Recent content by D Nguyen

Let I be the current through the circuit. The V to the drop at 2R and 6R, so 2RI + 6RI = 80 8RI = 80 RI = 10 For the whole circuit, U - 3RI - 2RI - 6RI - 4RI = 0 Try to take it from there!

Homework Statement Homework Equations Magnetic flux = integral(B dot dA) line integral(B dot ds) = (u_o)(i_enc) (Ampere's Law) The Attempt at a Solution I can't see how Ampere's law applies because all of the magnetic field vectors along the loop will be perpendicular to that stretch of...

Homework Statement Here is the problem: Homework Equations Typical RL circuit equations. The Attempt at a Solution My main question with this problem is figuring out what to do with the third resistor, the one on the right. Initially, I thought of using RL circuit equations with the left...

I don't think the circuit starts at steady state because the capacitor is initially uncharged.

OK, I think it's starting to make sense. The potential across E2 can't change when the switch it closed, so open or closed, I can ignore E1 and R1? Then I use Thevenin?

Can I use a Thevenin approach for a question like this? I'm trying to use information from this website: https://www.allaboutcircuits.com/textbook/direct-current/chpt-16/complex-circuits/ Thanks for all of your help and patience!

Hello! Thanks for the welcome! I think it will? E1 will push more current to the right of E2? I tried another approach using Thevenin equivalent circuits. I got E(Thevenin) = 3.33 V and R(Thevenin) = 13.33 ohms. Then I would just use an RC circuit with those values of E and R?

Homework Statement Homework Equations For a basic RC circuit during charging: q = CV(1-e-t/RC) i = V/R (e-t/RC) The Attempt at a Solution I can solve basic RC circuits but this is just out of my realm of understanding. I can also find the max charge of the capacitor. I'm not looking for...