Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
The next four questions refer to the figure below. Denote the current flowing through the leftmost resistor as I, middle resistor as IM, and rightmost resistor as IR. Use the sign convention that currents are positive if they flow in the directions indicated by the arrows. 20 18 V O 18 -2 IM + 2/₁ = 0 O 18-2/M-21₁ = 0 + Which is a correct application of Kirchoff's Loop Law applied to the leftmost loop involving only the 18 V battery, leftmost resistor, and middle resistor? O 18 -2 /M-2/L-2/R = 0 O 18 +2/M + 2/₁ = 0 O 18 +2 IM-2/L=0 2.2 203 -12V acer

The next four questions refer to the figure below. Denote the current flowing through the leftmost resistor as I, middle resistor as IM, and rightmost resistor as IR. Use the sign convention that currents are positive if they flow in the directions indicated by the arrows. 20 18 V O 18 -2 IM + 2/₁ = 0 O 18-2/M-21₁ = 0 + Which is a correct application of Kirchoff's Loop Law applied to the leftmost loop involving only the 18 V battery, leftmost resistor, and middle resistor? O 18 -2 /M-2/L-2/R = 0 O 18 +2/M + 2/₁ = 0 O 18 +2 IM-2/L=0 2.2 203 -12V acer

icon
Related questions
Question

B.) What is the current in the middle resistor

c.) imagine the resistors are lightbulbs, rank them from brightest to dimmest using leftmost as L then M for middle and R for right 

