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Chapter 12: Problem 1669

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

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

Step by step solution

01

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.
02

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: \(R_{2} = \frac{\rho_{2}L}{A}\)
03

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: \(R_{eq} = \frac{L}{A}(\rho_{1}+\rho_{2})\)
04

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) \(\left(\rho_{1}+\rho_{2}\right)\)

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Most popular questions from this chapter

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)

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In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)
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