To solve AC circuit problems, you need to understand the concept of the Power Triangle, which relates three types of power using trigonometry. Think of an AC circuit like a “latte”: the liquid is the useful energy doing work, while the foam represents extra power that does not perform useful work.
- Active Power (P): The real power that does actual work (heating, motor rotation), measured in Watts (W).
\( P = S \cos \phi \) - Reactive Power (Q): The power that oscillates between source and load due to inductors and capacitors, measured in VAR.
\( Q = S \sin \phi \) - Apparent Power (S): The total supplied power, measured in VA.
\( S = \sqrt{P^2 + Q^2} \)
The Power Factor (PF) represents how effectively the power is being used and is defined as:
\[
\text{PF} = \cos \phi = \frac{\text{Active Power (P)}}{\text{Apparent Power (S)}}
\]
When solving problems where \(Q\) and \(\cos \phi\) are given, and \(P\) is required, we use the tangent relationship from the power triangle:
\[
\tan \phi = \frac{Q}{P}
\]
Rearranging:
\[
P = \frac{Q}{\tan \phi}
\]
To compute step-by-step:
- Given: \( \cos \phi = 0.8 \)
- Using identity: \( \sin^2 \phi + \cos^2 \phi = 1 \)
- \( \sin \phi = \sqrt{1 – 0.8^2} = 0.6 \)
- \( \tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{0.6}{0.8} = 0.75 \)
- \( P = \frac{300}{0.75} = 400 \text{ W} \)
Final Answer: \( P = 400 \text{ W} \)
Quick exam tips:
- If \( \cos \phi = 0.8 \), then \( \sin \phi = 0.6 \) and \( \tan \phi = 0.75 \)
- If \( \cos \phi = 0.6 \), then \( \sin \phi = 0.8 \) and \( \tan \phi \approx 1.33 \)