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If R₁ = 63 Q and E₁ = 79 V, what is the power of R₁? (Round the FINAL answer to one decimal place.) & ww

If R₁ = 63 Q and E₁ = 79 V, what is the power of R₁? (Round the FINAL answer to one decimal place.) & ww

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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**Problem Statement:** If \( R_1 = 63 \, \Omega \) and \( E_1 = 79 \, \text{V} \), what is the power of \( R_1 \)? (Round the FINAL answer to one decimal place.) **Diagram Explanation:** The diagram shows an electrical circuit with three resistors, \( R_1 \), \( R_2 \), and \( R_3 \), arranged in parallel. The circuit is powered by a voltage source denoted as \( E_1 \). The resistors are connected such that each has its own loop with the same voltage across it due to the parallel configuration. **How to Solve:** To find the power across resistor \( R_1 \), use the formula for power dissipation in a resistor: \[ P = \frac{V^2}{R} \] where \( V \) is the voltage across the resistor and \( R \) is the resistance. Given: \[ R_1 = 63 \, \Omega, \quad E_1 = 79 \, \text{V} \] Apply the formula: \[ P = \frac{79^2}{63} \] **Calculation:** 1. \( 79^2 = 6241 \) 2. \( \frac{6241}{63} \approx 99.048 \) Round the answer to one decimal place: \[ P \approx 99.0 \, \text{W} \] **Conclusion:** The power dissipation across resistor \( R_1 \) is approximately \( 99.0 \, \text{W} \).
Transcribed Image Text:**Problem Statement:** If \( R_1 = 63 \, \Omega \) and \( E_1 = 79 \, \text{V} \), what is the power of \( R_1 \)? (Round the FINAL answer to one decimal place.) **Diagram Explanation:** The diagram shows an electrical circuit with three resistors, \( R_1 \), \( R_2 \), and \( R_3 \), arranged in parallel. The circuit is powered by a voltage source denoted as \( E_1 \). The resistors are connected such that each has its own loop with the same voltage across it due to the parallel configuration. **How to Solve:** To find the power across resistor \( R_1 \), use the formula for power dissipation in a resistor: \[ P = \frac{V^2}{R} \] where \( V \) is the voltage across the resistor and \( R \) is the resistance. Given: \[ R_1 = 63 \, \Omega, \quad E_1 = 79 \, \text{V} \] Apply the formula: \[ P = \frac{79^2}{63} \] **Calculation:** 1. \( 79^2 = 6241 \) 2. \( \frac{6241}{63} \approx 99.048 \) Round the answer to one decimal place: \[ P \approx 99.0 \, \text{W} \] **Conclusion:** The power dissipation across resistor \( R_1 \) is approximately \( 99.0 \, \text{W} \).
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