I found a problem in one of sample paper of jee main about accept or rejector My teacher said it is out of scope of mains but the way you teach I understood almost everything Thank you
Whew! NOW I can understand better! Teachers should start with the more complex stuff AFTER the fundamentals of a subject. THANK YOU! (It was a lot of stuff! My head hurts, but I think it's because Im hungry!)
at 13:11 how you put v/r bro this parallel rlc kya isme bhi z=r hota hai at resonance ? also uske 2 3 step pehle jab tumne sq aur root kiya to usme ek -2L/C ka term bhi aayega
Sir in my book it is mentioned that parallel LC circuit is rejector circuit not parallel LCR circuit. Is it wrongly written? Because at 3:31 you mentioned parallel LCR to be rejector not parallel LC.
When in the book, they say LC, it is actually the RLC circuit only. Because every inductor has some finite resistance. So, when one connects L in parallel with C, it is L in series with R (internal resistance of L) and that combination is in parallel with C. That circuit too acts as a rejector. But its resonance frequency is slightly different from the circuit which is discussed in the video. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thank you sir for your earliest response. However as you've mentioned at 6:47 that when Xl=Xc, resonant angular frequency is 1/√LC So even if we take an ideal LC, circuit, for its resonant angular frequency, there also we'll do Xl=Xc which will give 1/√LC. I couldn't understand why’d you mentioned in the previous reply that there'll be “slight” difference, please clarify?
@@umasrivastava4142 The case which is discussed in the video is when R, L and C are in parallel with each other. But here, I am talking about the case, where R and L are in series and the combination is in parallel with C. I hope it will clear your doubt. In this case, wo = sqrt [ 1/LC - (R^2/L^2)] I hope it will clear your doubt
In case of the parallel resonant circuit, when w < wr, the circuit is inductive. Watch the video from 6:49 onwards. In case of a series resonant circuit, when w < wr, the circuit is capacitive.
For parallel RC Circuit, the admittance (Y) is [1/R + j(wC -1/wL) ] And at resonant frequency, wL= 1/wC. i.e Y is minimum. Hence Z is maximum and due to that current, I will be minimum. I hope it will clear your doubt.
Sir u said, series RLC circuit is a acceptor circuit because this is accepting frequencies which are around resonant frequency. But this circuit is also rejecting frequencies which are far away from resonant frequency then why can't it be rejector circuit?
If the resonance frequency is to be developed in LC parallel circiut then the DC voltage must be applied during the first quarter of every cycle Rest of the wave form which is 3/4 th of the wave is self generating Thank you for your feed backMukundan
If there is no R, then the circuit is highly selective. Meaning that Q will be very high. Just put R equal to infinity in the Q-factor equation. Ideally Q would be infinite and It will act as an oscillator.
while calculating quality factor energy stored in capacitor is used why not energy stored in inductor , and why two of them give different equation. help plz
at 12.43 how third equation changes into 4th..........! square cannot be applied on individual item by applying whole under root............? or i am missing some thing
Could u please tell me why the Vp curve is upward pointed ... even though the Ip curve is downward point ? Coz according to ohms law Vp and Ip are proportional... so shouldn’t both be of the same shape ?
Let me explain it to you intuitively. When Xc > XL, more current will flow through the inductor. So, at low frequencies, in parallel resonance, the circuit is inductive in nature. At high frequencies, XL > Xc. Therefore more current will flow through the capacitor and hence the circuit is capacitive in nature. For series resonance, it is other way around. Because there, the current through all elements is the same. We are measuring the voltage across each element. Therefore at low frequencies when Xc> XL, the circuit is capacitive in nature, and when XL > Xc, it is inductive in nature. I hope it will help you.
ALL ABOUT ELECTRONICS now for parallel as you said if more current flowing through the inductor it is inductive and same on another side it is capacitive it is because in parallel current is same . Then please elaborate me for series where voltage is same then why we say in case of series that at lower frequency XC>XL i.e capacitive and p.f is leading and for hight frequency XL>XC i.e inductive and p.f is lagging (Now in series here leading means more current should flow from capacitor so according to that XL>XC at lower frequency and for higher frequency lagging means more current should have flown through inductive so XC>XL why it is not like that in series.) Please help me that. Thank you.
In the trainer kit, in parallel resonance circuit R is connected in series with L and C( L and C are in parallel). Do we call this configuration and the configuration shown in the video as parallel resonance circuit?
