Chapter 24: Problem 34
An ac generator with a frequency of \(1.34 \mathrm{kHz}\) and an rms voltage of \(24.2 \mathrm{V}\) is connected in series with a \(2.00-\mathrm{k} \Omega\) resistor and a \(315-\mathrm{mH}\) inductor. (a) What is the power factor for this circuit? (b) What is the average power consumed by this circuit?
Short Answer
Step by step solution
Calculate the Angular Frequency
Calculate the Inductive Reactance
Determine the Impedance
Calculate Power Factor
Calculate Average Power Consumed
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Angular Frequency
Deeper insight: Angular frequency acts as a bridge between time-based oscillation (frequency) and the physical aspects of the circuit.
Inductive Reactance
As seen in the solution, with \(\omega = 8425.57 \, \text{rad/s}\) and \(L = 0.315 \, \text{H}\), the inductive reactance is calculated to be \(2653.6 \, \Omega\).
Remember: A higher frequency or inductance increases the inductive reactance. This means faster oscillations or larger coils oppose changes in current more.
Power Factor
In our example, with \(R = 2000 \, \Omega\) and \(Z = 3308.8 \, \Omega\), the power factor is \(0.604\).
Why it matters: A lower power factor means more energy is wasted. Optimizing the power factor can lead to more efficient energy use, reducing energy costs.
Impedance
In the provided solution, this results in \(Z = 3308.8 \, \Omega\).
Significance: Impedance provides a complete picture of how much a circuit opposes AC. Understanding it helps in designing circuits with desired characteristics for energy transfer.
Average Power
The solution involves finding \(I_{\text{rms}}\) first: \(I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = 0.00731 \, \text{A}\). Hence, average power \(P = 0.106 \, \text{W}\).
Takeaway: Average power is crucial for quantifying the energy consumed. It reflects how much electrical energy is converted into useful work over time in the circuit.