Power Factor Correction ⭐ Example 4: Parallel Inductive Loads

In this video, we will discuss the concept of power factor correction using another example. This is example 4. This example will consider two parallel loads with different power factors. We will compensate these loads using a parallel element, in this case a capacitor. We will calculate the required value of the capacitor such that the power factor is increased from a leading factor to 0.95 lagging. The solutions are given step by step such that it clear for you what and why we carry out a specific action. 👉 See the complete playlist about AC Steady-State Power Electric Circuits: https://www.youtube.com/playlist?list=PLuUNUe8EVqlm6Pbyn8dGmohXKUs8YfmPl ⭐ If you have questions or comments, please let me know. Help us to reach more people. Like and share this video. Subscribe to our channel: https://www.youtube.com/canbijles/?sub_confirmation=1 ⚡ CAN Education - Tutoring in Electrical Engineering, Analog Electronics, Power Electronics, Control Systems, and Math Courses ⭐ For questions, collaboration or consulting 👇 📧 [email protected] ☎️ +31616179479 🌐 www.canbijles.nl #powerfactor #inductive #electriccircuits #inductor #capacitor #resistor