Exercises I solved (asking to see if they're right)

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newzed

Joined Jan 11, 2012
16
Pleaseeee help me! I already solved them, I'm only asking to you if they're right or not :) :) :)

So, First page:
1) Is the only one I wasn't able to solve.. I know I have to attach to the terminals 1V source but then? what am i supposed to do?
2) Laplace -> 1/2 e^-2Ts - 2/s e^-3Ts + 1/s e^-4Ts
3) T= L/(Req) where Req= (R // (R+R))+R)
4) No current right? I mean, it's an ideal opamp so no current in inverting input and no current in noninverting input so everything has 0 current! So 0!

Second page:
1) if part one has been open circuited we're looking for impedance parameter, so, for example Z22 will be= v2/i2 while i1=0
2) in order to compute phasor I
First i find, after summing in parallel 8-j2 with j6 I
Then, by current division i find it in the branch i'm looking for
I= 0.72 (phase 76.35°)

3) Having apparent power that is Vrms*Irms= 50
and pf=0.8 --> since V= 20 --> vrms = 20/√2 = 14.14
so Irms= 2.28
So z= 6.20 phase of 36.86
4) 1/(s^2CL + sRC + 1)

Last Page is about First Order Circuit
for t<0 vc(0) = 14/3
for t>0 what is going on?? I mean, the switch closes so the 7A current passes all on the right... so??

Thank you so much for your time, I really really appreciate it
 

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panic mode

Joined Oct 10, 2011
1,783
Ix is current flowing from outside into terminal A (and from terminal B out to external circuit)
Test voltage is same as Vx, and then equivalent resistance of the circuit is
Rab=Vx/Ix

Assuming that Ix flows from A to B through R, and assuming that Iy flows from +beta*Vx through 3R and R to negative terminal of beta*Vx, we can write Kirchoff equations:
Iy=(beta*Vx-Vx)/3R
or
Iy=Vx(beta-1)/3R

and
Vx=R(Ix+Iy)
or
Vx/R=Ix+Vx(beta-1)/3R
Vx([(1/R)+ (1/3R) - (beta/3R)]=Ix

finally
Vx/Ix=1/[(1/R)+ (1/3R) - (beta/3R)]
Vx/Ix=1/[(3/3R)+ (1/3R) - (beta/3R)]
Vx/Ix=3/(3+1-beta)
Vx/Ix=3/(4-beta)
 
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