AC Steady-State Power ⭐ Design Example 2

In this video (design example 2), we will determine the required value of the parallel element which is connected to two load with different real power and power factors. The design objective is to determine the value of this parallel such that the power factor of the total circuit is unity (pf = 1). Again, the concept of power factor and the difference between leading and lagging will be essential in this design example. 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, Electric Circuits, Control Systems, and Math Courses ⭐ For questions, collaboration or consulting 👇 📧 [email protected] ☎️ +31616179479 🔗 WhatsApp: https://whatsapp.com/channel/0029VajYlHu0bIdkguxq6p2e 🔗 Telegram: https://t.me/canedux 🌐 https://www.canbijles.nl #powerfactor #AC #steadystate #electriccircuits #inductor #capacitor #resistor #design