Question

What is the effect of the area of cross-section on the resistance of the conductor? How does the resistance of a wire change, when the area cross-section is

1. tripled
2. reduced to one-fourth?

Answer 1

Step 1- Relationship between resistance and area of cross-section

Resistance is a measure of opposition to the flow of current.

The equation for the resistance of a conductor is $R=\rho \frac{l}{A}$

Where $R$ is resistance, $\rho$ is resistivity, $l$ is the length of the wire and $A$ is the area of the cross section.

The resistance $R$ is inversely proportional to the area of the cross-section $A$ of the wire.

If the cross-section area increases, the resistance decreases. If the cross-section area decreases, the resistance increases.

Step 2- Finding the formula for the resistance after the cross-section area changes

Consider $R_i$ to be the initial resistance and $A_i$ to be the initial cross-section area.

Therefore, the expression for the initial resistance is

$R_i=\frac{\rho l}{A_i}...\left(1\right)$

Consider $R_f$ to be the final resistance and $A_f$ to be the final cross-section area.

$R_f=\frac{\rho l}{A_f}...\left(2\right)$

Divide equation $\left(2\right)$ with equation $\left(1\right)$ to obtain the expression for $R_f$.

$$\begin{gathered} \frac{R_f}{Ri}=\frac{{\displaystyle \frac{\rho l}{A_f}}}{{\displaystyle \frac{\rho l}{A_i}}} \\ \frac{R_f}{Ri}=\frac{A_i}{A_f} \\ {R}_f=R_i\left(\frac{A_i}{A_f}\right)...\left(3\right) \end{gathered}$$

Step 3- Finding the value of resistance after the area of cross-section is tripled

(a) Here, the area is tripled so, $A_f=3A_i$

Substitute $3A_i$ for $A_f$ into equation $\left(3\right)$.

$$\begin{gathered} R_f=R_i\left(\frac{A_i}{3A_f}\right) \\ {R}_f=\frac{1}{3}{R}_i \end{gathered}$$

Hence, the resistance reduces to one-third after the area of the cross-section is tripled.

Step 4- Finding the value of resistance after the area of cross-section is reduced to one-fourth

(b) Here, the area is reduced to one-fourth so, $A_f=\frac{1}{4}{A}_i$

Substitute $\frac{1}{4}{A}_i$ for $A_f$ into equation $\left(3\right)$.

$$\begin{gathered} R_f=R_i\left(\frac{A_i}{{\displaystyle \frac{1}{4}}{A}_i}\right) \\ {R}_f=4R_i \end{gathered}$$

Hence, the resistance becomes four times its initial value after the area of the cross-section is reduced to one-fourth.