i was stuck on one problem for 8 hours today, i didint even need to get halfway through your vid to completely understand how to do it now. After 10 years you are still helping students thanks
As someone who was sitting right next to him attempting to figure this out, thanks for the reply! We've been using your video pretty extensively to figure these things out, I have to say your cheat sheet is really useful.
I have my circuits exam in a little bit less than 3 weeks and this will certainly help with passing, as it is the most complicated subject we will have on the exam. Thanks a lot for posting this! You did an amazing job!
Learning more here than I did the whole semester! Nice job! Thanks! But would be much more helpful, if had harder and/or different examples... even if you go trough them a little faster.
he made a mistake which you can see but he corrects it by using a describtion....in the vid its B2 t e^(-5t) that you derivate....(not B2 e^(-5t) as written in the vid)... which equals to B2 t' e^(-5t) + B2 t e^(-5t)'....(same way when you derivate [f(x)g(x)] '= f'(x)g(x) + f(x)g'(x)
@ 17:22 "This means the circuit is critically damped." Indeed it does, but you still need to mind your initial conditions. I solved this circuit in the s-domain and found the expression: I(s) = 3(s + 5)/(s + 5)^2. The zero in the numerator cancels one of the poles in the denominator and you get I(s) = 3/(s+5). This is a 2nd order circuit that exhibits first order behavior. Performing the ILT on I(s), you obtain: i(t) = 3*e^(-5t) ∎
***** Capacitors and resistors can't be treated the same in transient circuits. There is a voltage on the cap, but there is no current flowing in that branch the resistor is still bound by ohm's law. No current, no voltage. The reason there is no current is that the capacitor is fully charged in this instance - therefore it acts as an open
"this is the cheatsheet that I've created, you can pause the video and look at it but we'll come back to it throughout the video" every single student: 📸📸📸📸
Hi Matthew, nice video. Just a little confusion from what you said at 23:43, "we're looking for an equation for ALL-TIME AFTER t=0". It that were the case, then why would there be a 5v across the capacitor? wouldn't it be -10v? did you mean just when t first =0? sorry, not trying to nitpick, I just have a really hard time with 2nd order RCL and trying to learn.
Really nice job, Matt. Do you mind telling me what software you are using? I'd love to be able to follow your example with digital signal processing problems.
do you imagine capacitors as open and inductors as shorts initially no matter what? does it make a difference if the switch is originally open or closed?
if the source is connected for the time such that the capacitor will be charged fully and inductor stores maximum energy, than we consider them as open and short cause they don't have the capacity anymore. That's my concept. Correct me if I am wrong..
At 13:55 i don't get why we were changing the sign of the capacitors voltage. Why does it not remain a minus? I mean the current is entering though the negative terminal so by my understanding it should remain that way. Thanks to anyone replying.
Hi friend , i have to get the initial conditions from a circuit with 2 storage devices (L & C) but it also has 2 power supplies and they are always present even before and after the switch closes or opens , how can i get the initial conditions ? may i use superposition ? i mean , passivate one power supply & get the initial condition , then passivate the other power supply & get the initial condition , afterwards my 2nd system initial condition will be the sum of those partial ones ? thanks
In the first kvl the voltage contribution from the capacitor is +vc and the last time you do kvl with one minute left the contribution is -vc. This is not consistent. Why does the "passivity" change? Thanks
man its a 9 year old video but imma ask anyway.. So I applied Laplace transform to this example circuit, solved it and got the same expression of i(t)= 3e^(-5t). Same answer for less work it seems like. So what are the cases where the Laplace approach cannot be used?
It’s probably not necessary, but how would you go about finding the voltage drop across the 5 ohm res in the middle branch in series with the inductor when t=(0-)
The initial and final conditions are dealt with in the beginning of the video. The characteristic equation is to get the natural response of the circuit. By superposition they will sum together to get the final response. Check out forced second order ODEs.
@@angelojunior9507 There are better forums to ask on. Try groups like r/ECE or r/Electrical Engineering on reddit. You can also try #electronics on the freenode IRC network.
