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For the lower loop of the circuit below, which equation is resulted from a correct use of Kirchhoff's loop rule? Assume the clockwise direction is used for the loop. 18.0 V 5.00 N 8.00 N 12.0 V 11.0 2 7.00 N 5.00 N 36.0 V O -11 I2 + 12 - 7 I2 + 36 + 5 I3 = 0 O -11 12 - 12 - 7 I2 - 36 - 5 I3 = o O 11 12 + 12 +7 I2 - 36 + 5 13 = 0 O -11 I2 + 12 - 7 12 - 36 + 5 13 = 0 O -11 I2 + 12 - 7 12 - 36 - 5 I3 = 0

For the lower loop of the circuit below, which equation is resulted from a correct use of Kirchhoff's loop rule? Assume the clockwise direction is used for the loop. 18.0 V 5.00 N 8.00 N 12.0 V 11.0 2 7.00 N 5.00 N 36.0 V O -11 I2 + 12 - 7 I2 + 36 + 5 I3 = 0 O -11 12 - 12 - 7 I2 - 36 - 5 I3 = o O 11 12 + 12 +7 I2 - 36 + 5 13 = 0 O -11 I2 + 12 - 7 12 - 36 + 5 13 = 0 O -11 I2 + 12 - 7 12 - 36 - 5 I3 = 0

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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### Educational Website Content on Kirchhoff's Loop Rule #### Problem Statement For the lower loop of the circuit shown below, which equation results from a correct use of Kirchhoff’s loop rule? Assume the clockwise direction is used for the loop. #### Circuit Diagram Overview - The circuit contains three separate loops with various resistors and voltage sources. - There is a voltage source of 18.0 V at the top left and a 36.0 V at the bottom right. - The resistances given in the circuit are: - 5.00 Ω on the top right - 11.0 Ω at the left center - 8.00 Ω on the left - 12.0 Ω in the center - 7.00 Ω at the center right - 5.00 Ω at the bottom - Currents are indicated with arrows I₁, I₂, and I₃. #### Equations from Kirchhoff's Loop Rule Choose the correct equation for the lower loop: - \( \bigcirc \) \(-11I_2 + 12 + 7I_2 + 36 + 5I_3 = 0\) - \( \bigcirc \) \(-11I_2 - 12 - 7I_2 + 36 - 5I_3 = 0\) - \( \bigcirc \) \(11I_2 + 7I_2 + 36 + 5I_3 = 0\) - \( \bigcirc \) \(-11I_2 + 12 - 7I_2 - 36 + 5I_3 = 0\) - \( \bigcirc \) \(11I_2 + 12 - 7I_2 - 36 + 5I_3 = 0\) #### Explanation of Diagrams and Equations The circuit is analyzed by applying Kirchhoff’s loop rule, which states that the sum of the potential differences (voltage) around any closed loop in a circuit must equal zero. The resistances and voltage sources must be accurately taken into account as you move through the loop in a specified direction (clockwise, in this instance). Reviewing the mentioned equations, substitute the resistances and the voltage sources accurately according to the loop rule for the specific path in question. ### Conclusion Apply Kirchhoff’s
Transcribed Image Text:### Educational Website Content on Kirchhoff's Loop Rule #### Problem Statement For the lower loop of the circuit shown below, which equation results from a correct use of Kirchhoff’s loop rule? Assume the clockwise direction is used for the loop. #### Circuit Diagram Overview - The circuit contains three separate loops with various resistors and voltage sources. - There is a voltage source of 18.0 V at the top left and a 36.0 V at the bottom right. - The resistances given in the circuit are: - 5.00 Ω on the top right - 11.0 Ω at the left center - 8.00 Ω on the left - 12.0 Ω in the center - 7.00 Ω at the center right - 5.00 Ω at the bottom - Currents are indicated with arrows I₁, I₂, and I₃. #### Equations from Kirchhoff's Loop Rule Choose the correct equation for the lower loop: - \( \bigcirc \) \(-11I_2 + 12 + 7I_2 + 36 + 5I_3 = 0\) - \( \bigcirc \) \(-11I_2 - 12 - 7I_2 + 36 - 5I_3 = 0\) - \( \bigcirc \) \(11I_2 + 7I_2 + 36 + 5I_3 = 0\) - \( \bigcirc \) \(-11I_2 + 12 - 7I_2 - 36 + 5I_3 = 0\) - \( \bigcirc \) \(11I_2 + 12 - 7I_2 - 36 + 5I_3 = 0\) #### Explanation of Diagrams and Equations The circuit is analyzed by applying Kirchhoff’s loop rule, which states that the sum of the potential differences (voltage) around any closed loop in a circuit must equal zero. The resistances and voltage sources must be accurately taken into account as you move through the loop in a specified direction (clockwise, in this instance). Reviewing the mentioned equations, substitute the resistances and the voltage sources accurately according to the loop rule for the specific path in question. ### Conclusion Apply Kirchhoff’s
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