What is the role of placing LC circuit?
If you mean the problem of post #113 -- Good! --- Let's see your solution and workok thanks
problem sort out and solved.
You know where I stand as regards silly questionsWhat is the role of placing LC circuit?
What is silly in it?You know where I stand as regards silly questions
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!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?
I'm referring to your 'pastime' of asking questions that you know the answers to just for 'fun' -- Honest questions are welcome! ---- Bogus questions waste my time and try my patience!!!I don't get you what is game in it?
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!ok let come to topic.
if we can get direct 10 A with 1ohm what is the need of adding LC circuit here?
Go back and this time READ posts #257 and #145!!!Is there RL present and we have to remove reactance by C??
@RRITESH KAKKARpretending=behave so as to make it appear that something is the case when in fact it is not.
no.
NO! --- It is to illustrate the manner in which reactive networks may transform impedance -- Recall that impedance may be regard the ratio of EMF to Current...OK, it is for understanding how reactance is taken.
The discussion in post #145 is intended as an aid to solution of the exercise of post #113...yes, the circuit is for learning how it work
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!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??
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.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!!!
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!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?
I understand this very well.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
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