+Nick Spoutz The name of the book is on the title screen in the first few seconds of the video. You can get that book online for less than $20 usually. It's pretty good.
Capacitor and Inductor will act differently when switch is on or off depending on circuit. For this when switch on at 0-/0+ inductor will be short and capacitor will be open. But for infinite the switch is open and the source is therefor not connected to circuit so inductor and capacitor discharge away after the battery after the inductor dies. The resistor on top left doesn't get affected because its open there so it doesn't matter. Now again inductor is short and capacitor is open and making the entire circuit not connected with anything else so its nothing, 0. But when it was 0-/0+ the circuit was still connected and power was running so there would still be current since current depends on resistor and voltage and we still had those so we can calculate it.
Marc Zapatero i(0+) is the instant right after the switch is flipped. i(infinity) is after a long time, which is when the circuit has reached steady state. For inductors: iL(0-)=iL(0+)
Your initial Vc is incorrect; since there are 2 supplies, you need to use superposition when you perform KVL. Total voltage should be 15V when you add the 10V from the 20V supply. Said another way, since the cap is an O.C., the right most resistor is removed from the circuit, leaving a 1/2 voltage divider in series with a single 30V supply.
This is wrong. First, Superposition is unecessary. KVL works fine. Second, the 10V source is pointing downward, so (20Vs - 10vs) or 10V - 5V = 5V. I also checked this with a simulation (5V), not sure what clifford ayer did, except maybe make the same mistake.
KVL left loop: 20 = iL(5) + iL(5) - 10 (iL = 3) => 20 = 15 + 15 - 10 == 0 Voltage at 3-resistor node (Vc) using Nodal: [Vc - (-10)] / 5 = 3; Vc = 5V KVL in the left loop has nothing to do with it.
I stop watching at 16:48. I’m upset that while you were deriving your characteristic equations that you didn’t show all the steps. Now I’m left wondering how you got the expression you found. You should have brought the equation you were working with over. If you are teaching people show all the steps. Everything is important to people trying to learn.
In the first loop, you referenced the technique as the "KVL" when you should have said "KCL". Using KVL=Kirchhoff's Voltage Law means that you are using the node voltage analysis to calculate voltage.
Mesh uses KVL (adding up the voltages) and Nodal uses KCL (adding up the currents). He was correct when saying he was using KVL. I'm not sure what you're referring to, because when using Nodal analysis, your equations are set up as V/R + V/R . . . etc, which is current. Mesh uses R(I) + R(I) . . . etc, which is voltage. You may have gotten that backwards if I interpreted your comment correctly.
A little late now, but I believe it is correct the way he wrote it. If you do a mesh current for i2 (which is going to equal 0 A), you will get an equation of: 10V + 5(i2-i1) + 5(i2) + Vc = 0 Since i2 is 0A, this equation goes to: 10V + 5(0-i1) + 5(0) + Vc = 0 ----> 10V -5i1 + Vc = 0 Vc = 5i1 - 10V = [5ohms x (3A)] - 10V = 15V - 10V = 5V
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
I love how he says "Good luck" at the end, as if he knows that we're watching this to prepare for a test.
i was stuck on one problem for 8 hours today, i didint even need to get halfway through your vid to completely understand how to do it now. After 10 years you are still helping students thanks
As someone who was sitting right next to him attempting to figure this out, thanks for the reply!
We've been using your video pretty extensively to figure these things out, I have to say your cheat sheet is really useful.
I have my circuits exam in a little bit less than 3 weeks and this will certainly help with passing, as it is the most complicated subject we will have on the exam. Thanks a lot for posting this! You did an amazing job!
I have my midterm tomorrow and I am still lost.
@@BrownieX001 same :/
You should provide that cheat sheet in the description :)
you do an awesome job of demystifying first and second order circuits. I've watched both videos and you just saved my test grade for tomorrow. Thanks!
literally me rn, 9 year old comment
Great video Matthew! You literally saved me... I didn't understand a thing from lecture and the book
"Just a happy coincidence"
Officially the Bob Ross of circuits.
Learning more here than I did the whole semester! Nice job! Thanks! But would be much more helpful, if had harder and/or different examples... even if you go trough them a little faster.
