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GETTING STEADY

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niga

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hello there,

Here are a few things i wud like to GET cleared.Can someone just check in and tell me ? :D

1. WHEN AN R C IS IN SERIES,ACROSS DC SUPLY IT TAKES 5 TIME CONSTANTS TO REACH THE STEADY STATE.BUT IN RC PARALLEL CONNT ITS FASTER.THE TIMECONSTANT(TC) IS SMALLER.

HOWEVER THE STEADYSTATE AND NATURAL RESPONSE IN PARALEL AND SERIES CONNT FOLLOW THE SAME PATTERN.THE ONLY DIFFERENCE BEING THE LESSER TC IN PARALLEL.
AM I CORRECT?

2. THE SAME APPROACH GOES FOR RL SERIES ACROSS DC.RIGHT?

3.I READ THAT THE STEADY STATE IN RL PARALLEL ACROSS DC IS ZERO STATE.L ACTS AS A SHORT AFTER THE TRANSIENT PERIOD.IS THAT WHY ITS CALLED ZERO STATE?BUT IN RC, C ACTS OPEN AFTER SOMETIME THEN IS THAT NOT ZERO STATE?

PLEASE EXPLAIN WHAT IS ZERO STATE AND IF ITS THERE FOR BOTH RL AND RC ACROSS DC.

4.TO FIND THE COMPLETE SOLUTION OF RL OR RC SERIES ACROSS DC

A.APPLY KVL,KCL

B. ASSUME SOLUTION K1 + K2 EXP(st)

C. K1 REFERS TO STEADY STATE Vs/R

D. K2 REFERS TO STEADY STATE FOUND BY APPLYING INITIAL CONDITIONS AT t=0.

IS THIS RIGHT UNDERSTANDING?
IS THIS APPLICABLE FOR RL AND RC PARALLEL ACROSS DC AS WELL?

THANX FOR ALL HELP :p
 
1. It takes 5 TC to approach the steady state, an infinite time to reach SS.

RC in parallel across a DC source would have a 0 TC since an ideal DC source has 0 source resistance. In practice, the source would have some resistance say Rs. So TC = C Rs

2. TC = L/Rs and the current will approach V/Rs as t approaches infinity.

3. I have not heard the term "zero state" Do you mean a zero in the S plane?

4. I don't know what you mean by "APPLY KVL,KCL ".

Are you trying to solve for the current through, or the voltage across, one of the components? If it is the latter, the solution will depend upon which component is connected to the source.

If the solution is "K1 + K2 EXP(st)" then the initial condition (ie. at t = 0) is K1 + K2 and the final condition is K2 (assuming that s is negative)

Len
 
HEY LEN, U R REALLY BEING A GREAT HELP.
THANKYOU.

It takes 5 TC to approach the steady state, an infinite time to reach SS.??

1.OKAY SO I UNDERSTAND THAT IT TAKES 5 TC TO REACH SS IN SERIES CONNT AND INFINITE TIME FOR SS IN PARALLEL CONNT?BUT tc IS VERY SMALL RIGHT SO IT SHUD REACH SS FASTER?


2.I GOT ZEROSTATE FROM THIS PRESENTATION.I DIN UNDERSTAND HOW THAT HAPPENED.

THANX FOR ALL UR HELP.
 

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niga said:
1. It takes 5 TC to approach the steady state, an infinite time to reach SS.??

2.OKAY SO I UNDERSTAND THAT IT TAKES 5 TC TO REACH SS IN SERIES CONNT AND INFINITE TIME FOR SS IN PARALLEL CONNT?

3. BUT tc IS VERY SMALL RIGHT SO IT SHUD REACH SS FASTER?

1. 5 TC is an approximation only. It is a asymtote which means in theory that it never reaches ss.

2. No, 5 TC is as I said above. The parallel case also takes 5 TC to approximate ss. But in this case, the TC is much smaller since the R is the source resistance.

3. Yes

I suggest you draw the graphs of y = e^(-t/T) and y = {1 - e^(-t/T)}

t is time and T is the time constant.

What is y at 4T? What is y at 5T? What is y at 6T?

Len
 
okay i get it!
so there is no zero state business in RL parallel to dc,then?
wat abt rl and rc paralel across ac ...am i getting all mixed up?

