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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 resistors when connected in parallel have an equivalent of \(2 \Omega\) and when in series of \(9 \Omega\) The values of the two resistors are. (A) \(2 \Omega\) and \(9 \Omega\) (B) \(3 \Omega\) and \(6 \Omega\) (C) \(4 \Omega\) and \(5 \Omega\) (D) \(2 \Omega\) and \(7 \Omega\)

The potential difference between the terminals of a battery is $10 \mathrm{~V}\( and internal resistance \)1 \Omega\( drops to \)8 \mathrm{~V}$ when connected across an external resistor find the resistance of the external resistor. (A) \(40 \Omega\) (B) \(0.4 \Omega\) (C) \(4 \mathrm{M} \Omega\) (D) \(4 \Omega\)

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

If \(\sigma_{1}, \sigma_{2}\), and \(\sigma_{3}\) are the conductance's of three conductor then equivalent conductance when they are joined in series, will be. (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) $\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 / \sigma_{3}\right)$ (C) $\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right) /\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}$ (D) None of these.

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)
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