Since it is a parallel resonant circuit, when Xc > XL, then more current will flow through the inductor. In parallel, the voltage will remain the same. But more current is flowing through the inductor. That's why it is inductive. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS mean in which component the more current will flow will decide the behaviour of the circuit whether it will be inductive or capacitive and thankyou for clearing my doubt😇😇😇🥰
Yes, atleast in parallel circuit. In series RLC circuit, its other way around. Because in series, the current is same through each component. So when XC > XL, the circuit will be more capacitive, since there is more voltage drop across capacitor. I hope it will clear your doubt.
Clarification needed!!!! If we want to listen to all the other frequencies as in listening to songs, why can't we use rejector circuit or band stop filter instead of band pass filter?
This is parallel RLC circuit. So, when Xc>XL, more current flow through the inductor than the capacitor. So, nature of the circuit will be inductive. For inductive circuit voltage leads the current or we can say that current lags the voltage. I hope it will clear your doubt.
Yes, That is correct. Because in series RLC circuit, when Xc>XL, there will be more voltage drop across capacitor than across the inductor, And circuit will be capacitive in nature. So, current leads the voltage.
Sir at resonance XL=XC,it also says that current and voltage are in phase so it is purely resistive,so Impedance should be minimum and current should be maximum
When Xc is greater than XL, then inductor draws more current. Hence, it's inductive in nature. ( In parallel RLC CIRCUIT). In, series RLC circuit, it's other way around. Means when Xc is greater than XL,then circuit is capacitive in nature. I hope it will clear your doubt.
Sir, what will happen if we have a current source in place of the voltage source? Sir, I feel that in such case the current through the resistor should be maximum at resonance since the capacitor and inductor would become open. Sir, is my guess correct? please help me ...
In the parallel RLC circuit, when the voltage is applied to the circuit, then the voltage across each element remains the same. But at the resonant frequency, the admittance becomes minimum or impedance becomes maximum. I hope it will clear your doubt.
It will remain the same. The thing is in the parallel RLC circuit, we are measuring the current, as the voltage across each element is same. And if you see the response of the current through the resistor then it will remain the same.
So... the resonant frequency are the same in parallel and series RLC circuit? Okay, I got it (the part XC=XL, then omega^2=1/LC). But I stumbled upon a circuit which R is series to L, then both R and L is parallel to C. When I saw the solution manual, the frequency at resonance is something like omega^2=(1/LC)*(1-(R^2*C/L)). Do you know where the R from (1-(R^2*C/L) come from? All the resources in the internet and your videos I read and watched said that omega^2=1/LC and XC=XL, but no other further info about R at resonance. Did I miss something? :' Btw, awesome explanation 👏
when XL=0, then inductor will act as a short circuit. So, entire current will flow through So, at that time it doesn't matter what is R and XC in the circuit because the entire current is flowing through the inductor. So, in general, at lower frequencies, inductive reactance will be dominant as the majority of the current is flowing through the inductor. I hope it will clear your doubt.
and i could see that you have used this formula when calculating I1 for bandwidth. i.e a+jb = (a^2+b^2)^(1/2) could you please elaborate this?i am stuck here.
Sir, will this come in JEE exam.. Pls answer soo sir.. I have JEE tommorow..... Its not given inmy module but i have seen this somewhere bfore so i came here....did this concept of parallel circuit ever came in JEE? Or its just electrical engg stuff
I don't understand how you got from the second step, into the third step where you have the square root(see 13:02)...I think that's mathematically incorrect..
Its the amplitude or modulus of the complex number. Let's say we have one complex number A+ jB. Then its amplitude or modulus will be sqrt (A^2 +B^2) I hope it's clear to you now.
Ah yes, I understand now. I have another question though. Why is the energy stored per cycle only considers one factor, either the inductor or the capacitor? Why not the total of them both?
Because the R is constant At low frequency the capacitor is open and the inductor is short The inductor will have the lowest reactance this causes a low impedance At high frequency capacitor will act short circuit and that will provide a lower reactance this also cause low impedance At resonance both will be equal So the impedance will be the highest
12:30 I know this is just maths but can someone clear me out this thing? It just knocks my dumb brain out when I saw this suddenly. How the imaginary j disappear in a sudden?
