Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Resistors Combined.

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Presentation transcript:

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Resistors Combined Both in Parallel and in Series Many complex circuits can be understood by isolating segments that are in series or in parallel and simplifying them to their equivalent resistances. Work backward to find the current in and potential difference across a part of a circuit.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 18 Analysis of Complex Circuits Section 3 Complex Resistor Combinations

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem Equivalent Resistance Determine the equivalent resistance of the complex circuit shown below.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Equivalent Resistance Reasoning The best approach is to divide the circuit into groups of series and parallel resistors. This way, the methods presented in Sample Problems A and B can be used to calculate the equivalent resistance for each group.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Equivalent Resistance 1. Redraw the circuit as a group of resistors along one side of the circuit. Because bends in a wire do not affect the circuit, they do not need to be represented in a schematic diagram. Redraw the circuit without the corners, keeping the arrangement of the circuit elements the same. TIP: For now, disregard the emf source, and work only with the resistances.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Equivalent Resistance 2. Identify components in series, and calculate their equivalent resistance. Resistors in group (a) and (b) are in series. For group (a): R eq = 3.0 Ω Ω = 9.0 Ω For group (b): R eq = 6.0 Ω Ω = 8.0 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Equivalent Resistance 3. Identify components in parallel, and calculate their equivalent resistance. Resistors in group (c) are in parallel.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Equivalent Resistance 4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single equivalent resistance.The remainder of the resistors, group (d), are in series.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem Current in and Potential Difference Across a Resistor Determine the current in and potential difference across the 2.0 Ω resistor highlighted in the figure below.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Current in and Potential Difference Across a Resistor Reasoning First determine the total circuit current by reducing the resistors to a single equivalent resistance. Then rebuild the circuit in steps, calculating the current and potential difference for the equivalent resistance of each group until the current in and potential difference across the 2.0 Ω resistor are known.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Current in and Potential Difference Across a Resistor 1. Determine the equivalent resistance of the circuit. The equivalent resistance of the circuit is 12.7 Ω, as calculated in the previous Sample Problem.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Current in and Potential Difference Across a Resistor 2. Calculate the total current in the circuit. Substitute the potential difference and equivalent resistance in ∆V = IR, and rearrange the equation to find the current delivered by the battery.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued 3. Determine a path from the equivalent resistance found in step 1 to the 2.0 Ω resistor. Review the path taken to find the equivalent resistance in the figure at right, and work backward through this path. The equivalent resistance for the entire circuit is the same as the equivalent resistance for group (d). The center resistor in group (d) in turn is the equivalent resistance for group (c). The top resistor in group (c) is the equivalent resistance for group (b), and the right resistor in group (b) is the 2.0 Ω resistor.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued Current in and Potential Difference Across a Resistor 4. Follow the path determined in step 3, and calculate the current in and potential difference across each equivalent resistance. Repeat this process until the desired values are found.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued 4. A. Regroup, evaluate, and calculate. Replace the circuit’s equivalent resistance with group (d). The resistors in group (d) are in series; therefore, the current in each resistor is the same as the current in the equivalent resistance, which equals 0.71 A. The potential difference across the 2.7 Ω resistor in group (d) can be calculated using ∆V = IR. Given: I = 0.71 A R = 2.7 Ω Unknown: ∆V = ? ∆V = IR = (0.71 A)(2.7 Ω) = 1.9 V

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued 4. B. Regroup, evaluate, and calculate. Replace the center resistor with group (c). The resistors in group (c) are in parallel; therefore, the potential difference across each resistor is the same as the potential difference across the 2.7 Ω equivalent resistance, which equals 1.9 V. The current in the 8.0 Ω resistor in group (c) can be calculated using ∆V = IR. Given: ∆V = 1.9 V R = 8.0 Ω Unknown: I = ?

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Complex Resistor Combinations Chapter 18 Sample Problem, continued 4. C. Regroup, evaluate, and calculate. Replace the 8.0 Ω resistor with group (b). The resistors in group (b) are in series; therefore, the current in each resistor is the same as the current in the 8.0 Ω equivalent resistance, which equals 0.24 A. The potential difference across the 2.0 Ω resistor can be calculated using ∆V = IR. Given: I = 0.24 A R = 2.0 Ω Unknown: ∆V = ?