Sadly i missed the question. For some reason I am getting different numbers everytime i try to calculate. I've been doing homework for hours so i must jst be tired. The final answer is -4.91*10^-2, but i don't know how they got that answer
That's why I was thinking the negative wouldn't work. Thanks for clearing that up! Now I have to find the current in the inductor at that time given in part (e)
I tried using i = -ωQmax(sin(ωt+∅)
but it's telling me that it's not the right answer. I got -4.896*10^-9
I had something like this in my notes but I didn't understand what it was saying. I'm just tryng to make sense out of it and I don't know if I'm explaining it the right way.
The capacitor starts discharging through the inductor but it doesn't just instantly discharge to zero. It discharges as current builds up to its maximum since current is not instantaneous. So if the current doesn't change automatically, the capacitor will start absorbing charge in the opposite...
Homework Statement
A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.0 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.50 H .
What is the charge on the capacitor after a time...
Can someone explain to me why that highlighted force is 0?
Homework Equations
∑Fx = 0
The Attempt at a Solution
I found FDC to be 780lb and I am still trying to solve for FDE
Oh okay! Thanks :smile:
So if I were to assign three loops like this
would these be the right eqns?
Eqn #1: (loop with 14V, starting after the 14V)
14V - (I2+I1)*(1Ω) - (I1+I2+I3)*(2Ω) = 0
Eqn #2: (upper loop, starting above I2+I1)
(I2+I1)*(1Ω)+(I2)*(2Ω) - (I3)*(1Ω) = 0
Eqn #3: (lower loop...