Thanks for the quick reply and sorry for wasting your time with that one. Really appreciate the videos, keep up the good work!
YOU JUST SAVED MY LIFE!!! YOU ARE THE BEST!!!!
Really helpful video. However I have this question: in the derivative di / dt that you get at 21:57, where do you get the second term +B2e^(-5t)?
he made a mistake which you can see but he corrects it by using a describtion....in the vid its B2 t e^(-5t) that you derivate....(not B2 e^(-5t) as written in the vid)... which equals to B2 t' e^(-5t) + B2 t e^(-5t)'....(same way when you derivate [f(x)g(x)] '= f'(x)g(x) + f(x)g'(x)
This is great man. Look forward to watching more of your videos for help as I go through uni.
Im in a Elements of Electrical Eng. class and we dont get a cheat sheet........... life would be so much easier with a cheat sheet
My last quiz is exactly that question even the values was same i take 100 👍 thank you
@ 17:22 "This means the circuit is critically damped."
Indeed it does, but you still need to mind your initial conditions. I solved this circuit in the s-domain and found the expression: I(s) = 3(s + 5)/(s + 5)^2. The zero in the numerator cancels one of the poles in the denominator and you get I(s) = 3/(s+5). This is a 2nd order circuit that exhibits first order behavior. Performing the ILT on I(s), you obtain: i(t) = 3*e^(-5t) ∎
This video was awesome, you just clearified a lot of things for me and surely helped me with my test tomorrow :)
for the V_cap equation @8:15,10-5i+Vc=0. Why do we not consider the 5 ohms resistor
Because there's no current passing through it :)
***** Capacitors and resistors can't be treated the same in transient circuits. There is a voltage on the cap, but there is no current flowing in that branch the resistor is still bound by ohm's law. No current, no voltage. The reason there is no current is that the capacitor is fully charged in this instance - therefore it acts as an open
sebaogal1 are you an engineer now
+Abdulla Omar still on it :)
so i think it was because the cap was fully charged
"this is the cheatsheet that I've created, you can pause the video and look at it but we'll come back to it throughout the video"
every single student: 📸📸📸📸
Awesome video! This really helped out a lot.
Hi Matthew, nice video. Just a little confusion from what you said at 23:43, "we're looking for an equation for ALL-TIME AFTER t=0". It that were the case, then why would there be a 5v across the capacitor? wouldn't it be -10v? did you mean just when t first =0? sorry, not trying to nitpick, I just have a really hard time with 2nd order RCL and trying to learn.
Really nice job, Matt. Do you mind telling me what software you are using? I'd love to be able to follow your example with digital signal processing problems.
notice that i(l) = 3. when you plug this into the purple equation at 26:00 you get di/dt = 15 - 30
Great video, thank's a million!!!
Hi. At 26:00 why is there a current through the resistors at t=0? Also, what if the question asks us to find v(t) across capacitor for t>0 instead?
Find vc(0-/0+) , then find dvc/dt , and same equation , btw it's ur comment from almost 3 years ago, 😅 you know it
nice video
21:10 you missed a t between B2 and e. but overall great vid!
That t=0 so he shouldn't have written B2 at all.
@@peronkop yes but he wrote
thank you for share aroujomatt . its gonna help me
Thanks for the feedback!
much appreciated help..just subscribed
Thank you sir I understand very well...
At around 26:19, when di/dt is isolated, what happened to the value of L?
Eqn: -10 - 5+10i + Ldi/dt = 0;
shouldn't it be di/dt = (-15)/L?
L = 1. Look at the circuit. -15/1 = -15
Oh, I see. My bad. Thanks!
do you imagine capacitors as open and inductors as shorts initially no matter what? does it make a difference if the switch is originally open or closed?
if the source is connected for the time such that the capacitor will be charged fully and inductor stores maximum energy, than we consider them as open and short cause they don't have the capacity anymore. That's my concept. Correct me if I am wrong..
Hello sir can i ask? On how can you determine the polarity of a conductor if not given
Very useful!