Yes, That's true. At these points impedance will be maximum and the current through the circuit will be minimum. But when XL = Xc, impedance of the circuit will be minimum and current will be maximum.
@@ALLABOUTELECTRONICS But then the graph at 7:17 says that current will be minimum and impedance maximum at Wr . Also, when the circuit is capacitative (Xc>XL), shouldn't I lead V and lag behind V in an inductive (XL>Xc) circuit ?
Oh, I beg your pardon. What I said is true for series RLC circuit. For parallel RLC circuit, the impedance will maximum and current will be minimum at resonant frequency. Let me explain in this way, it is parallel RLC circuit. So, at the given frequency from which circuit component (inductor or capacitor) maximum current is flowing will decide whether a circuit is inductive or capacitive. At w=0, XL = 0 and XC = infinite. (i.e XC>XL). But as XL =0, entire current will flow through inductor. Or it can be said that circuit is inductive. Likewise at w= infinity, XC= 0 and XL= infinite. So, at a very high-frequency majority of current will flow through capacitor and circuit will be capacitive. So, if you move away from the resonant frequency then the current will increase. But at resonant frequency current will minimum or it can be said that impedance will be maximum. I hope now it will clear your doubt.
Below resonant frequency, since Xc is greater than XL, so more current flows through inductor. Thats why the circuit is inductive. ( Even though Xc > XL). The reverse happens above resonance. Because above resonance, the XL > Xc. So, for the same voltage, more current will flow through the capacitor ( because of lower reactance). And therefore, the circuit is capacitive. In case of series resonance, it's totally opposite.
I was stuck at the same problem, but the fact is, XL and Xc are in parallel, take the equivalent of reactance or impedance of these two branches, then compare at different frequencies.
Basically, it represents the phase information. It represents the leading and lagging of current and voltage. For more info, please check the video on Phasor and Phasor diagram. Here is the link: ruclips.net/video/zlmwmvijn1Y/видео.html
But by putting C= 1/√(w2L) in Q factor formula Q=wRC, we get Q= R/wL which is opposite to that derived by taking inductor. By taking inductor in place of capacitor Q= wL/R
Because the curve is for the current. If you see the impedance, curve at the resonance, the impedance is maximum. But since the current is V/Z, so curve gets inverted.
Cleared all my doubts no body explained this topic in so depth thank you very much
I found a problem in one of sample paper of jee main about accept or rejector
My teacher said it is out of scope of mains but the way you teach I understood almost everything
Thank you
Indians are like what more formulas should I eat up to get more marks
@@abhijithanilkumar4959 please do not generalise and atleast search jee exam it is a conceptual exam not all about formulas
Super sir i have doubt on this topic ihave seen so many videos but I can't understand them u made me to understand clearly thanks sir
Thank you so much for making this great video in English. I have seen very good videos, but in Hindi, which I cannot understand.
BEST VIDEO ON THIS TOPIC. I WANT TO BUY YOU A MEAL
Nice and in ful depth .. I really enjoyed it.. and note down lots of point in my book in between :)
One of the best channels for electronics students
at the resonance the impedance should be minimum because xL=xC, and the current is maximum
A
Current is maximum in series rlc Circuit this is parallel rlc Circuit bro🥱
Esi video ki talash thi✌️✌️
Awesome explanation 👍👍🙏🙏🙏🙏🙏
Whew! NOW I can understand better!
Teachers should start with the more complex stuff AFTER the fundamentals of a subject.
THANK YOU!
(It was a lot of stuff! My head hurts, but I think it's because Im hungry!)
at 13:11 how you put v/r bro this parallel rlc kya isme bhi z=r hota hai at resonance ? also uske 2 3 step pehle jab tumne sq aur root kiya to usme ek -2L/C ka term bhi aayega
AT 18:16, it was mentioned that Q=R/XL but actually Q=XL/R=XC/R
in seriec RLC circuit Q=XL/R, while in parallel RLC circuit Q= R/XL
Very lucid and clear explanation.Thanks
Thanks. Also spelling of Summary is wrong at 19:05 .
Well explanation. So many helpful
Sir in my book it is mentioned that parallel LC circuit is rejector circuit not parallel LCR circuit. Is it wrongly written? Because at 3:31 you mentioned parallel LCR to be rejector not parallel LC.
