Working of Inductor and Capacitor LC circuit with a DC supply, a graphical study

[@pedrobolsi8366]

Good video my friend, good animations too. It was a good review for me of this fundamental behavior. For those curious to what would happen after, since now the capacitor voltage is larger than source voltage (Vc > Vs), there is a difference in potential, meaning the capacitor will discarge, until it is equal to source voltage (Vc = Vs). As the capacitor is discharging its voltage, the current in the inductor is also building up to Imax, as is in the video, but in the opposite direction (-Imax), since energy is going back to the source. The inductor current is at -Imax at the same moment that capacitor voltage is at Vc = Vs. Now the capacitor current wants to go to zero, because the voltage is in equilibrium, but the inductor current is still at -Imax. Therefore, as in the video (but in opposite direction), the current will continue to flow into the source until the inductor's current is back at zero. Meanwhile, as the inductor is maintaining the current flow in the negative direction, the capacitor voltage is still discharging, and will reach Vc=0 at the same time as the inductor current Il=Ic=0. After this, we are back at the beggining. The capacitor will charge up to 2*Vs again, and this will happen over and over... This is ressonance.

[@raloed.363]

so why when we connect this dc circuit and measure the voltage on the capacitor it is not even greater than the supply voltage. I even added a diode in series to prevent the capacitor from discharging through the inductor and still no greater supply voltage. I'm assuming my multimeter is not sensitive enough to detect the changes or its because practical components are not ideal. could someone please do this simple experiment with an oscilloscope and post the results. please it will really help

[@shoutitallloud]

@@raloed.363 Because this whole explanation in video is nonsense. The capacitor voltage will be the same as battery. The inductor in circuit would change the time to fully charge a capacitor

[@raloed.363]

@@shoutitallloud ok so here is another question. When the capacitor is connected to power source, the charging current starts at its maximum. but when inductor connected to power source the current flowing cannot rise to maximum instantly so it starts at minimum. So what happens when the series inductor capacitor is connected to the power source? Does the current start at maximum or minimum?

[@shoutitallloud]

@@raloed.363 It starts to increase from 0. Have a look at this video. It's much more clear. kzbin.info/www/bejne/bHS9pYOegpxpp8U

[@potencializadordeenergiagr7551]

I was forced to stop the video to comment again on the way of explaining your video, very correct and easy to understand.

[@malwaysfine]

This is what I was looking for. Short and to the point to what happens to V and I in this setup. Thank you!

[@gargibasu5111]

Excellent. Looking forward to more such splendid lectures.

[@kanthakborkar710]

That's what I was looking for.. thanks a lot

[@jacksonsee2746]

This problem makes me sick many times , until watched your video, I am free now. Thank you, Sir.

[@Prateekpushp]

Exactly what i was hounding since a long time...thankyou so much bless you

[@ProfeARios]

Thank you so much for sharing!!!! Regards from the Republic of Panama.

[@omakarthakare4397]

Great ,your thought of making video on such a basic n important concept is really awesome

[@Xplanator]

Thank you so much! Your Appreciation means a lot!

[@anjalikankarwal9975]

Amaze!!! Perfectly explained. Thanks buddy

[@amp_001]

Very simple explanation I needed some detailed explanation with laplace transform

[@akshay_garje]

Good explanation brother. Well done.👍👍👍

[@tcoo1999]

Hope this isn't a stupid question. If the Capacitor was positioned before the inductor, would current flow backwards to the capacitor? Great video also.

[@bandaruramana2119]

Excellent explanation with animation kindly please explain for DC parallel combination

[@artichoudhary-lz6em]

Love you Bhai great explanation

[@Saveindian]

I didn't get such type explanation yet. But you provide us

[@StanHunt007]

So beautifully explained sir. Thank you.

[@Xplanator]

Thank you :)

[@ridenhour1]

Thank you for making this video

[@dawoodaltai6113]

this video helps me thank you so much

[@fantasywarriors278]

Jaber dast yahi to chahiye tha

[@R945-l6f]

This is the simplest definition thanks

[@Sarkaari_Shayar]

Plz make a video on behavior of LC circuit on ac supply.

[@Piyushthote0867]

In LC circuit with AC voltage. 1 voltage and current have phase 90°. No disppation of energy.

[@Peepstep]

how did you make this video..what tools and softwares you used

[@tanyatripti9855]

Will it show oscillation after current becomes zero every time...because them there will be a potential difference between capacitor and battery every time the current becomes zero

[@ambujkr]

Yes Tanya. This will keep on going. You can also verify it by calculating the voltage across the capacitor using inverse Laplace transform. It will be sinusoidal.

[@knightofmordor]

Clearly explained!

[@Xplanator]

Thank You :) Do subscribe to stay updated :)

[@jamilajamila3658]

is the induced back emf goes to maximum point instantly, i mean when the switch is on? and then decrease with increasing flux?in case of DC supply.

[@tab5668]

Thank you so much sir very informative

[@Xplanator]

Welcome :)

[@rameezjadoon2958]

Good explanation......

[@anuragmittal9816]

if a diode is connected in series with the arrangement , ie half wave rectifier ..how will the voltage waveform across diode look like?