At 13:55 i don't get why we were changing the sign of the capacitors voltage. Why does it not remain a minus? I mean the current is entering though the negative terminal so by my understanding it should remain that way. Thanks to anyone replying.
I also wondered that but cant explain why he does that
He has assumed a ccw direction of the current, so all polarities of the components will switch signs to follow the passive sign convention.
could you please make more videos about how to solve problems?
aren't you missing a particular solution, since we have the 10v voltage source which produces a forced response?
Great video! Just a happy coincidence haha
So is the initial closed or open ?
Thank you for this vid!
could you do a video, solving it using laplace transforms?
Many thanks
I understand it thank to u
PLEASE, where did the Theta in case 2 of the Cheat sheet come from?? What is it?
I don't get how you derive i(t) to get di/dt. why does the middle term of di/dt seems unchanged from i(t)?
Hi friend , i have to get the initial conditions from a circuit with 2 storage devices (L & C) but it also has 2 power supplies and they are always present even before and after the switch closes or opens , how can i get the initial conditions ? may i use superposition ? i mean , passivate one power supply & get the initial condition , then passivate the other power supply & get the initial condition , afterwards my 2nd system initial condition will be the sum of those partial ones ? thanks
I knew from the start it going to be critically dumb !
how ... ?
Thanks Matthew!
In the first kvl the voltage contribution from the capacitor is +vc and the last time you do kvl with one minute left the contribution is -vc. This is not consistent. Why does the "passivity" change? Thanks
man its a 9 year old video but imma ask anyway.. So I applied Laplace transform to this example circuit, solved it and got the same expression of i(t)= 3e^(-5t). Same answer for less work it seems like. So what are the cases where the Laplace approach cannot be used?
It’s probably not necessary, but how would you go about finding the voltage drop across the 5 ohm res in the middle branch in series with the inductor when t=(0-)
Hello, why in the characteristic equation you didn't consider the initial capacitor voltage?
The initial and final conditions are dealt with in the beginning of the video. The characteristic equation is to get the natural response of the circuit. By superposition they will sum together to get the final response. Check out forced second order ODEs.
Thanks
Can you help me with a second order circuit? If so, how can I send you?
@@angelojunior9507 There are better forums to ask on. Try groups like r/ECE or r/Electrical Engineering on reddit. You can also try #electronics on the freenode IRC network.
At 26:00 when di/dt is solved for -15, shouldn't it be positive 15?
why does i(0+) = i(L) at the 4:50 mark
the current is the same throughout the loop, and there is only one loop in the initial conditions.
but current 3 amp is will be after a long time not at 0+.
Could you upload a pdf of the cheatsheet? Thanks for the video
i see in some questions that you dont have to solve till i(infinity) you just stop at i(0) please can you tell me when to go till i(infinity)
Shouldn't the second kvl be -10 + 5 + 10 iL + L di/dt =0 ??
great
Thank you so much!
The voltage in a capacitor can change instantaneously if there is a pulse (deltax). So this is right only in some situations
Looking to do more problems like this. What book is this problem out of? Anyone know?
+Nick Spoutz The name of the book is on the title screen in the first few seconds of the video. You can get that book online for less than $20 usually. It's pretty good.
+Thomas Henley Duh! Thanks for pointing that out! Didn't even think to look there lol
Well explained! XD
isn't i(0+) suppose to be the same as i(infinity)? and shouldnt iL equal to i(o-) which is before changing the switch?
Capacitor and Inductor will act differently when switch is on or off depending on circuit. For this when switch on at 0-/0+ inductor will be short and capacitor will be open. But for infinite the switch is open and the source is therefor not connected to circuit so inductor and capacitor discharge away after the battery after the inductor dies. The resistor on top left doesn't get affected because its open there so it doesn't matter. Now again inductor is short and capacitor is open and making the entire circuit not connected with anything else so its nothing, 0. But when it was 0-/0+ the circuit was still connected and power was running so there would still be current since current depends on resistor and voltage and we still had those so we can calculate it.
Marc Zapatero
i(0+) is the instant right after the switch is flipped. i(infinity) is after a long time, which is when the circuit has reached steady state.
