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For the circuit shown, the bottom node is chosen as the reference node. What is the Node Voltage Method equation at node a? Va Va j3 Ω 12-45° A 4Ω -j10 4Ω Ve j2 n ↑ ) 4215° A -j5 N Va-Va j3 - (12 – 45°) = 0 Va-Va j3 - (12 – 45°) = 0 O Va-Va - 0 j3 V-Va j3 + (12 – 45°) = 0

For the circuit shown, the bottom node is chosen as the reference node. What is the Node Voltage Method equation at node a? Va Va j3 Ω 12-45° A 4Ω -j10 4Ω Ve j2 n ↑ ) 4215° A -j5 N Va-Va j3 - (12 – 45°) = 0 Va-Va j3 - (12 – 45°) = 0 O Va-Va - 0 j3 V-Va j3 + (12 – 45°) = 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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**Transcription for Educational Website** **Circuit Analysis Using Node Voltage Method** For the circuit shown, the bottom node is chosen as the reference node. What is the Node Voltage Method equation at node \( a \)? **Circuit Description:** - The circuit features three nodes: \( V_a \), \( V_b \), and \( V_c \). - There is a current source of \( 1 \angle -45^\circ \) A pointing upwards at node \( V_a \). - The impedances connected between nodes are as follows: - Between \( V_a \) and \( V_b \): \( j3 \, \Omega \) - Between \( V_a \) and the reference node: \( -j1 \, \Omega \) - Between \( V_a \) and \( V_c \): \( 4 \, \Omega \) - Between \( V_b \) and the reference node: \( j2 \, \Omega \) - Between \( V_c \) and the reference node: \( -j5 \, \Omega \) - Node \( V_c \) has a current source of \( 4 \angle 15^\circ \) A pointing towards the bottom node (reference node). **Answer Choices for Node \( a \):** 1. \(\frac{V_a - V_d}{j3} - (1 \angle -45^\circ) = 0\) 2. \(\frac{V_d - V_a}{j3} - (1 \angle -45^\circ) = 0\) 3. \(\frac{V_a - V_d}{j3} = 0\) 4. \(\frac{V_a - V_d}{j3} + (1 \angle -45^\circ) = 0\) To solve the problem, apply the node voltage method at node \( a \) considering the impedance and current sources.
Transcribed Image Text:**Transcription for Educational Website** **Circuit Analysis Using Node Voltage Method** For the circuit shown, the bottom node is chosen as the reference node. What is the Node Voltage Method equation at node \( a \)? **Circuit Description:** - The circuit features three nodes: \( V_a \), \( V_b \), and \( V_c \). - There is a current source of \( 1 \angle -45^\circ \) A pointing upwards at node \( V_a \). - The impedances connected between nodes are as follows: - Between \( V_a \) and \( V_b \): \( j3 \, \Omega \) - Between \( V_a \) and the reference node: \( -j1 \, \Omega \) - Between \( V_a \) and \( V_c \): \( 4 \, \Omega \) - Between \( V_b \) and the reference node: \( j2 \, \Omega \) - Between \( V_c \) and the reference node: \( -j5 \, \Omega \) - Node \( V_c \) has a current source of \( 4 \angle 15^\circ \) A pointing towards the bottom node (reference node). **Answer Choices for Node \( a \):** 1. \(\frac{V_a - V_d}{j3} - (1 \angle -45^\circ) = 0\) 2. \(\frac{V_d - V_a}{j3} - (1 \angle -45^\circ) = 0\) 3. \(\frac{V_a - V_d}{j3} = 0\) 4. \(\frac{V_a - V_d}{j3} + (1 \angle -45^\circ) = 0\) To solve the problem, apply the node voltage method at node \( a \) considering the impedance and current sources.
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