When in the book, they say LC, it is actually the RLC circuit only. Because every inductor has some finite resistance. So, when one connects L in parallel with C, it is L in series with R (internal resistance of L) and that combination is in parallel with C.
That circuit too acts as a rejector.
But its resonance frequency is slightly different from the circuit which is discussed in the video.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS
thank you sir for your earliest response.
However as you've mentioned at 6:47 that when Xl=Xc, resonant angular frequency is 1/√LC
So even if we take an ideal LC, circuit, for its resonant angular frequency, there also we'll do Xl=Xc which will give 1/√LC.
I couldn't understand why’d you mentioned in the previous reply that there'll be “slight” difference, please clarify?
@@umasrivastava4142 The case which is discussed in the video is when R, L and C are in parallel with each other.
But here, I am talking about the case, where R and L are in series and the combination is in parallel with C.
I hope it will clear your doubt. In this case, wo = sqrt [ 1/LC - (R^2/L^2)]
I hope it will clear your doubt
@@ALLABOUTELECTRONICS yes sir it's perfectly clear now. Thank you so much sir 🙏
Clarification needed!!!!
Can anyone please tell me @12:41 which formula did he use?
sqrt (a+b)^2 OR {(a+b)(a-b)}
which one?
none of them is matching.
Its the magnitude of the A+ jB . i.e sqrt (A^2 + B^2)
@@ALLABOUTELECTRONICS Thanks a lot.
Sir but we have anti resonant circuit ,.is this parallel resonance same as anti resonant circuit
Parallel rlc circuit is also termed as anti resonance circuit
U are a life saver😭
Very detailed explanation great sir
💯 percent valuable video.thank u very much sir
At 15:49 How you get the value of omega2?
At 14:36, by solving the second equation, you can find w2.
If resonance angular frequency is wr
if w < wr so circuit act as capacitive or inductive ? for parallel resonance and serise resonance
In case of the parallel resonant circuit, when w < wr, the circuit is inductive. Watch the video from 6:49 onwards.
In case of a series resonant circuit, when w < wr, the circuit is capacitive.
thanxx
Sir how it became wc-1/wl at 4.45?
1:44 If there's a resonance, the Z = sqrt (R^2 + (XL - XC)^2)
= sqrt (R^2 + 0), so won't the impedance minimum?
For parallel RC Circuit, the admittance (Y) is [1/R + j(wC -1/wL) ]
And at resonant frequency, wL= 1/wC.
i.e Y is minimum. Hence Z is maximum and due to that current, I will be minimum.
I hope it will clear your doubt.
Sir u said, series RLC circuit is a acceptor circuit because this is accepting frequencies which are around resonant frequency. But this circuit is also rejecting frequencies which are far away from resonant frequency then why can't it be rejector circuit?
The series RLC circuit acts as a bandpass filter. So, it accepts a certain band of frequencies and rejects the remaining ones.
@@ALLABOUTELECTRONICS Thank you sir
Why it is Wc-1/WL when take common from j... Please explain that part
At 12:26 Xc=1/wc and XL=wl why have you substituted in reverse?
As Xc>XL then It should be [ 1/wc - wL ]
Sab reverse hojayega
what is this software you use to draw circuits and write everything?
Will this come in JEE
Sir in case of band width we use frequency f why u use omega w??
w is angular frequency. w= 2*pi*f.
So, w and f can be used interchangeably.
If the resonance frequency is to be developed in LC parallel circiut then the DC voltage must be applied during the first quarter of every cycle Rest of the wave form which is 3/4 th of the wave is self generating Thank you for your feed backMukundan
If it’s just parallel L C, does it work the same?
If there is no R, then the circuit is highly selective. Meaning that Q will be very high. Just put R equal to infinity in the Q-factor equation. Ideally Q would be infinite and It will act as an oscillator.
@@ALLABOUTELECTRONICS sir in LC circuit, R is almost zero...(?)
Thank u so much sir for ur excellent explanation ..... 💐💐💐💐💐💐
Are in college or in 12?
5:27, shouldn't be that 1/wL = wC?
Mathematically, both are the same. isn't it ??
GREAT EXPLANATION SIR .
Great help .... thank you so much sir
At 5:00 the step may be wrong. How you can take j common? Where in case of wc it is in numerator and in case of wl it is in denominator.
1/j = -j, If you closely observe, there is a negative sign in the next step.
@@ALLABOUTELECTRONICS How 1/j becomes -j? Help me to understand this.
