Question

Two wires of equal lengths, equal diameters and having resistivities \(\rho_{1}\) and \(\rho_{2}\) are connected in series The equivalent resistivity of the combination is.... (A) \(\left(\rho_{1}+\rho_{2}\right)\) (B) \((1 / 2)\left(\rho_{1}+\rho_{2}\right)\) (C) \(\left\\{\left(\rho_{1} \rho_{2}\right) /\left(\rho_{1}+\rho_{2}\right)\right\\}\) (D) \(\left.\sqrt{(} \rho_{1} \rho_{2}\right)\)

Short answer

The short answer is: The equivalent resistivity of the combination is (A) \(\left(\rho_{1}+\rho_{2}\right)\).

Step-by-step solution

Understand the given information

We are given that two wires have equal lengths (L) and equal diameters. Their resistivities are ρ₁ and ρ₂. The wires are connected in series.

Calculate the resistance of each wire

Since the two wires have the same length and diameter, let's denote their length as L, and cross-sectional area as A. To find the resistance (R) of each wire, we can use the formula for resistivity: \(R = \frac{\rho L}{A}\) For Wire 1: \(R_{1} = \frac{\rho_{1}L}{A}\) For Wire 2: \(R_{2} = \frac{\rho_{2}L}{A}\)

Calculate the equivalent resistance in series

When resistors (or in this case, wires) are connected in series, their equivalent resistance is simply the sum of the individual resistances. So, we have: \(R_{eq} = R_{1} + R_{2}\) Substitute the values of \(R_{1}\) and \(R_{2}\) from Step 2: \(R_{eq} = \frac{\rho_{1}L}{A} + \frac{\rho_{2}L}{A}\) Now, take the common factor L/A out: \(R_{eq} = \frac{L}{A}(\rho_{1}+\rho_{2})\)

Calculate the equivalent resistivity

Since we now have the equivalent resistance, we can find the equivalent resistivity (ρ_eq) using the resistivity formula: \(\rho_{eq} = \frac{R_{eq}A}{L}\) Substitute the earlier found value of \(R_{eq}\): \(\rho_{eq} = \frac{(\frac{L}{A}(\rho_{1}+\rho_{2}))A}{L}\) Simplify the equation: \(\rho_{eq} = (\rho_{1}+\rho_{2})\) Hence, the equivalent resistivity of the combination is: (A) \(\left(\rho_{1}+\rho_{2}\right)\)

Key concepts

Resistivity

Resistivity is a fundamental property of materials that measures how strongly a material opposes the flow of electric current. It is denoted by \( \rho \) (rho) and is intrinsic to the material, irrespective of its shape or size. The unit of resistivity is ohm-meter (\( \Omega \cdot m\)). The lower the resistivity, the better the material conducts electricity.
The formula for resistivity is given by:

• \( R = \frac{\rho L}{A} \)

where:

• \( R \) is the resistance of the material
• \( L \) is the length of the material
• \( A \) is the cross-sectional area

Understanding resistivity is vital for determining how different materials can be used in electrical applications. Materials with high resistivity, such as rubber, are typically used as insulators, while materials with low resistivity, like copper, are used as conductors.

Series Resistance

Series resistance refers to the total resistance experienced by a current as it passes through multiple resistive components lined up end-to-end in a single path. In a series circuit, the total resistance \( R_{eq} \) is simply the sum of the individual resistances. This is because the current has only one path to follow and has to pass through each resistor sequentially.
The formula is:

• \( R_{eq} = R_1 + R_2 + R_3 + \ldots + R_n \)

Using series resistance simplifies calculations but can also mean a higher total resistance compared to parallel configurations. A key takeaway is that in a series circuit, any increase in one resistor directly increases the overall resistance. This principle is useful in controlling the overall resistance in circuits and is essential in designing electrical devices where precise resistance values are needed.

Resistance Calculation

To calculate resistance, whether individual or equivalent, you need the resistivity \( \rho \), the length \( L \), and the cross-sectional area \( A \). The formula \( R = \frac{\rho L}{A} \) allows you to determine the resistance of any given material setup.
In the example problem, each wire resistance was calculated as:

• For Wire 1: \( R_1 = \frac{\rho_1 L}{A} \)
• For Wire 2: \( R_2 = \frac{\rho_2 L}{A} \)

Calculating the equivalent resistance in series involved adding both resistances:

• \( R_{eq} = R_1 + R_2 \)

Which was simplified to:

• \( R_{eq} = \frac{L}{A}(\rho_1 + \rho_2) \)

And from the equivalent resistance, the equivalent resistivity was derived to be \((\rho_1 + \rho_2)\). This calculation is crucial when designing circuits to ensure they have the desired resistance characteristics.

JEE Physics

The Joint Entrance Examination (JEE) Physics section challenges students to apply physics concepts to real-world problems. It covers various topics, including electricity and magnetism, where understanding concepts like resistivity and series resistance is crucial.
The JEE Physics section often requires students to solve problems involving circuit analysis. Competency in these topics helps students perform well on questions involving:

• Resistance and resistivity problems
• Circuit configurations and their equivalent calculations
• Application of formulas in innovative ways

Thorough preparation and mastery of these areas not only aid in JEE but also in further studies and applications in electrical engineering and physics. Understanding and effectively applying these principles is a testament to one's problem-solving skills and conceptual comprehension.