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The circuit below has three parallel resistors R₁= 4KS, R₂=6 KQ and R3= 12 KS, and a supply voltage of 24 Volts. What the power dissipated across resistor R₂? 0.036Watt 0.072Watt 0.096Watt 0.024Watt Ovsuppl VSupply R₁ R₂ R3

The circuit below has three parallel resistors R₁= 4KS, R₂=6 KQ and R3= 12 KS, and a supply voltage of 24 Volts. What the power dissipated across resistor R₂? 0.036Watt 0.072Watt 0.096Watt 0.024Watt Ovsuppl VSupply R₁ R₂ R3

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
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**Problem Statement:** The circuit below has three parallel resistors \( R_1 = 4 \, \text{K}\Omega \), \( R_2 = 6 \, \text{K}\Omega \), and \( R_3 = 12 \, \text{K}\Omega \), with a supply voltage of 24 volts. What is the power dissipated across resistor \( R_2 \)? **Circuit Diagram:** The schematic shows a power supply (Vsupply) connected in parallel with three resistors: \( R_1 \), \( R_2 \), and \( R_3 \). The circuit is configured to distribute the same voltage across each resistor. **Question Options:** - \( 0.036 \, \text{Watt} \) - \( 0.072 \, \text{Watt} \) - \( 0.096 \, \text{Watt} \) - \( 0.024 \, \text{Watt} \) **Explanation:** To find the power dissipated across \( R_2 \), use the formula \( P = \frac{V^2}{R} \) for a resistor in parallel: - \( V = 24 \, \text{V} \) (voltage across \( R_2 \)) - \( R = 6 \, \text{K}\Omega = 6000 \, \Omega \) Substitute into the formula: \[ P = \frac{24^2}{6000} \] \[ P = \frac{576}{6000} \] \[ P = 0.096 \, \text{Watt} \] Thus, the power dissipated across resistor \( R_2 \) is \( 0.096 \, \text{Watt} \).
Transcribed Image Text:**Problem Statement:** The circuit below has three parallel resistors \( R_1 = 4 \, \text{K}\Omega \), \( R_2 = 6 \, \text{K}\Omega \), and \( R_3 = 12 \, \text{K}\Omega \), with a supply voltage of 24 volts. What is the power dissipated across resistor \( R_2 \)? **Circuit Diagram:** The schematic shows a power supply (Vsupply) connected in parallel with three resistors: \( R_1 \), \( R_2 \), and \( R_3 \). The circuit is configured to distribute the same voltage across each resistor. **Question Options:** - \( 0.036 \, \text{Watt} \) - \( 0.072 \, \text{Watt} \) - \( 0.096 \, \text{Watt} \) - \( 0.024 \, \text{Watt} \) **Explanation:** To find the power dissipated across \( R_2 \), use the formula \( P = \frac{V^2}{R} \) for a resistor in parallel: - \( V = 24 \, \text{V} \) (voltage across \( R_2 \)) - \( R = 6 \, \text{K}\Omega = 6000 \, \Omega \) Substitute into the formula: \[ P = \frac{24^2}{6000} \] \[ P = \frac{576}{6000} \] \[ P = 0.096 \, \text{Watt} \] Thus, the power dissipated across resistor \( R_2 \) is \( 0.096 \, \text{Watt} \).
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