You would have studied in the high school that, j^2= -1. So, 1= -j^2.
Hence, 1/j = -j
@@ALLABOUTELECTRONICS Thanks a lot. 🙏🙏
These series for electrical circuit analysis are very useful
while calculating quality factor energy stored in capacitor is used why not energy stored in inductor , and why two of them give different equation. help plz
Nicely explained. Cheers :)
At 17:53 how omega r is written root c/L
Very helpful!..😊😊😊😊😊
Thanku soo much sir
Very nicely explained 👏👏😃
you must cover every topic..what frequency it wil discard low or high ?
at 12.43 how third equation changes into 4th..........!
square cannot be applied on individual item by applying whole under root............?
or i am missing some thing
It's correct try to watch it again
I thought in a parallel RLC circuit the equation for resonant frequency w=1/sqrt(LC) drops out and doesnt work? Is this not true then?
Could u please tell me why the Vp curve is upward pointed ... even though the Ip curve is downward point ? Coz according to ohms law Vp and Ip are proportional... so shouldn’t both be of the same shape ?
what about Power Factor at Resonance RLC parallel circuit ?
At resonance, the power factor will be Unity because at resonance, Xc= -XL.
what are the variations in the parallel LCR circuit?
Ye jee me aayega bhai?
@@starscream1457 nhi aayega bhai, jha tk mujhe pta h
6:54 it should come XL > XC therefore lagging portion . And the way you have written XC > XL it comes in series resonance circuit.
Let me explain it to you intuitively. When Xc > XL, more current will flow through the inductor. So, at low frequencies, in parallel resonance, the circuit is inductive in nature. At high frequencies, XL > Xc. Therefore more current will flow through the capacitor and hence the circuit is capacitive in nature. For series resonance, it is other way around. Because there, the current through all elements is the same. We are measuring the voltage across each element. Therefore at low frequencies when Xc> XL, the circuit is capacitive in nature, and when XL > Xc, it is inductive in nature.
I hope it will help you.
ALL ABOUT ELECTRONICS now for parallel as you said if more current flowing through the inductor it is inductive and same on another side it is capacitive it is because in parallel current is same . Then please elaborate me for series where voltage is same then why we say in case of series that at lower frequency XC>XL i.e capacitive and p.f is leading and for hight frequency XL>XC i.e inductive and p.f is lagging (Now in series here leading means more current should flow from capacitor so according to that XL>XC at lower frequency and for higher frequency lagging means more current should have flown through inductive so XC>XL why it is not like that in series.) Please help me that. Thank you.
In parallel voltage across each element is same while in series current through each element is same.
If W< Wr then circuit is capacitive means current should lead voltage while W>Wr circuit is inductive and current should lag voltage
In the trainer kit, in parallel resonance circuit R is connected in series with L and C( L and C are in parallel). Do we call this configuration and the configuration shown in the video as parallel resonance circuit?
At 7:05 As Xc is greater than XL then circuit should behave like capacitive then how it is inductive please clarify this #allaboutelectronics
Since it is a parallel resonant circuit, when Xc > XL, then more current will flow through the inductor. In parallel, the voltage will remain the same. But more current is flowing through the inductor. That's why it is inductive. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS mean in which component the more current will flow will decide the behaviour of the circuit whether it will be inductive or capacitive and thankyou for clearing my doubt😇😇😇🥰
Yes, atleast in parallel circuit. In series RLC circuit, its other way around. Because in series, the current is same through each component. So when XC > XL, the circuit will be more capacitive, since there is more voltage drop across capacitor.
I hope it will clear your doubt.
Clarification needed!!!!
If we want to listen to all the other frequencies as in listening to songs, why can't we use rejector circuit or band stop filter instead of band pass filter?
1. Current value is less at the resonating frequency of parallel LCR
2. Q factor will be affected
how current lags voltage when Xc>Xl?
This is parallel RLC circuit. So, when Xc>XL, more current flow through the inductor than the capacitor. So, nature of the circuit will be inductive. For inductive circuit voltage leads the current or we can say that current lags the voltage. I hope it will clear your doubt.
ok sir..thankx a lot..bt in series resonance current leads when Xc>Xl..is this correct??
Yes, That is correct. Because in series RLC circuit, when Xc>XL, there will be more voltage drop across capacitor than across the inductor, And circuit will be capacitive in nature. So, current leads the voltage.