[@rodericksibelius8472]

So How do we find out the current in the circuit using only 'algebra'?, or can you explain the mathematical electrical engineering physics how we derive all the equation behaviors of this dc battery/ inductor / capacitor circuit? Thank You for the qualitative explanation.

[@bandaruramana2119]

What happened when L & C are connected to ac supply how they exchange power

[@ratnadeeppatra4252]

what is the value of i max??? n wht will happen after i has reached 0,will it rise again (oscillations)??

[@neminathnarule2257]

Thank u so much really infinity thanx ..thats what i was looking for great work sir ...👌👍👍👏..one question please answer in inductor as current is going from zero to max value so the shape of waveform is convex from zero to maximum valu of current ,but capacitor is also charging from zero to Vs so capacitor is not having that particular shape of waveform from zero to Vs why that bend?

[@joshuak.n5553]

Hello sir, this question was asked in IIT-JEE 2010 Physics. But they have given answer as V(capacitor)= V and V(inductor)=0 at steady state? Is this wrong

[@veronicanoordzee6440]

@ 0:34 I never understand why people say that energy is stored in the magnetic field. When the current builds up the inductor creates an opposite voltage. Then this voltage drops. When the current is maximum the voltage is zero (because dI/dt = 0) and the magnetic field reaches a maximum (because the current reaches a maximum). Why should this field contain energy??? In my opinion, the only energy in the circuit is the energy of the electrons being pushed around by the voltage. Faraday's law of induction is about changing magnetic fields creating a voltage, not an energy.

[@avishekmishra8177]

Same question, we read that in DC circuit an inductor act as a nothing but resistor, then how it will store energy, also di/dt = 0 here !!!!!

[@thevinhpham4594]

Love thiss ! thanks !

[@adityasingh-pu1kd]

What happens after the capacitor charged to twice the voltage..???

[@sarathchandra4323]

it will discharge into source

[@rahulbhavsar1402]

I don't know the diode with inductive circuits please understand me if you know that

[@pallerlacherishma985]

After discharging whether capacitor polarity gets reversed

[@pandu7820]

In DC frequency is zero So XL and Xc will be zero Then how we analysis LC circuit using dc supply

[@melplishka5978]

Very good. Ty

[@ShivaM6]

Thanks mate.

[@prashantshahu5686]

Bhai bhai 👍👍👍👍

[@BS-eu9do]

What does happen with AC?

[@omsingharjit]

Your way of explanation is different

[@l-_._-l]

A thumbs up from 🇵🇰

[@IronDuh]

wouldn't the reverse current fry the battery?

[@everythingaccount9619]

gonzalo samaniego it could. Depends mainly on the inductance and capacitance values.

[@bhaisahab538]

Sir if capacitor block dc then why current flow through it

[@llarry2009]

killer chora Capacitors store energy until it is discharged, the current is driven towards the positive side of the battery but stays in the capacitor mostly. That's what i got from it.

[@mrakjunior]

Excellent.

[@sonalsingh6046]

sir please explain why capacitor voltage becomes 2Vs

[@Xplanator]

Please refer the graph. The transfer of energy when current stops flowing in the circuit makes it charge to 2Vs.

[@sathyakashaboina2570]

@@Xplanator it violates KVL

[@shobra123a6]

Wonderful

[@hardikjain-brb]

nice

[@rahulbhavsar1402]

At t=0+ inductor act as open circuit than how capacitor charged at the= 0+

[@jaspreetsingh-yu4cp]

Or jab capacitor 2Vs charge hoga uske baad kya hoga ,Kya battery charge hogi vapis

[@Xplanator]

it is not a standalone circuit by itself. It is a circuit used to turn off thyristor. Forced commutation mein iska istamaal hota hai.

[@skh4113]

The voltage diagram of the capacitor and battery, is completly wrong! Brother. Check it up again using the diagrams on the internet.

[@ahmadeldesokey9844]

Thanks

[@Durga189]

Ok now there is a potential difference between source and capacitor. Now what happens? The answer is simple. Now capacitor has high potential,because of that current flows through inductor from higher potential to lower potential. Once the source voltage equal to capacitor voltage again inductor starts release energy. Like this the process continues.

[@mirzaataullahbaig413]

what in ac

[@antoines1341]

air conditioning

[@akashbaveja4045]

Bht ache...

[@shoutitallloud]

This is complete bullshit. You are confusing people with that nonsense

[@akakingcash55]

This is false knowledge

[@lukiepoole6701]

False knowledge your head. This is simplicity of LC tank circuit.

[@carlo697y]

This is false. If enough time has passed, this circuit will be in steady state (because this is a dc circuit). The capacitor will act as an open circuit and the inductor will act as a short circuit. Then, you will find out that the voltage across the capacitor will be equal to the voltage source (as there is no voltage drop across the inductor).

[@lukiepoole6701]

@@carlo697y I know lol. However the explanation above is transient state. He is saying that transient state is false knowledge. This is what happen when people neglect electrical engineering :v

[@nguyencuong-ye1gn]

can you show me the equation of capacitor's voltage in this circuit, please sir

[@Muck-qy2oo]

Why should it charge to twice the supply voltage?

[@Traderhood]

Excellent.