Electronic Circuit Question.

What is silly in it?
having or showing a lack of common sense or judgement

If source is giving 10V with 1ohm 10Amp from where LC comes?
Taken seriously, your question indicates that you've neither read nor comprehended anything posted on this thread over the past fortnight! -- You know better and so do I!
Let me know when you're done playing games...
 
ok let come to topic.
if we can get direct 10 A with 1ohm what is the need of adding LC circuit here?
The 'LC circuit' comprises the transformation network - as we've been discussing all along! -- Thus it is that we derived≈62.83V from a 10V supply!

Either you've missed the entire point while pretending to 'follow along' OR you're pretending to have missed the entire point for reasons all your own! --- Either way it's deceitful, rude, disrespectful to those of us attempting to assist you -- and, oh yea - NOT FUNNY!!!:mad::mad::mad:
 

Thread Starter

RRITESH KAKKAR

Joined Jun 29, 2010
2,829
OK, it is for understanding how reactance is taken.
Here I present an example of transformation of 10v to over 60V via a series resonant RLC circuit --- While not directly applicable to the exercise, this is intended to set you upon the right path!
 
Everything is fine but we can get 10A with 1ohm and 10V sin wave so, why to add LC here are they both giving any advantages??
Because, as per the scenario of the example (in post #145), we don't want 10V!!! --- We want ≈ 62.83V! - But 10V is all we have to 'work with' -- Hence our use of an LC network to transform the impedance such that a drop of ≈ 62.83V appears across each reactor!!! --- Had you read posts #145 and #257 you'd understand!


Ya know -- It's pretty sad (and, I might add, suspect) when the hints, clues and dead giveaways confuse you more than the exercise itself!!!

Tolerantly
HP
 

Thread Starter

RRITESH KAKKAR

Joined Jun 29, 2010
2,829
Because, as per the scenario of the example (in post #145), we don't want 10V!!! --- We want ≈ 62.83V! - But 10V is all we have to 'work with' -- Hence our use of an LC network to transform the impedance such that a drop of ≈ 62.83V appears across each reactor!!! --- Had you read posts #145 and #257 you'd understand!

Ya know -- It's pretty sad (and, I might add, suspect) when the hints, clues and dead giveawaysconfuse you more than the exercise itself!!!
OK, Now i got you that we want help of inductor to increase the voltage and capictor will cancel it reactive power which has no use.
as 10V is less capacitor will increase it, RIGHT?
 
OK, Now i got you that we want help of inductor to increase the voltage and capictor will cancel it reactive power which has no use.
as 10V is less capacitor will increase it, RIGHT?
WRONG! As previously stated - the circuit in post #145 is NOT representative of a 'self-inductance' topology!!! --- 'back EMF', 'inductive kicks', etc... have nothing to do with it!

One last time!
By definition; At resonance Xc=Xl ergo Xc in series with Xl =0Ω → -2πΩ +2πΩ =0Ω
1Ω resistance in series with a (net) reactance of 0Ω=1Ω → 1Ω+0Ω=1Ω
The current attending an EMF of 10V across a resistive impedance of 1Ω = 10A → 10V/1Ω = 10A
The drop across either reactor (without regard to phase angle) = I*|X| → 10A*2πΩ≈62.83V

@RRITESH KAKKAR -- If achievement is truly your aim you must learn to attend to instruction and cease your attempts to 'form reality' to fit your expectations/experience... --- This isn't that complicated!!!:mad::mad::mad:
 

Thread Starter

RRITESH KAKKAR

Joined Jun 29, 2010
2,829
One last time!
By definition; At resonance Xc=Xl ergo Xc in series with Xl =0Ω → -2πΩ +2πΩ =0Ω
1Ω resistance in series with a (net) reactance of 0Ω=1Ω → 1Ω+0Ω=1Ω
The current attending an EMF of 10V across a resistive impedance of 1Ω = 10A → 10V/1Ω = 10A
The drop across either reactor (without regard to phase angle) = I*|X| → 10A*2πΩ≈62.83V
I understand this very well.
I am asking what to do with it now?
 
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