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Problem #2A E E, = E, = XL XL = XC Z = R = kvar - kvar kVA = kW: E, 240V/60HZ kW = 4.92 7.82 Line %3D kvar kvar = 4.39 %3D

Problem #2A E E, = E, = XL XL = XC Z = R = kvar - kvar kVA = kW: E, 240V/60HZ kW = 4.92 7.82 Line %3D kvar kvar = 4.39 %3D

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 #2A** - Symbols represent components in an electrical circuit diagram, which includes: - A series circuit with a source, resistor, inductor, and capacitor. **Parameters for Each Component:** 1. **Source Voltage and Current:** - \( E_t = \) - \( I_t = \) 2. **Resistor:** - Voltage across resistor, \( E_r = \) - Current through resistor, \( I_r = \) - Resistance, \( R = \) - Power in kilowatts, \( \text{kW} = \) 3. **Inductor:** - Voltage across inductor, \( E_{XL} = \) - Current through inductor, \( I_{XL} = \) - Inductive reactance, \( XL = \) - Reactive power in kilovars, \( \text{kvar}_{XL} = \) 4. **Capacitor:** - Voltage across capacitor, \( E_{XC} = \) - Current through capacitor, \( I_{XC} = \) - Capacitive reactance, \( XC = \) - Reactive power in kilovars, \( \text{kvar}_{XC} = \) **Overall Circuit:** - Total impedance, \( Z = \) - Apparent power in kilovolt-amperes, \( \text{kVA} = \) **Given Values:** - Line Voltage, \( E_{\text{Line}} = 240 \text{V/60Hz} \) - Power across load, \( \text{kW} = 4.92 \) - Reactive power for inductor, \( \text{kvar}_{XL} = 7.82 \) - Reactive power for capacitor, \( \text{kvar}_{XC} = 4.39 \)
Transcribed Image Text:**Problem #2A** - Symbols represent components in an electrical circuit diagram, which includes: - A series circuit with a source, resistor, inductor, and capacitor. **Parameters for Each Component:** 1. **Source Voltage and Current:** - \( E_t = \) - \( I_t = \) 2. **Resistor:** - Voltage across resistor, \( E_r = \) - Current through resistor, \( I_r = \) - Resistance, \( R = \) - Power in kilowatts, \( \text{kW} = \) 3. **Inductor:** - Voltage across inductor, \( E_{XL} = \) - Current through inductor, \( I_{XL} = \) - Inductive reactance, \( XL = \) - Reactive power in kilovars, \( \text{kvar}_{XL} = \) 4. **Capacitor:** - Voltage across capacitor, \( E_{XC} = \) - Current through capacitor, \( I_{XC} = \) - Capacitive reactance, \( XC = \) - Reactive power in kilovars, \( \text{kvar}_{XC} = \) **Overall Circuit:** - Total impedance, \( Z = \) - Apparent power in kilovolt-amperes, \( \text{kVA} = \) **Given Values:** - Line Voltage, \( E_{\text{Line}} = 240 \text{V/60Hz} \) - Power across load, \( \text{kW} = 4.92 \) - Reactive power for inductor, \( \text{kvar}_{XL} = 7.82 \) - Reactive power for capacitor, \( \text{kvar}_{XC} = 4.39 \)
Expert Solution
Step 1

Referring to the given circuit, all the elements are connected in parallel, hence the voltage across them will be equal.

As given:

The voltage across them are 240V with the frequency of the 60Hz.

The real power or the active power of the circuit is 4.92kW.

Inductive power of the circuit is 7.82.

Capacitive power of the circuit is 4.39.

 

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