I would help you, but I'm signed up to take my first signals class next quarter!
Sorry!
Doesn't time domain mean just convert it from phasors to regular coordinates? but you already have it in terms of t so yah I wouldn't be able to help you yet :|
Yours is right I believe, but here:
and I1 = I2 * N2/N1I believe it should be:
I1 = I2 * -N2/N1
Because the current i2 is flowing OUT of the transformer at the dot, so shouldn't it be Negative value?
Thx again
http://imgur.com/4yxwi
Here is the original picture.
Ibasically did KCL setting the nodes from left to right
Va, Vb, Vc, Vd
You can zoom in and it should be more clear.
I understand what you did but i still would appreciate it if you could explain afew things to me.
What exactly HAPPENS to this circuit when the switch is open? Also what does it mean to have a discharge path.
So basically I understand this:
Since the circuit has been at rest while switch is...
OK that makes a lot more sense
SO since voltage across capacitor is 2V
We can set this equation for after short is open1/c integral (0 to infinity) of i(t) = 2when i integrate the first one.. I stil have 2 unknow variables, C and R1. how do I get rid of one of them?
i don't know the value of...
You are right, it should be C(R+R1), i had a typo. But also we are given that R = 1 ohm, so I just replaced it in the equation to get C(R1+1).
So now that we have i(t),
we need another equation to relate R and C after switch is open.
We know that before switch is open, 2V is stored in...
my value for i(t) is actually K*e^(-t/(R1(C+1)) where K = 1.
Not sure how you got C(R1+R) under the divider?
ANyways, can you look at my previous post and see if I did it correctly. thanks
Ok that makes a lot more sense..
So basically I think i figured it out. Ths is what I did:
After opening the switch: I applied 2V to the right hand lop like you said and i get THIS equation:
2 + R1 + i(t) + 1/c integral(0-> infinity) i(t) dt = 0
BUT WE KNOW that i(T) =...
Charges up to 2V? So C = 2? How many Farads would that be ?
IF C = 2, Then we can just plug it into the euqationC = (-2*10^-6)/(R+1) and solve for R?
Can you explain to me why entire 2V is stored into the Capacitor?
Wow that was simple enough.. Thanks a lot for your help!
So my final answer I get is:
v1(t) = -N1 flux w cos (wt)
i1(t) = (N2)^2 flux w cos (wt)
v1(t) / i1(t) = -N1 / (N2)^2
Is this correct?
Voltage = d(flux)/dt for 1 turn in the coil (thats what my TA said today)
so if i have these 2 equations:
v(t) = 4 i2 * (n1/n2)
and v(t) = -4 i1 ( n1^2 / n2^2)
I set v(t) = flux initial omega cos(omega t) and then what.. ?
i need to solve for i1, and v1..
so is that my answer?
v(t)...
Source of i(t) is the Capacitor.. where Vc = 1/c integral (i dt)
We set that equal to 2 because that's how much voltage is stored in the capacitor at before switch is open for a very long time. If we do that we get ::
2 = 1/c integral idt
2c = integral idt
take derivative of both sides...