Recent content by Tofuwu6

My apologies, this was my first time posting anything and this is all the question gave me above it states "The capacitor in (Figure 1) is initially uncharged. The switch is closed at t=0" which I guess I should have included.

Ohhhhhhhhhhhhh, so the current of the circuit after the 3 ohm, 4 mu-f leg is blocked off is 3 because I = ΔV/R = 42/(8+6) = 3 then you find the voltage across the 6 ohm resistor which is 6 * 3 = 18V and since this leg is parallel to the leg with the capacitor it would take the same amount of...

Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?

I had two trains of thought. One is that the capacitor will fully charge when t = infinity, so when you plug t = infinity into the equation of charge as a function of time you get 1.68E-4, which you also could've gotten from Q = CΔV where ΔV = 42V. My other train of thought was that when t =...