Likes for both of you.
Sir at resonance XL=XC,it also says that current and voltage are in phase so it is purely resistive,so Impedance should be minimum and current should be maximum
Is law of conservation of current not valid in parallel rlc circuit
If XC is greater than XL before the resonance frequency then it should be capacitive in nature ...why it is inductive???
When Xc is greater than XL, then inductor draws more current. Hence, it's inductive in nature. ( In parallel RLC CIRCUIT).
In, series RLC circuit, it's other way around. Means when Xc is greater than XL,then circuit is capacitive in nature. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thanks sir 🙏😊
Please describe for mix circuit
Sir, what will happen if we have a current source in place of the voltage source? Sir, I feel that in such case the current through the resistor should be maximum at resonance since the capacitor and inductor would become open. Sir, is my guess correct? please help me ...
In the parallel RLC circuit, when the voltage is applied to the circuit, then the voltage across each element remains the same. But at the resonant frequency, the admittance becomes minimum or impedance becomes maximum.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Sir, will the response will be the same if we replace the voltage source with the current source?
It will remain the same. The thing is in the parallel RLC circuit, we are measuring the current, as the voltage across each element is same. And if you see the response of the current through the resistor then it will remain the same.
@@ALLABOUTELECTRONICS Thank you so much for your kind help sir ! ❣
I think at resonance condition current is maximum,
so the impedence is minimum..
Crt me if i am wrong..
It's for series RLC circuit, for parallel RLC , impedance is max.
Can you explain how the impedance in parallel RLC circuit is maximum
So... the resonant frequency are the same in parallel and series RLC circuit? Okay, I got it (the part XC=XL, then omega^2=1/LC). But I stumbled upon a circuit which R is series to L, then both R and L is parallel to C. When I saw the solution manual, the frequency at resonance is something like omega^2=(1/LC)*(1-(R^2*C/L)). Do you know where the R from (1-(R^2*C/L) come from? All the resources in the internet and your videos I read and watched said that omega^2=1/LC and XC=XL, but no other further info about R at resonance. Did I miss something? :' Btw, awesome explanation 👏
omega^2=1/LC is based on the assumption that coil doesn't have any resistance of its own in parallel RLC.
vlo laglo vai
100%
Thank you very much
why is it that when XL=0 the inductuctive reactance is dominant?
when XL=0, then inductor will act as a short circuit. So, entire current will flow through So, at that time it doesn't matter what is R and XC in the circuit because the entire current is flowing through the inductor. So, in general, at lower frequencies, inductive reactance will be dominant as the majority of the current is flowing through the inductor.
I hope it will clear your doubt.
oh i get it. thanks
and i could see that you have used this formula when calculating I1 for bandwidth. i.e a+jb = (a^2+b^2)^(1/2)
could you please elaborate this?i am stuck here.
tayenjam jeneetaa hello friends
@@tayenjamjeneetaa494 that is root of A Square and Bsquare
Super! Thanks!
Effective explanation!!!!!
What is the application? Please give example.
if RLC circuit are connected in parallel with a dc voltage source then no of node ?
Apart from the reference node (ground), it has only one node.
What is quality factor only in terms of L and R sir?
For parallel resonant circuit, Q = R / ωL.
I have already mentioned in the description of the video.
Sir, will this come in JEE exam.. Pls answer soo sir.. I have JEE tommorow..... Its not given inmy module but i have seen this somewhere bfore so i came here....did this concept of parallel circuit ever came in JEE? Or its just electrical engg stuff
I don't understand how you got from the second step, into the third step where you have the square root(see 13:02)...I think that's mathematically incorrect..
Its the amplitude or modulus of the complex number.
Let's say we have one complex number A+ jB. Then its amplitude or modulus will be sqrt (A^2 +B^2)
I hope it's clear to you now.
Ah yes, I understand now. I have another question though. Why is the energy stored per cycle only considers one factor, either the inductor or the capacitor? Why not the total of them both?
At 4:51 why it is wc-1/wL
Because 1/ j = -j
Due to vector addn ....i.e using phasor diagram they lag by 180degree hence they are substracted ...correct me if i am wrong
why will the impedance be maximum?
