2 kohms (1) 70 nF 13 H Load #3 Find the average power and the reactive power absorbed by the load in the circuit shown above if ig=20*cos(1250*t) mA 1) Average Power (Unit: W) Submit Answer Tries 0/3 2) Reactive Power (Ignore the sign; Unit: W) Submit Answer Tries 0/3
2 kohms (1) 70 nF 13 H Load #3 Find the average power and the reactive power absorbed by the load in the circuit shown above if ig=20*cos(1250*t) mA 1) Average Power (Unit: W) Submit Answer Tries 0/3 2) Reactive Power (Ignore the sign; Unit: W) Submit Answer Tries 0/3
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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Transcribed Image Text:**Circuit Analysis: Average and Reactive Power Calculation**
**Circuit Description:**
The circuit features:
- A current source labeled \( i_g = 20 \times \cos(1250t) \, \text{mA} \).
- An inductor of \( 13 \, \text{H} \) in parallel with a load.
- The load is a series combination of a resistor \( 2 \, \text{k}\Omega \) and a capacitor \( 70 \, \text{nF} \).
**Task:**
Calculate the following for the given circuit:
1. **Average Power** absorbed by the load (Unit: Watts).
Input your answer in the provided field and press "Submit Answer". You have 3 attempts.
2. **Reactive Power** absorbed by the load (Ignore the sign; Unit: Watts).
Input your answer in the provided field and press "Submit Answer". You have 3 attempts.
**Additional Notes:**
- The average power involves calculating the real power dissipated in the resistor.
- Reactive power involves calculations from both the inductor and capacitor as they store energy temporarily.
This exercise is designed to enhance your understanding of AC circuit analysis and power calculation in electrical engineering studies.
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