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For the circuit shown, what is the reactive power, Q, consumed by the 1+ j5 N load impedance? (Assume the current source is in 2Ω 10 Is 3 20° A -j10 j50 wwHH

For the circuit shown, what is the reactive power, Q, consumed by the 1+ j5 N load impedance? (Assume the current source is in 2Ω 10 Is 3 20° A -j10 j50 wwHH

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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**Title:** Calculating Reactive Power in AC Circuits **Description:** In the given circuit diagram, we aim to determine the reactive power \( Q \) consumed by the \( 1 + j5 \, \Omega \) load impedance. The problem assumes the current source is in RMS (Root Mean Square) value. The key components and their configurations in the circuit are described below: **Circuit Components:** 1. **Current Source (\( I_s \)):** - \( I_s = 3 \angle 0^\circ \, \text{A} \) 2. **Resistors and Reactances in Parallel:** - A \( 2 \, \Omega \) resistor is connected in series with a \( -j1 \, \Omega \) reactance. - Another resistor of \( 1 \, \Omega \) is connected to this network, forming the first branch. 3. **Impedance in Parallel:** - A load impedance represented by \( 1 + j5 \, \Omega \) is connected in parallel to the aforementioned branch. **Objective:** To calculate the reactive power \( Q \) consumed by the complex load \( 1 + j5 \, \Omega \). **Solution Approach:** - Use concepts of impedance, apparent power, and the relation between current, voltage, and reactive power in AC circuits. - Reactive power \( Q \) is calculated using the formula: \[ Q = V^2 / |Z| \times \sin(\theta) \] where: - \( V \) is the voltage across the load, - \( |Z| \) is the magnitude of the impedance, and - \( \theta \) is the phase angle of the impedance.
Transcribed Image Text:**Title:** Calculating Reactive Power in AC Circuits **Description:** In the given circuit diagram, we aim to determine the reactive power \( Q \) consumed by the \( 1 + j5 \, \Omega \) load impedance. The problem assumes the current source is in RMS (Root Mean Square) value. The key components and their configurations in the circuit are described below: **Circuit Components:** 1. **Current Source (\( I_s \)):** - \( I_s = 3 \angle 0^\circ \, \text{A} \) 2. **Resistors and Reactances in Parallel:** - A \( 2 \, \Omega \) resistor is connected in series with a \( -j1 \, \Omega \) reactance. - Another resistor of \( 1 \, \Omega \) is connected to this network, forming the first branch. 3. **Impedance in Parallel:** - A load impedance represented by \( 1 + j5 \, \Omega \) is connected in parallel to the aforementioned branch. **Objective:** To calculate the reactive power \( Q \) consumed by the complex load \( 1 + j5 \, \Omega \). **Solution Approach:** - Use concepts of impedance, apparent power, and the relation between current, voltage, and reactive power in AC circuits. - Reactive power \( Q \) is calculated using the formula: \[ Q = V^2 / |Z| \times \sin(\theta) \] where: - \( V \) is the voltage across the load, - \( |Z| \) is the magnitude of the impedance, and - \( \theta \) is the phase angle of the impedance.
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