Most Important RC Circuit Concept, Analog Interviews

Here me and my friend have discussed a very important concept for analog interviews

[@ytaccount9420]

Please continue making these videos!!! this was extremely helpful

[@analogeasy-peasy7559]

Watch @ 1.25X also @ start my mic is quite away from my mouth so please embrace the low sound.

[@shashankmourya4366]

Cool concept 👍

[@analogeasy-peasy7559]

Cooler than AC or Desert Cooler?

[@shashankmourya4366]

May be like Antarctica 😜

[@soumil6135]

excellent Sir, please upload more videos on these topics and questions asked in TI, INTEL, SAMSUNG AND OTHER MNCs

[@analogeasy-peasy7559]

Well noted sir

[@soumil6135]

May I get your contact information? Sir

[@analogeasy-peasy7559]

[e-mail]

[@anujagrawal1036]

God question, having great time with your videos.Pls upload more videos. Is there any way i can contribute to your initiative?

[@analogeasy-peasy7559]

Thanks man! Yeah rn you can share some questions maybe [e-mail]

[@RaviKumarlk-r6y]

What about voltage across resistence can we neglect

[@JennyTheBear]

In both cases capacitor are parallel from vx, how come 2nd came in series

[@saurabhtrivedi6181]

They came in series because of Infinite current. Think of capacitor being a wire, so no current will flow through the resistance. If still not able to understand follow Nagendra Krishnapura lectures on the same. :)

[@jastinaveen4128]

At t=0+, there will be sudden change(jump)... Sudden change is an high frequency phenomenona right...So for high frequency phenomena, the capacitance will exhibit almost low impedance.. The resistor in parallel with low impedance gets open circuited and hence both capacitors are in series...

[@sujayvikramgs8588]

have an interview test in some time. im spamming these and im at so at peace

[@akanchhakumari5666]

could you tell, where i can get more RC circuits questions? which are being asked in interview ..any resources

[@analogeasy-peasy7559]

There r not many actually. I got all from my fellow students who appeared for the Analog Interviews.

[@akanchhakumari5666]

@@analogeasy-peasy7559 okay

[@akanchhakumari5666]

@@analogeasy-peasy7559 sir please continue this series ❤️

[@jastinaveen4128]

Could u solve the same circuit considering some C1 and C2 has initial charge rather than uncharged...This makes the circuit more challenged🔥 than the previous one...

[@analogeasy-peasy7559]

Surely that will be hard 🥵🙈

[@goks_cool]

@@analogeasy-peasy7559 But won't that be the same except that Vx(0+) would be the Initial potential across C2?

[@98505177229850590818]

7:21 i think you are missing bracket closing in equation

[@govindreddy1810]

Sir can u explain the interpretation of bode plot?

[@analogeasy-peasy7559]

That's quite a task to do in comment section.

[@hemantj675]

Isn't it a second order circuit?

[@aakashraathod1996]

why won't the capacitor voltage does not change suddenly concept follow, the capacitors were discharged at 0 minus and still at 0 plus due to sc voltage will be 0...the real reason should be something else

[@aakashraathod1996]

i think it is related to charge sharing between capacitors in series

[@ameenpasha3913]

Hi, The transient equation(v(t)=v(Infinite)+(V(0)-V(Infinite)*e^-(t/Tou))) which you used to solve the circuit is valid only for 1st order systems where you can replace the two or more capacitors with a single capacitor with its equivalent value(This is applicable to 2nd question where 2 Capacitors are in series and they can be combined and represented as a single Capacitor of equivalent value). But in the 1st circuit, the capacitors are neither in series nor in parallel, So its means that the circuit is a second order system and hence the transient equation you used is invalid to solve the question. So we go with Laplace transform method to solve second order equation in order to get the correct answers. And the explaination which you provided for Voltage across the capacitor will not be the same when infinite current flows has nothing to do with the concept of Capacitors. If both the capacitors get's short circuited, then voltage source will get short circuited which will violate the KVL and the basic property of the Voltage source will be violated which is a voltage source cannot be short circuited. Regarding the Calculation of Time constant, it can only calculated directly by using the circuit when the circuit is containing only one Capacitor/Inductor or when you have multiple (capacitors/Inductors) which can be replaced by its equivalent single Capacitor. Otherwise for 2nd order systems, the time constant will be calculated only after taking the inverse Laplace transformation and finding the voltage across capacitor. You cannot find it directly as you did in the 1st question. And Time constant is given by (Tou= Equivalent resistance of the circuit* Equivalent Capacitance of the circuit) for First order RC circuit and (Tou= Equivalent Inductance circuit/Equivalent Resistance of the circuit) for first order RL circuits The explaination you provide with the equation you used holds good for first order system not for second order system. I hope this explanation is helpful. Have a great day!!

[@shridharambady2069]

Nope, it is a 1st order system. There is only 1 *independent* energy storage component. If you define the voltage across one capacitor, the voltage across the other is automatically defined by Vin and the first capacitor. Try solving out the transfer function for yourself, you'll see there is only one pole.

[@tanmayvadhera4250]

Thats a first order system !

[@motivationalvideous6060]

Order defined by energy storing elements

[@motivationalvideous6060]

Second order laplace direct