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I₁ 4Ω 12 V. 5 02- -41₁ +6+713 = 0 O 13 + 12 + 11 = 0 Consider the circuit in the figure and identify which one of the following is not a correct equation generated with Kirchhoff's Rules. 792 ww 12 -713-6+512 - 12 = 0 O 12-512-411 = 0 O-41₁ +51₂ -12 = 0 6 V

I₁ 4Ω 12 V. 5 02- -41₁ +6+713 = 0 O 13 + 12 + 11 = 0 Consider the circuit in the figure and identify which one of the following is not a correct equation generated with Kirchhoff's Rules. 792 ww 12 -713-6+512 - 12 = 0 O 12-512-411 = 0 O-41₁ +51₂ -12 = 0 6 V

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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**Transcription for Educational Website:** --- **Circuit Analysis with Kirchhoff's Rules** The diagram features an electrical circuit consisting of three resistors and two voltage sources. Here's a detailed description of the components and their arrangement: - **Resistors**: - \(4 \, \Omega\) (top left) - \(7 \, \Omega\) (top right) - \(5 \, \Omega\) (center) - **Voltage Sources**: - \(12 \, \text{V}\) (connected across the \(5 \, \Omega\) resistor) - \(6 \, \text{V}\) (right side, connected to \(I_3\)) - **Currents**: - \(I_1\) (entering through \(4 \, \Omega\) resistor) - \(I_2\) (through \(5 \, \Omega\) resistor, opposite \(12 \, \text{V}\) source) - \(I_3\) (exiting through \(6 \, \text{V}\) source) **Task:** Consider the circuit in the figure and identify which one of the following is not a correct equation generated with Kirchhoff's Rules. **Options:** 1. \( -4I_1 + 6 + 7I_3 = 0 \) 2. \( I_3 + I_2 + I_1 = 0 \) 3. \( -7I_3 - 6 + 5I_2 - 12 = 0 \) 4. \( 12 - 5I_2 - 4I_1 = 0 \) 5. \( -4I_1 + 5I_2 - 12 = 0 \) --- This exercise requires analyzing the provided circuit and determining the equation that does not align with Kirchhoff’s Voltage and Current Laws, fundamental tools used in circuit analysis.
Transcribed Image Text:**Transcription for Educational Website:** --- **Circuit Analysis with Kirchhoff's Rules** The diagram features an electrical circuit consisting of three resistors and two voltage sources. Here's a detailed description of the components and their arrangement: - **Resistors**: - \(4 \, \Omega\) (top left) - \(7 \, \Omega\) (top right) - \(5 \, \Omega\) (center) - **Voltage Sources**: - \(12 \, \text{V}\) (connected across the \(5 \, \Omega\) resistor) - \(6 \, \text{V}\) (right side, connected to \(I_3\)) - **Currents**: - \(I_1\) (entering through \(4 \, \Omega\) resistor) - \(I_2\) (through \(5 \, \Omega\) resistor, opposite \(12 \, \text{V}\) source) - \(I_3\) (exiting through \(6 \, \text{V}\) source) **Task:** Consider the circuit in the figure and identify which one of the following is not a correct equation generated with Kirchhoff's Rules. **Options:** 1. \( -4I_1 + 6 + 7I_3 = 0 \) 2. \( I_3 + I_2 + I_1 = 0 \) 3. \( -7I_3 - 6 + 5I_2 - 12 = 0 \) 4. \( 12 - 5I_2 - 4I_1 = 0 \) 5. \( -4I_1 + 5I_2 - 12 = 0 \) --- This exercise requires analyzing the provided circuit and determining the equation that does not align with Kirchhoff’s Voltage and Current Laws, fundamental tools used in circuit analysis.
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Step 1: Data given

An electric circuit is given

We can divide it into two loops and can apply kirchoff voltage rule for two loops separately.


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