can determine the value of resistance. Then, we can rank the value of resistances from greatest to lowest.

Formula:

The resistance of the material,$R=\frac{\mathrm{ÒÏ}L}{A}$ …(¾±)

$\mathrm{ÒÏ}$is the resistivity, L is the length of cross section,A is the area of the cross section

First, we calculate cross section area of each wire given as,

Area of wire (a) is given using the data as:

$$\begin{gathered} A_a=I^2 \\ =4^2 \\ =16{\mathrm{mm}}^2 \end{gathered}$$

Area of the wire (b) is given using the data as follows:

$$\begin{gathered} A_b=2×5 \\ =10m{m}^2 \end{gathered}$$

Similarly, area of the wire (c) is given using the data as follows:

$$\begin{gathered} A_c=3×6 \\ =18{\mathrm{mm}}^2 \end{gathered}$$

Now, from equation (i), the given wires are of same material and length so the resistivity and length are same for all wires. Thus, the resistance will be proportional to only the area of cross section according to equation (i) as follows:

$$\begin{gathered} R_a=\frac{\mathrm{ÒÏ}L}{16} \\ =0.0625\mathrm{ÒÏ}L \end{gathered}$$

$$\begin{gathered} R_b=\frac{\mathrm{ÒÏ}L}{10} \\ =0.1\mathrm{ÒÏ}L \end{gathered}$$

$$\begin{gathered} R_c=\frac{\mathrm{ÒÏ}L}{18} \\ =0.055\mathrm{ÒÏ}L \end{gathered}$$

Therefore, the ranking of resistances is,$R_b>R_a>R_c$