Question

Figure 26-15 shows cross sections through three long conductors of the same length and material, with square cross sections of edge lengths as shown. Conductor Bfits snugly within conductor A, and conductor Cfits snugly within conductor B. Rank the following according to their end-to-end resistances, greatest first: the individual conductors and the combinations of A+B(Binside A),B+C(Cinside B), and A+B+C(Binside Ainside C).

Short answer

The rank of resistances from greatest to lowest is $R_A=R_B=R_C>R_B+R_C=R_A+R_B>R_A+R_B+R_C$.

Step-by-step solution

The given data

Figure 26-15 is the cross sections of conductors.

Understanding the concept of the resistance

We use the formula of resistance related to resistivity, length, and area. Here, we have given the material and length as the same for the given conductors, so the resistivity and length are the same for all. So, the resistance will be proportional to only the area of the cross-section of the conductor. Using the proportionality relation, we can rank the resistances for given combinations.

Formula:

The resistance value of a material,$R=\frac{\rho L}{A}$ …(i)

Calculation of the ranking of the resistances

We can calculate the area of cross section of each conductor as,

Area of conductor C is$l^2$

Area of conductor B is given as:

$$\begin{gathered} {\left(\sqrt{2}l\right)}^2-l^2=2l^2-l^2 \\ =l^2 \end{gathered}$$

Similarly, area of conductor A is given as follows:

$$\begin{gathered} {\left(\sqrt{3}l\right)}^2-l^2-l^2=3l^2-2l^2 \\ =l^2 \end{gathered}$$

This gives that the area of cross section of the given conductors as the same.

Now using equation (i) for resistance, we see that thegiven conductors are of same material and length so the resistivity and length are the same for each conductor. Thus, the resistance depends on only the area of cross section of the conductor as,

role="math" localid="1661412007577" $R\propto \frac{1}{A}$

The resistance of given combination of conductors are thus given in terms of length as follows:

For combination of$A+B$we get,

role="math" localid="1661412127838" $$\begin{gathered} R_A+R_B\propto \frac{1}{l^2}+\frac{1}{l^2} \\ \propto \frac{2}{l^2} \end{gathered}$$

For$B+C$we get,

$$\begin{gathered} R_B+R_C\propto \frac{1}{l^2}+\frac{1}{l^2} \\ \propto \frac{2}{l^2} \end{gathered}$$

For$A+B+C$we get, the resistance value as follows:

$$\begin{gathered} R_A+R_B+R_C\propto \frac{1}{l^2}+\frac{1}{l^2}+\frac{1}{l^2} \\ \propto \frac{3}{l^2} \end{gathered}$$

From this, we can rank the resistances from greatest to lowest as follows,

$R_A=R_B=R_C>R_B+R_C=R_A+R_B>R_A+R_B+R_C$