Because the
R is constant
At low frequency the capacitor is open and the inductor is short
The inductor will have the lowest reactance this causes a low impedance
At high frequency capacitor will act short circuit and that will provide a lower reactance this also cause low impedance
At resonance both will be equal
So the impedance will be the highest
current maginify ka reason koi explain kar do plz...
12:30 I know this is just maths but can someone clear me out this thing? It just knocks my dumb brain out when I saw this suddenly. How the imaginary j disappear in a sudden?
a+jb = √(a^2+b^2) complex conversion
Xc>XL then the circuit would be Capcitive than inductive and the current lead the voltage ??? If yes. so plz again explain the sec graph of the vidio
But at w=0, we get Xc=infinity and XL=infinity at w=infinity, so doesn't that make the impedance maximum at these points ?
Yes, That's true. At these points impedance will be maximum and the current through the circuit will be minimum. But when XL = Xc, impedance of the circuit will be minimum and current will be maximum.
@@ALLABOUTELECTRONICS But then the graph at 7:17 says that current will be minimum and impedance maximum at Wr .
Also, when the circuit is capacitative (Xc>XL), shouldn't I lead V and lag behind V in an inductive (XL>Xc) circuit ?
Oh, I beg your pardon. What I said is true for series RLC circuit. For parallel RLC circuit, the impedance will maximum and current will be minimum at resonant frequency.
Let me explain in this way, it is parallel RLC circuit. So, at the given frequency from which circuit component (inductor or capacitor) maximum current is flowing will decide whether a circuit is inductive or capacitive.
At w=0, XL = 0 and XC = infinite. (i.e XC>XL). But as XL =0, entire current will flow through inductor. Or it can be said that circuit is inductive.
Likewise at w= infinity, XC= 0 and XL= infinite. So, at a very high-frequency majority of current will flow through capacitor and circuit will be capacitive.
So, if you move away from the resonant frequency then the current will increase. But at resonant frequency current will minimum or it can be said that impedance will be maximum.
I hope now it will clear your doubt.
@@ALLABOUTELECTRONICS Yes, it did .
Thank you so much !
I love your channel !
7.42 correction.
Frequency below the resonant frequency circuit is capacitive.
Above resonant frequency circuit behaves like inductive.
Below resonant frequency, since Xc is greater than XL, so more current flows through inductor. Thats why the circuit is inductive. ( Even though Xc > XL). The reverse happens above resonance. Because above resonance, the XL > Xc. So, for the same voltage, more current will flow through the capacitor ( because of lower reactance). And therefore, the circuit is capacitive.
In case of series resonance, it's totally opposite.
I was stuck at the same problem, but the fact is, XL and Xc are in parallel, take the equivalent of reactance or impedance of these two branches, then compare at different frequencies.
Can you provide pdf of this?
Thank u so much sir
Good content.thank u
Subscribed !
Is this is in jee mains syllbs
XL=jwL
Sir, why you write j with wL? Kindly tell me what is j ?
Basically, it represents the phase information. It represents the leading and lagging of current and voltage.
For more info, please check the video on Phasor and Phasor diagram.
Here is the link: ruclips.net/video/zlmwmvijn1Y/видео.html
What does j means in jomegal
Can anyone help me with a parallel rlc circuit numerical...???
Solving Q with inductor equations..
Q=[w*L*(Irms)^2]/(Irms)^2*R
Cancelling Irms^2 we get
Q=wL/R=XL/R
Correct me if I'm wrong
But by putting C= 1/√(w2L) in Q factor formula Q=wRC, we get Q= R/wL which is opposite to that derived by taking inductor. By taking inductor in place of capacitor Q= wL/R
Nailed it...
Where did he nailed it,Is it perfectly stuck now,Or is it shaking,?
excellent...keep doing
Explain the term "j"?
thanks sir !
At 04:54, you take "j" common,
There is a mistake, recheck it Sir.
1/j=-j, there's no mistake
nice work
w1 and w2 are called half-power frequencies, but for parallel circuits, you calculated them by doubling the power. why?
Because the curve is for the current. If you see the impedance, curve at the resonance, the impedance is maximum. But since the current is V/Z, so curve gets inverted.
@@ALLABOUTELECTRONICS i am asking that why are w1 and w2 called 'half' power frequencies when they are obtained by 'doubling' the power?
Thank you
I think it better to use a current source for the parallel circuit.
omg, thanks a lot!
Thank you!!
i have a problem sir...