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Find the total equivalent resistance of the circuit diagrammed below: The battery is hooked up between the open leads at points "a" and "d". R1 = 11.0 Ω, R2 = 3.00 Ω, R3 = 1.00 Ω, R4 = 12.0 Ω, R5 = 7.00 Ω, R6 = 4.00 Ω, and R7 = 4.00 Ω

Find the total equivalent resistance of the circuit diagrammed below: The battery is hooked up between the open leads at points "a" and "d". R1 = 11.0 Ω, R2 = 3.00 Ω, R3 = 1.00 Ω, R4 = 12.0 Ω, R5 = 7.00 Ω, R6 = 4.00 Ω, and R7 = 4.00 Ω

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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  1. Find the total equivalent resistance of the circuit diagrammed below: The battery is hooked up between the open leads at points "a" and "d". R1 = 11.0 Ω, R2 = 3.00 Ω, R3 = 1.00 Ω, R4 = 12.0 Ω, R5 = 7.00 Ω, R6 = 4.00 Ω, and R7 = 4.00 Ω

     

      a.

    18.7 Ω
      b.

    24.7 Ω
      c.

    15.7 Ω
      d.

    12.7 Ω
      e.

    21.7 Ω
**Problem: Equivalent Resistance Calculation** Find the total equivalent resistance of the circuit diagrammed below. The battery is connected between the open leads at points "a" and "d". **Resistor Values:** - \( R_1 = 11.0 \, \Omega \) - \( R_2 = 3.00 \, \Omega \) - \( R_3 = 1.00 \, \Omega \) - \( R_4 = 12.0 \, \Omega \) - \( R_5 = 7.00 \, \Omega \) - \( R_6 = 4.00 \, \Omega \) - \( R_7 = 4.00 \, \Omega \) **Circuit Diagram:** The circuit is arranged as follows: - \( R_1 \) connects directly from point "a" to point "b". - \( R_2 \) and \( R_3 \) are in series between points "b" and the top of \( R_4 \). - \( R_4 \) connects from this junction to point "c". - \( R_6 \) and \( R_7 \) are in parallel, connecting between point "b" and "c". - \( R_5 \) connects between point "c" and point "d". **Multiple Choice Answers:** - a. 18.7 \(\Omega\) - b. 24.7 \(\Omega\) - c. 15.7 \(\Omega\) - d. 12.7 \(\Omega\) - e. 21.7 \(\Omega\) **Explanation:** The circuit consists of a mix of series and parallel resistors arranged between points "a" and "d". Calculate the equivalent resistance by strategically reducing complex parts of the circuit step by step—starting from the simplest series and parallel combinations and moving toward the larger sections.
Transcribed Image Text:**Problem: Equivalent Resistance Calculation** Find the total equivalent resistance of the circuit diagrammed below. The battery is connected between the open leads at points "a" and "d". **Resistor Values:** - \( R_1 = 11.0 \, \Omega \) - \( R_2 = 3.00 \, \Omega \) - \( R_3 = 1.00 \, \Omega \) - \( R_4 = 12.0 \, \Omega \) - \( R_5 = 7.00 \, \Omega \) - \( R_6 = 4.00 \, \Omega \) - \( R_7 = 4.00 \, \Omega \) **Circuit Diagram:** The circuit is arranged as follows: - \( R_1 \) connects directly from point "a" to point "b". - \( R_2 \) and \( R_3 \) are in series between points "b" and the top of \( R_4 \). - \( R_4 \) connects from this junction to point "c". - \( R_6 \) and \( R_7 \) are in parallel, connecting between point "b" and "c". - \( R_5 \) connects between point "c" and point "d". **Multiple Choice Answers:** - a. 18.7 \(\Omega\) - b. 24.7 \(\Omega\) - c. 15.7 \(\Omega\) - d. 12.7 \(\Omega\) - e. 21.7 \(\Omega\) **Explanation:** The circuit consists of a mix of series and parallel resistors arranged between points "a" and "d". Calculate the equivalent resistance by strategically reducing complex parts of the circuit step by step—starting from the simplest series and parallel combinations and moving toward the larger sections.
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