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3. For the circuit shown in Figure 7.3, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 120 volts. j 150 200 b Figure 7.3 E -j 50

3. For the circuit shown in Figure 7.3, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 120 volts. j 150 200 b Figure 7.3 E -j 50

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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 For the circuit shown in Figure 7.3, determine apparent power S, real power P, reactive power Q
and power factor PF. Also, draw the power triangle. The source is 120 volts.

NOTE: All measurements are in base units (ohms, farads, etc.)

--- ### AC Power Analysis - Example Problem #### Example 3 **Problem Statement:** For the circuit shown in Figure 7.3, determine the apparent power \( S \), real power \( P \), reactive power \( Q \), and power factor \( PF \). Also, draw the power triangle. The source is 120 volts. ![Circuit Diagram](image-url) **Figure 7.3** **Given Data:** - Voltage Source \( E \) = 120 V - Inductive Reactance \( j150 \) ohms - Resistor \( 200 \) ohms - Capacitive Reactance \( -j50 \) ohms **Solution:** 1. **Impedance Calculation:** \[ Z_{total} = 200 + j(150 - 50) \] \[ Z_{total} = 200 + j100 \] 2. **Magnitude of Impedance:** \[ |Z_{total}| = \sqrt{200^2 + 100^2} \] \[ |Z_{total}| = \sqrt{40000 + 10000} \] \[ |Z_{total}| = \sqrt{50000} \] \[ |Z_{total}| = 100\sqrt{5} \] 3. **Current Calculation:** \[ I = \frac{V}{Z} = \frac{120}{100\sqrt{5}} \] \[ I = \frac{120}{100\sqrt{5}} \] \[ I = \frac{12\sqrt{5}}{50} \] \[ I = \frac{2.4}{\sqrt{5}} \] \[ I = 0.48\sqrt{5} \approx 1.07 \text{ A} \] 4. **Apparent Power \( S \):** \[ S = VI \] \[ S = 120 \times 1.07 \] \[ S = 128.4 \text{ VA} \] 5. **Real Power \( P \):** \[ P = I^2 R \] \[ P = (1.07)^2 \times 200 \] \[ P = 1.1449 \times 200 \] \
Transcribed Image Text:--- ### AC Power Analysis - Example Problem #### Example 3 **Problem Statement:** For the circuit shown in Figure 7.3, determine the apparent power \( S \), real power \( P \), reactive power \( Q \), and power factor \( PF \). Also, draw the power triangle. The source is 120 volts. ![Circuit Diagram](image-url) **Figure 7.3** **Given Data:** - Voltage Source \( E \) = 120 V - Inductive Reactance \( j150 \) ohms - Resistor \( 200 \) ohms - Capacitive Reactance \( -j50 \) ohms **Solution:** 1. **Impedance Calculation:** \[ Z_{total} = 200 + j(150 - 50) \] \[ Z_{total} = 200 + j100 \] 2. **Magnitude of Impedance:** \[ |Z_{total}| = \sqrt{200^2 + 100^2} \] \[ |Z_{total}| = \sqrt{40000 + 10000} \] \[ |Z_{total}| = \sqrt{50000} \] \[ |Z_{total}| = 100\sqrt{5} \] 3. **Current Calculation:** \[ I = \frac{V}{Z} = \frac{120}{100\sqrt{5}} \] \[ I = \frac{120}{100\sqrt{5}} \] \[ I = \frac{12\sqrt{5}}{50} \] \[ I = \frac{2.4}{\sqrt{5}} \] \[ I = 0.48\sqrt{5} \approx 1.07 \text{ A} \] 4. **Apparent Power \( S \):** \[ S = VI \] \[ S = 120 \times 1.07 \] \[ S = 128.4 \text{ VA} \] 5. **Real Power \( P \):** \[ P = I^2 R \] \[ P = (1.07)^2 \times 200 \] \[ P = 1.1449 \times 200 \] \
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