For inductors: iL(0-)=iL(0+)
Marc Zapatero sorry for being seven months late lol
at 24:28 the di/dt should be -5B1(e^-5t)-5B2(e^-5t) ||
+narpatraj gehlot in equation of i(t) at 20:20 you do not write there t in 2nd term so i get confused.
+narpatraj gehlot I realised that as well. I just inserted it. He actually took it into account in the line below thoat though.
Your initial Vc is incorrect; since there are 2 supplies, you need to use superposition when you perform KVL. Total voltage should be 15V when you add the 10V from the 20V supply. Said another way, since the cap is an O.C., the right most resistor is removed from the circuit, leaving a 1/2 voltage divider in series with a single 30V supply.
Ran circuit on Multisim, Voltage across the cap is indeed: 15V
This is wrong. First, Superposition is unecessary. KVL works fine. Second, the 10V source is pointing downward, so (20Vs - 10vs) or 10V - 5V = 5V. I also checked this with a simulation (5V), not sure what clifford ayer did, except maybe make the same mistake.
Jarrett Montgomery Well if you do kvl on the left loop then you'll see that 20v and 10v are in the same orientation. So... it is in fact 15v..
KVL left loop: 20 = iL(5) + iL(5) - 10 (iL = 3) => 20 = 15 + 15 - 10 == 0
Voltage at 3-resistor node (Vc) using Nodal: [Vc - (-10)] / 5 = 3; Vc = 5V
KVL in the left loop has nothing to do with it.
Jarrett Montgomery no
I love this... It's beautiful.
I find it to be quite the opposite.
Thanks!
Can you post your cheat sheet?
speed 1.5 (=
I did the same
Rookie numbers. You can understand him at 2.5x as well.
thank you
Thanks
GUYS I CONFUSE AT 22.41 HOW B2e^-5T COME?
I stop watching at 16:48. I’m upset that while you were deriving your characteristic equations that you didn’t show all the steps. Now I’m left wondering how you got the expression you found. You should have brought the equation you were working with over. If you are teaching people show all the steps. Everything is important to people trying to learn.
electrical engineer, but cant spell characteristic, great video regardless ;)
Missing t at B2e^(-5t), should be B2te^(-5t). This cracks my head when seeing the differentiation
when you wrote the voltage of the capacitor in the characteristic equation you forgot to put the initial condition
+David Rodriguez Since he took the derivative it would of became zero anyways so it wouldn't have changed the final answer.
ok, but I think it is necessary to write it, because sometimes it does not become zero
Hey could you try this software? Pin Point: 'Circuit Solver' by Phasor Systems on Google Play.
In the first loop, you referenced the technique as the "KVL" when you should have said "KCL". Using KVL=Kirchhoff's Voltage Law means that you are using the node voltage analysis to calculate voltage.
Mesh uses KVL (adding up the voltages) and Nodal uses KCL (adding up the currents). He was correct when saying he was using KVL. I'm not sure what you're referring to, because when using Nodal analysis, your equations are set up as V/R + V/R . . . etc, which is current. Mesh uses R(I) + R(I) . . . etc, which is voltage. You may have gotten that backwards if I interpreted your comment correctly.
Hey Batman.
You made a mistake while writing homogenious equation.
Can anyone test this out and give feedback? Pin Point: 'Circuit Solver' by Phasor Systems on Google Play.
Kvl 2 should be 10 -5il +5il +vc = 0
A little late now, but I believe it is correct the way he wrote it.
If you do a mesh current for i2 (which is going to equal 0 A), you will get an equation of:
10V + 5(i2-i1) + 5(i2) + Vc = 0
Since i2 is 0A, this equation goes to:
10V + 5(0-i1) + 5(0) + Vc = 0 ----> 10V -5i1 + Vc = 0
Vc = 5i1 - 10V = [5ohms x (3A)] - 10V = 15V - 10V = 5V
math is difficult
If anyone could test this app i'd be grateful! Search: 'Circuit Solver' by Phasor Systems on Google Play.
I solved this using Laplace and got B2 = -5 not 0
"Just a happy coincidence"
Officially the Bob Ross of circuits.
I'll take that as a compliment ;)