Suppose in a circuit, a silver wire of length L and cross-section area A is replaced by a chromium wire of length L/3 and cross section area is 3A. The resistance of the Circuit will ______.

(Given that $$\rho_{silver} = 1.6 \times 10^{-8}\Omega$$m and $$\rho_{chromium} = 12.9 \times 10^{-8}\Omega m$$)

Correct option is A

Correct Answer: A (decrease to 0.89 times of itself)

Explanation:

The resistance (R) of a conductor is given by the formula:

​$$R = \rho \frac{L}{A}$$​​

Where:

• ρ: Resistivity of the material (Ω·m)
• L: Length of the conductor (m)
• A: Cross-sectional area (m²)

Step 1: Initial Resistance of Silver Wire

For the silver wire:

​$$R_\text{silver} = \rho_\text{silver} \frac{L}{A}$$​​

Substitute the given values:

​$$R_\text{silver} = (1.6 \times 10^{-8}) \frac{L}{A}$$​​

Step 2: Resistance of Chromium Wire

For the chromium wire:

​$$R_\text{chromium} = \rho_\text{chromium} \frac{L_\text{chromium}}{A_\text{chromium}}$$​​

The length of the chromium wire is Lchromium=$$L_\text{chromium} = \frac{L}{3}$$​​, and the cross-sectional area is A_chromium=3A. Substituting these values:

​$$R_\text{chromium} = (12.9 \times 10^{-8}) \frac{\frac{L}{3}}{3A}$$​​

Simplify:

​$$R_\text{chromium} = (12.9 \times 10^{-8}) \frac{L}{9A}$$​​

Step 3: Ratio of Resistances

The ratio of chromium resistance to silver resistance is:

​$$\frac{R_\text{chromium}}{R_\text{silver}} = \frac{(12.9 \times 10^{-8}) \frac{L}{9A}}{(1.6 \times 10^{-8}) \frac{L}{A}}$$​​

Simplify:

​$$\frac{R_\text{chromium}}{R_\text{silver}} = \frac{12.9}{1.6} \cdot \frac{1}{9}\\ \ \\\frac{R_\text{chromium}}{R_\text{silver}} = \frac{12.9}{14.4} \approx 0.89$$​​

Thus, the resistance of the circuit decreases to 0.89 times the original value.

Information Booster:

● Resistance decreases because the chromium wire's shorter length (L/3) and increased cross-sectional area (3A) outweigh its higher resistivity.
● Silver is highly conductive due to its low resistivity (1.6×10⁻⁸Ω m), making it an excellent material for electrical circuits.
● Chromium's higher resistivity (12.9×10⁻⁸ Ω m) is mitigated by its adjusted dimensions in this setup.
● When replacing a conductor in a circuit, both resistivity and geometric factors (length and area) must be considered.
● Lower resistance generally results in better electrical efficiency.