**Text Transcription and Diagram Explanation for Educational Website** --- **Problem Statement:** The next four questions refer to the figure below. Denote the current flowing through the leftmost resistor as \( I_L \), middle resistor as \( I_M \), and rightmost resistor as \( I_R \). Use the sign convention that currents are positive if they flow in the directions indicated by the arrows. **Circuit Diagram:** - There is a series circuit with the following components: - An 18 V battery with the positive terminal connected to a 2 Ω resistor on the left. - The leftmost 2 Ω resistor follows the 18 V battery. - A middle 2 Ω resistor is connected after the leftmost resistor. - The current flow is indicated by arrows pointing counter-clockwise through these two resistors. - A separate loop on the right consists of a 12 V battery connected to a 2 Ω resistor on the right. - The arrows indicate the current direction from the positive terminal of the 12 V battery, flowing clockwise through the rightmost resistor. **Question:** Which is a correct application of Kirchhoff's Loop Law applied to the leftmost loop involving only the 18 V battery, leftmost resistor, and middle resistor? 1. \( 18 - 2I_M + 2I_L = 0 \) 2. \( 18 - 2I_M - 2I_L = 0 \) 3. \( 18 + 2I_M + 2I_L = 0 \) 4. \( 18 + 2I_M - 2I_L = 0 \) **Diagram Explanation:** The circuit diagram illustrates a basic schematic with two loops. The left loop includes the 18 V battery and two resistors (both 2 Ω), while the right loop involves a 12 V battery and a single 2 Ω resistor. Arrows in the diagram specify the presumed direction of current flow for solving the problem using Kirchhoff’s Loop Law, which involves summing the electromotive forces and voltage drops around a closed circuit loop. --- This transcription provides a clear and detailed understanding of the problem and diagram, facilitating educational use of the content related to Kirchhoff's laws in electrical circuits.
Transcribed Image Text:**Text Transcription and Diagram Explanation for Educational Website** --- **Problem Statement:** The next four questions refer to the figure below. Denote the current flowing through the leftmost resistor as \( I_L \), middle resistor as \( I_M \), and rightmost resistor as \( I_R \). Use the sign convention that currents are positive if they flow in the directions indicated by the arrows. **Circuit Diagram:** - There is a series circuit with the following components: - An 18 V battery with the positive terminal connected to a 2 Ω resistor on the left. - The leftmost 2 Ω resistor follows the 18 V battery. - A middle 2 Ω resistor is connected after the leftmost resistor. - The current flow is indicated by arrows pointing counter-clockwise through these two resistors. - A separate loop on the right consists of a 12 V battery connected to a 2 Ω resistor on the right. - The arrows indicate the current direction from the positive terminal of the 12 V battery, flowing clockwise through the rightmost resistor. **Question:** Which is a correct application of Kirchhoff's Loop Law applied to the leftmost loop involving only the 18 V battery, leftmost resistor, and middle resistor? 1. \( 18 - 2I_M + 2I_L = 0 \) 2. \( 18 - 2I_M - 2I_L = 0 \) 3. \( 18 + 2I_M + 2I_L = 0 \) 4. \( 18 + 2I_M - 2I_L = 0 \) **Diagram Explanation:** The circuit diagram illustrates a basic schematic with two loops. The left loop includes the 18 V battery and two resistors (both 2 Ω), while the right loop involves a 12 V battery and a single 2 Ω resistor. Arrows in the diagram specify the presumed direction of current flow for solving the problem using Kirchhoff’s Loop Law, which involves summing the electromotive forces and voltage drops around a closed circuit loop. --- This transcription provides a clear and detailed understanding of the problem and diagram, facilitating educational use of the content related to Kirchhoff's laws in electrical circuits.
The image features an electrical circuit consisting of two voltage sources and three resistors. Here's a detailed description: **Circuit Diagram:** - There are two voltage sources in the circuit: an 18V battery on the left and a 12V battery on the right. - Three resistors, each labeled as 2Ω, are positioned within the circuit: one on the left, one in the middle, and one on the right. - The circuit is closed, forming two loops. Arrows indicate the current direction in each loop. **Question:** The challenge involves applying Kirchhoff's Loop Law to the rightmost loop of the circuit, which includes the 12V battery, the middle resistor, and the rightmost resistor. The goal is to determine which of the provided equations correctly represents Kirchhoff's Loop Law for this section of the circuit. **Options:** 1. \(12 + 2I_M + 2I_R = 0\) 2. \(12 + 2I_M - 2I_R = 0\) 3. \(12 - 2I_M - 2I_R = 0\) 4. \(12 + 2I_M + 2I_R + 2I_L = 0\) 5. \(12 - 2I_M + 2I_R = 0\) Here, \(I_M\) represents the current through the middle resistor, \(I_R\) represents the current through the rightmost resistor, and \(I_L\) might indicate current related to a left section of the circuit, but only options specific to the rightmost loop apply.
Transcribed Image Text:The image features an electrical circuit consisting of two voltage sources and three resistors. Here's a detailed description: **Circuit Diagram:** - There are two voltage sources in the circuit: an 18V battery on the left and a 12V battery on the right. - Three resistors, each labeled as 2Ω, are positioned within the circuit: one on the left, one in the middle, and one on the right. - The circuit is closed, forming two loops. Arrows indicate the current direction in each loop. **Question:** The challenge involves applying Kirchhoff's Loop Law to the rightmost loop of the circuit, which includes the 12V battery, the middle resistor, and the rightmost resistor. The goal is to determine which of the provided equations correctly represents Kirchhoff's Loop Law for this section of the circuit. **Options:** 1. \(12 + 2I_M + 2I_R = 0\) 2. \(12 + 2I_M - 2I_R = 0\) 3. \(12 - 2I_M - 2I_R = 0\) 4. \(12 + 2I_M + 2I_R + 2I_L = 0\) 5. \(12 - 2I_M + 2I_R = 0\) Here, \(I_M\) represents the current through the middle resistor, \(I_R\) represents the current through the rightmost resistor, and \(I_L\) might indicate current related to a left section of the circuit, but only options specific to the rightmost loop apply.
Expert Solution
Step 1

Given Data:
Let all the resistance = R=2 Ω.
Voltage on left side (VL)=18 V.
Voltage on right side (VR)=12 V.

steps

Step by step

Solved in 5 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer