Electricity: Resistance and Cross-Sectional Area
Question: If the resistance of wire A is four times the resistance of wire B, find the ratio of their cross sectional areas.
Solution:
Let the resistance of wire A be $R_1$ and that of wire B be $R_2$.
According to the given problem:
$$R_1 = 4 R_2$$
Rearranging this gives us the ratio of their resistances:
$$\frac{R_1}{R_2} = \frac{4}{1} \quad \text{-------- Equation (1)}$$
We know that resistance is inversely proportional to the cross-sectional area, assuming the wires are of the same length and material:
$$R \propto \frac{1}{A}$$
Using the formula for resistance ($R = \rho \frac{l}{A}$), and assuming unit length ($l=1$) for comparison:
$$R = \rho \frac{1}{A}$$
(Where $\rho$ is a constant called resistivity)
Writing this for both wires:
$$R_1 = \rho \frac{1}{A_1}$$
$$R_2 = \rho \frac{1}{A_2}$$
Now, divide $R_1$ by $R_2$:
$$\frac{R_1}{R_2} = \frac{\rho \frac{1}{A_1}}{\rho \frac{1}{A_2}}$$
The resistivity ($\rho$) cancels out, leaving:
$$\frac{R_1}{R_2} = \frac{A_2}{A_1}$$
Substitute the ratio from Equation (1):
$$\frac{4}{1} = \frac{A_2}{A_1}$$
Inverting both sides to find the ratio of $A_1$ to $A_2$:
$$A_1 : A_2 = 1 : 4$$
Answer: The ratio of the cross-sectional areas of the wires is 1:4.