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Ch 26: Direct-Current Circuits
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 26, Problema 1

A uniform wire of resistance R is cut into three equal lengths. One of these is formed into a circle and connected between the other two (Fig. E26.1). What is the resistance between the opposite ends a and b?

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First, understand the configuration of the circuit. The wire of resistance R is divided into three equal parts, each with resistance R/3. One part is formed into a circle, and the other two parts are connected in series with the circle.
Identify the resistances in the circuit. The circle formed by one part of the wire has a resistance of R/3. The other two parts, each with resistance R/3, are connected in series with the circle.
Calculate the equivalent resistance of the circle. Since the circle is connected between two points, it acts as a parallel resistor to the other two resistors. The equivalent resistance of the circle is R/3.
Determine the total resistance between points a and b. The two resistors in series (each R/3) have a combined resistance of 2R/3. The circle (R/3) is in parallel with this series combination.
Use the formula for parallel resistors to find the total resistance: \( R_{total} = \frac{1}{\frac{1}{R_{series}} + \frac{1}{R_{circle}}} \), where \( R_{series} = 2R/3 \) and \( R_{circle} = R/3 \). Calculate the equivalent resistance using this formula.

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