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Determine the following for the network of bulbs that are purely resistive. 60 W 0 E + 240 V 45 W Total Power dissipation (PT) = Total reactive power (QT) = Total apparent power (|ST|) = Total current (Is) = 14 25 W units units units untis 171

Determine the following for the network of bulbs that are purely resistive. 60 W 0 E + 240 V 45 W Total Power dissipation (PT) = Total reactive power (QT) = Total apparent power (|ST|) = Total current (Is) = 14 25 W units units units untis 171

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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### Electrical Circuit Analysis for Purely Resistive Bulbs **Objective**: Determine the following for the network of bulbs that are purely resistive. #### Circuit Description The circuit consists of a 240 V AC source connected to three resistive bulbs with power ratings of 60 W, 45 W, and 25 W respectively. The components are wired in parallel. - **Voltage Source**: 240 volts - **Bulb 1**: 60 W - **Bulb 2**: 45 W - **Bulb 3**: 25 W #### Parameters to Determine 1. **Total Power Dissipation (Pₜ)**: - Formula: \( P_T = P_1 + P_2 + P_3 \) 2. **Total Reactive Power (Qₜ)**: - For purely resistive loads, \( Q_T = 0 \). 3. **Total Apparent Power (\(|S_T|\))**: - Formula: \(|S_T| = \sqrt{P_T^2 + Q_T^2}\) - Simplifies to \( |S_T| = P_T \) for purely resistive loads. 4. **Total Current (Iₛ)**: - Formula: \( I_s = \frac{P_T}{E} \) #### Input Fields - **Total Power Dissipation (Pₜ)** = _______ units - **Total Reactive Power (Qₜ)** = _______ units - **Total Apparent Power (\(|S_T|\))** = _______ units - **Total Current (Iₛ)** = _______ units ### Instructions Calculate each parameter using the given formulas and fill in the blanks with the correct units. This exercise provides foundational understanding of power analysis in circuits with purely resistive elements.
Transcribed Image Text:### Electrical Circuit Analysis for Purely Resistive Bulbs **Objective**: Determine the following for the network of bulbs that are purely resistive. #### Circuit Description The circuit consists of a 240 V AC source connected to three resistive bulbs with power ratings of 60 W, 45 W, and 25 W respectively. The components are wired in parallel. - **Voltage Source**: 240 volts - **Bulb 1**: 60 W - **Bulb 2**: 45 W - **Bulb 3**: 25 W #### Parameters to Determine 1. **Total Power Dissipation (Pₜ)**: - Formula: \( P_T = P_1 + P_2 + P_3 \) 2. **Total Reactive Power (Qₜ)**: - For purely resistive loads, \( Q_T = 0 \). 3. **Total Apparent Power (\(|S_T|\))**: - Formula: \(|S_T| = \sqrt{P_T^2 + Q_T^2}\) - Simplifies to \( |S_T| = P_T \) for purely resistive loads. 4. **Total Current (Iₛ)**: - Formula: \( I_s = \frac{P_T}{E} \) #### Input Fields - **Total Power Dissipation (Pₜ)** = _______ units - **Total Reactive Power (Qₜ)** = _______ units - **Total Apparent Power (\(|S_T|\))** = _______ units - **Total Current (Iₛ)** = _______ units ### Instructions Calculate each parameter using the given formulas and fill in the blanks with the correct units. This exercise provides foundational understanding of power analysis in circuits with purely resistive elements.
**Calculate the resistances of the light bulbs** - \( R_{60W} = \) ______ units - \( R_{45W} = \) ______ units - \( R_{25W} = \) ______ units (Note: The third line contains a typo; "units" is misspelled as "untis.")
Transcribed Image Text:**Calculate the resistances of the light bulbs** - \( R_{60W} = \) ______ units - \( R_{45W} = \) ______ units - \( R_{25W} = \) ______ units (Note: The third line contains a typo; "units" is misspelled as "untis.")
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