Find v₂ and the power supplied by the source 9 V( + 1 KQ Lw 1.5 KQ 500 $2 Click here for the answer VOLTAGE 500 1000+500 +1500 V₂ = −9. POWER R eq : 3kQ = p= R 9² 3000 = = -1.5V 27 mW
Find v₂ and the power supplied by the source 9 V( + 1 KQ Lw 1.5 KQ 500 $2 Click here for the answer VOLTAGE 500 1000+500 +1500 V₂ = −9. POWER R eq : 3kQ = p= R 9² 3000 = = -1.5V 27 mW
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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KVL and KCL
KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.
Sign Convention
Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.
Question
Please explain the solution in detail? (Step-by-step solution).
Such as why 9 is negative and etc.
![## Problem Statement
Find \( v_2 \) and the power supplied by the source.
### Circuit Diagram
The circuit includes:
- A 9V voltage source
- Three resistors:
- \( 1k\Omega \)
- \( 500\Omega \)
- \( 1.5k\Omega \)
### Solution Steps
#### Voltage Calculation
The voltage \( v_2 \) across the 500\(\Omega\) resistor is calculated using the following formula:
\[
v_2 = -\frac{9 \times 500}{1000 + 500 + 1500} = -1.5V
\]
#### Power Calculation
The total equivalent resistance (\( R_{eq} \)) is calculated as:
\[
R_{eq} = 3k\Omega
\]
The power (\( p \)) supplied by the source is calculated using:
\[
p = \frac{v^2}{R} = \frac{9^2}{3000} = 27mW
\]
### Interactive Element
A button labeled "Click here for the answer" provides the solution details on interaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1da7c4d-4e7c-4d93-850d-d41f99bc457d%2F1f24ccd5-0f74-42c2-960c-9605e3385492%2F4g0rlaf_processed.png&w=3840&q=75)
Transcribed Image Text:## Problem Statement
Find \( v_2 \) and the power supplied by the source.
### Circuit Diagram
The circuit includes:
- A 9V voltage source
- Three resistors:
- \( 1k\Omega \)
- \( 500\Omega \)
- \( 1.5k\Omega \)
### Solution Steps
#### Voltage Calculation
The voltage \( v_2 \) across the 500\(\Omega\) resistor is calculated using the following formula:
\[
v_2 = -\frac{9 \times 500}{1000 + 500 + 1500} = -1.5V
\]
#### Power Calculation
The total equivalent resistance (\( R_{eq} \)) is calculated as:
\[
R_{eq} = 3k\Omega
\]
The power (\( p \)) supplied by the source is calculated using:
\[
p = \frac{v^2}{R} = \frac{9^2}{3000} = 27mW
\]
### Interactive Element
A button labeled "Click here for the answer" provides the solution details on interaction.
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