Question

Finding steady state voltage.

Answer 1

@ganeshie8 The resistance i got was 1/3 + 2/3j

Answer 2

Initially, i made the capacitor as a short, and the inductor as an open. Calculated initial conditions. Then redid with capacitor is open and inductor as short. am i going about it right?

Answer 3

Can I help?

Answer 4

hah ofc! it's just that it's almost always, it had been ganeshie. haha I'd love for anyone to help.

Answer 5

you're trying to find the thevinin equivilant first, right ?

Answer 6

I'm getting \(z_{th} = 0+3j\)

Answer 7

hmm, I first turned off the source. So that's open circuit. Then I added inductor and capacitor, cause they are in series. Then I did capacitor and resistor in parallel. THen I did the previous 2 terms in parallel again.

Answer 8

am i going wrong in my math?

Answer 9

whats the reactance of inductor ?

Answer 10

\(j\omega L = j*1*2\) right ?

Answer 11

Here \(\omega = 1\) because the angular frequency of the source \(i(t) = \cos t+2\sin t\) is \(1\)

Answer 12

\[z_{th} = (j*1*2 + \dfrac{1}{j*1*1}) ~~||~~ \dfrac{1}{j*1*\dfrac{2}{3}}\]

Answer 13

do you have the same expression ?

Answer 14

yes

Answer 15

simplifying that gives \(3j\)

Answer 16

oh dear, ok, let me go back and check my math.

Answer 17

how did u get
1/3 + 2/3j

Answer 18

ok im also gonna check my math :)

Answer 19

In that second term, shouldn't it be in parallel with (3j/2 + 3)/ 9j/2

Answer 20

cause it's 1/(j*1*2/3) || 3

Answer 21

sorry, i missing a negative sign, in that 2nd to previous message.

Answer 22

while calculating \(z_{th}\), you need to treat the terminals as open circuit

remove 3 ohm resistor

Answer 23

OOO! that's where I was going wrong... ok ok. that makes more sense. thank you.

Answer 24

taking that into consideration, I get 3j

Answer 25

Looks good! what next ?

Answer 26

you want to find \(V_{th}\) ?

are you familiar with phasors ?

Answer 27

uhm, yes, but i do not have a strong grasp on it.

Answer 28

or maybe i do, and I don't realize what is actually the phasor. Cause i've solved these circuits very easily in the past, we're just revisiting them in order to step into laplace transform and fourier transform

Answer 29

consider below sinusoid :
\(y = 2\sin(\omega t+\phi)\)

Answer 30

amplitude is \(2\), angular frequency is \(\omega\) and the phase angle is \(\phi\)

Answer 31

that equation need not be a voltage or current,
it can represent any sinusoid motion, like SHM

Answer 32

familiar with simple harmonic motion, right ?

Answer 33

sorry, i had to do something real quick. And yes, i am familiar with that.

Answer 34

I'm getting \(V_{th} = \dfrac{6}{11}\angle -\frac{\pi}{2}\)

Answer 35

Can you pelase walk me through that?

Answer 36

oh wait, that is only from i(t)=cost

need to add 2sint part too

Answer 37

I am using superposition as it looks simpler here

Answer 38

\[V_{th} = (\dfrac{6}{11}\angle -\frac{\pi}{2} )+ (\dfrac{12}{11}\angle -\pi)\]

Answer 39

still here ?

Answer 40

yes, i'm trying to figure out how you got that.

Answer 41

First, notice that the given circuit is linear. Therefore the angular frequency of voltages across any element will be same.

The only things that change are the "amplutude" and "phase angle"

Answer 42

i(t) = cos t + 2sin t

first consider i(t) = cos t

whats the phasor representation of this source ?

Answer 43

well it doesn't have a single amplitude. So how would i evaluate that?

Answer 44

that is the reason we're using superposition

first, only consider i(t) = cos t

Answer 45

1e^j ?

Answer 46

and the other would 2e^j

Answer 47

i(t) = cos t isn't the phasor just \(1e^{j0}\) ?

Answer 48

ooo, wait yes. it's the phase angle that is the coeffecient of j. so yes, cost = 1. 2sint = 2e^(j * (pi/2)) right?

Answer 49

By applying KCL at the two nodes, do we get \[-1e^{j0}+\dfrac{V_a-V_b}{\frac{1}{j}}+\dfrac{V_a}{2j}=0\\~\\\dfrac{V_b-V_a}{\frac{1}{j}}+\dfrac{V_b}{\dfrac{1}{(2/3)j}}=0\]

Answer 50

solve them for \(V_a\) and \(V_b\)

\(V_b\) is the voltage across right side capacitor, which is also the \(V_{th}\) that we need

Answer 51

rest should be easy :)

Answer 52

oooo! ok ok, i see

Answer 53

those equations are only for i(t) = cost

you need to setup similar equations for i(t) = 2sint, solve V_b

Answer 54

in the end, add V_b for both the sources to get V_th

Answer 55

hmm ok, that makees much more sense. thanx for the guidance.

Answer 56

np :) if possible please post your solution/final answer when you're done