:? niga
 
niga said:
okay i get it!
1. so there is no zero state business in RL parallel to dc,then?
2. wat abt rl and rc paralel across ac ...am i getting all mixed up?

:? niga

1. As I said, I don't know what "zero state" means.

2. In the case of an AC source, there is a transient period followed by a steady state period. You should study the analysis of AC circuits.

Len
 
hi len

yes i have studied the analysis of each to Ac supply and dc and i quite understand it.
i think my problem is with the differential equations to find transient and ss.
i am quite confused if the equations of parallel RL(for eg) will give same answer as that of series RL (whether conntd to dc or ac. )
 
niga said:
hi len

1. yes i have studied the analysis of each to Ac supply and dc and i quite understand it.
2. i think my problem is with the differential equations to find transient and ss.
3. i am quite confused if the equations of parallel RL(for eg) will give same answer as that of series RL (whether conntd to dc or ac. )

1. Good.

2. The solution of the DEs gives both the transient and ss solutions. The ss solution is when t approches infinity.

3. I did not say that the equations of parallel RL will give same answer as that of series RL.

Parallel RL (connected to a DC source voltage) will result in the current approaching infinity as t approaches infinity.

In the case of series RL , the current will tend towards V/R as t approaches infinity.

Len
 
Thanku Len,
Yup i have figured out RC in parallel,series across dc.
But i am getting stuck with RL parallel across dc.

I=I(r) + I(L)

Ldi/dt= V(L)

I=V/R + V/L


but somehow i cant move ahead.i am okay with RC parallel across dc kinda wierd why i am stuck here.

:roll: can u help?
 
niga said:
Thanku Len,
Yup i have figured out RC in parallel,series across dc.
But i am getting stuck with RL parallel across dc.

I=I(r) + I(L)

Ldi/dt= V(L)

I=V/R + V/L


but somehow i cant move ahead.i am okay with RC parallel across dc kinda wierd why i am stuck here.

:roll: can u help?
When you connect a voltage source across an inductor, you need to look at the equation for current:
Code:
       ⌠
I=(1/L)⌡Vdt
Note that, since V is a step to a DC voltage, the current will be a ramp (the integral of a step) that goes to infinite current at infinite time. The current in the parallel resistor is, of course, V/R, and is independent of the inductor.

Ron
 
hey Ron, that was a very useful bit of info to me.Thanx.
But then when u connt R and L in parallel. the voltage across is the same so whether u connect voltage source or current source don we look at the current equation?and arrive at the same answer -the ramp.....
 
niga said:
hey Ron, that was a very useful bit of info to me.Thanx.
But then when u connt R and L in parallel. the voltage across is the same so whether u connect voltage source or current source don we look at the current equation?and arrive at the same answer -the ramp.....

No the current source and resistor can be replaced by its Norton (if I remember correctly) equivalent. This a voltage source in series with a resistor. The source voltage will be = IR ie. the open circuit voltage and the series resistance will be R.

Len
 
okay len,so if u have current source in parallel RL we convert it to equivalent voltage source in series with R.
So we have 2 resistors and an L in parallel with one of them.
Yup?am i getting it right?
 
See the diagram

Len
 

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You have helped me tremendously.
You are a great teacher i must add.
Thankou for all your help and time. :p
 
heheh just to be sure :lol:

is it right when i say

Its easier to convert RL parallel when connected across current source to the voltage source equivalent andthen solve the differential equation?

is this applicable to ac supply as well?

:mrgreen:
 
It would not make any difference. It is as easy to write the equations for a current source as it is for a voltage source.

The technique is general. It can be used for DC or AC sources.

Len
 
looks like my doubts never end,eh?am sorry abt that really..
:roll:
so when i write
for a parallel ckt with current source

I=Ir+IL
I=Vs/R + 1/L int V dt
(int is integral)

but int V dt is a ramp that means current is going to infinity.
so how do jot it in equation form then?
 
Your maths is wrong. See below

At t = 0, all of the current goes through the resistor.

But as t -> infinity, all of the current goes through the inductor. (we have assumed that it is a pure inductance, ie. it has no resistance). Therefore, as t -> infinity, the inductor represents a short circuit acroos the resistor.

Len
 

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