yeah, we are learning phasors and impedance soon. I've just started to solve basic inductive circuits by replacing the source with a complex forcing function. I think the person who wrote the question wants us to go through the hard way of solving it, so that when we do learn phasors, we can...
Hey gneill.I'm solving for the inductor current IL. The current source is 0.2IL and i should have mentioned before the dependent current flows in the opposite direction to the inductor current.
Hello CWatters. Are you suggesting that I can model the dependent source as an inductor parallel to...
an independent source with a forcing function modeled by cos(500t) is in series with an 100 ohm resistor. A dependent (current controlled) source is placed across the remaining resistor terminal and ground. A 0.3mH inductor is in parallel with the dependent source. The value of the dependent...
you have two independent nodes so you only need two equations which you have. and by the looks of it you have almost perfect equations. there is one mistake though.
What is the fundamental law that you are applying when you do node analysis?
Homework Statement
referring to the attatchment. the current consists of two semi circles. the question asks me to find the voltage across a 47-uF capacitor when t = 2ms.
Homework Equations
v(t) = 1/C ∫ i(t)dt +v(t0) {I realize their has to be limits for the integrand,I just can't type...
Yes I do know I must place a +ve charge between them, sorry I should have mentioned that however it's the magnitudes. It must be closer to the smaller charge.
Homework Statement
Q1.Two Charges, -Q and -3Q, are a distance l apart. The two charges are free to move but don't because there is a third charge nearby. What must the third charge be and where must it be placed for the first two to be in equilibrium?
Homework Equations
Coulombs Law: F =...
okay, firstly you have got the max displacement through Hooke's law i.e F = kx = 0.13m.
Thus you have got the Amplitude A.
Okay the total mechanical energy of the system is E = 1/2kA^2 = 2.535 Joules.
When the spring is fully pulled back the total energy is potential i.e E = 1/2kx^2 =...