@Michel van Biezen thanks a lot for your great lectures but please can you explain properly why current doesn't flow through the 1 ohm resistor when the switch is not yet opened ?
On the branch with the 1 ohm resistor, there is a capacitor in series. Once the capacitor is charged, no current will flow through that branch, because the capacitor will not allow current to flow through it.
Why didn't you include the 1 ohm resistor in the first circuit? isn't the resistor also part of the circuit and has an affect on the voltage that passes through the capacitor? Thank you so much for the great lectures.
We are assuming that before switch was open it was close For a long time so in steady state capacitor will act like a open circuit so there will be no current flow in capacitor. So we can neglect that resistance
@@fatihinal9165 he said that initialy the circuit was closed. Then he said that once teh capacitor is filled the current thru it stops, it stops bcus teh capacitor acts like an open circuit once is filled. So there is no current flowing on that side of the circuit, and that side includes teh 1ohm resistor, not the 9ohms bcus that one is also part of the other part of the resistor.
Before the switch is opened at t = 0, the capacitor would have been filled with charge and therefore, no currentl will pass through the capacitor and therefore no current will flow through the 1 ohm resistor.
sir, if we happen to have much bigger resistor on our 1 ohm, does it make any difference? or regardless how big the resistance it will have no function?
Changing resistors will always change the function of the circtuit. The best way to become familiar with that is to change the value a few times (like 2, 5, 10, 20....) and see how it changes the currents in the circuit and the potential across the load each time.
at 1.40, why do we not consider the 1ohm resistance in equivalent circuit? when the switch is closed, isn't the current flow into the 1ohm resistance as well? im so confused here
Before the switch opens, there is no current flow through the branch with the capacitor, so there is no current flowing through the 1 ohm resistor (and it can be removed from the equivalent circuit.
if i have to determine the voltage across a condenser can i use ( Vc * E^(-t/rc)) or (Vc * (1- E^(-t/rc) ? sorry i am a bit confuse about those equations!..
At t = 0, (when the switch first opens), the capacitor is filled with charge and there is no current flowing through 1 ohm resistor and all the voltage is across the capacitor. A moment later when current begins to flow through the resistor, the voltage will be shared.
Sir, I've seen this equation : V(t)=V(open circuit)+ [V(t=0) - V(open circuit)]e^-t/tow. Why doesn't V(open circuit) apply in your videos ? Or this equation does not exist ?
Before the switch is opened (at t < 0) the capacitor is fully charged and there is no current flowing through the 1 ohm resistor. Welcome to the channel!
@@MichelvanBiezen sad to hear it. I am a beginner Electric Engineering student. My teachers are so weak, so I do not get they explanations. May you introduce some books for me??
@@MichelvanBiezen thanks for your quick response. Sir may you provide me with your email? Cuz I will share some materials and you suggest me your playlist related to them😊
Sir, if the voltage accross the 9 ohm resistors is equal to 7.5 volt, why the V of 9 ohm equal to V capacitor? isnt the correct equatiion is (V 9ohm = V 1ohm + Vc)
i understand in the first statement that the circuit is in steady state , so the current accross the voltage is equal to zero, but i couldnt understant why you put V9 ohm = Vc, should the correct equation is V9ohm = v 1ohm + Vc?
At the moment the switch opens (at t = 0), the capacitor will discharge through both the 1 ohm and 9 ohm resistors. Since they are then in series, they add up to 10 ohms.
@@preciousa.4384 You can let the 1 Ohm resistor on the circuit, but remember that for DC circuits, the capacitor behaves as an open circuit, and thus the current on that mesh is zero. If the current is zero, there won't be a voltage drop on the 1 Ohm resistor (V = iR = 0 * 1 = 0) and therefore, the voltage on the capacitor terminals will be the one measure on the 9 Ohm resistor, which corresponds to the voltage given by the voltage division as explained in the video, 10V(9/(9+3)). Hope this makes sense now.
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thank you professor Michel van Biezen for helping me survive university...
We are glad we could help. Thanks for the note.
You have helped me pass my test in my network theory class for engineering. Thank you!
Your teaching is too great ,I really thank what you talked
inductors becomes a short circuit when connected to source.
conductors becomes a open circuit when connected to source.
ahh I love you buddy lol I just realized now that this is what I have been missing sometimes
why wouldnt the 7.5volts be shared between the 1 ohm and the capacitor??
Thanks a lot, you are father of science;i like U and also ur way of explanation
Thank you. Glad you found our videos! 🙂
Hi profassor thank you alot I am having a final exam and your teretoryal videos are very help full
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@Michel van Biezen thanks a lot for your great lectures but please can you explain properly why current doesn't flow through the 1 ohm resistor when the switch is not yet opened ?
On the branch with the 1 ohm resistor, there is a capacitor in series. Once the capacitor is charged, no current will flow through that branch, because the capacitor will not allow current to flow through it.
@@MichelvanBiezen thank you sir because I totally understand it now, much appreciated.
Thanks. I hope you release lessons on RLC circuits some day.
Until we complete the EE videos on that, you can take a look at this playlist which already covers some of these topics: PHYSICS 49 RCL CIRCUITS
Why didn't you include the 1 ohm resistor in the first circuit? isn't the resistor also part of the circuit and has an affect on the voltage that passes through the capacitor? Thank you so much for the great lectures.
Since there is no current through that part of the circuit, the one ohm resistor has no effect on the circuit.
Are we assuming that the capacitor is charged at t=0s? Because otherwise, wouldn't wouldn't the capacitor allow for a current to pass through?
We are assuming that before switch was open it was close
For a long time so in steady state capacitor will act like a open circuit so there will be no current flow in capacitor. So we can neglect that resistance
@@MichelvanBiezen How do we know that. I can't explain why there is no current on one ohm resistor.
@@fatihinal9165 he said that initialy the circuit was closed. Then he said that once teh capacitor is filled the current thru it stops, it stops bcus teh capacitor acts like an open circuit once is filled. So there is no current flowing on that side of the circuit, and that side includes teh 1ohm resistor, not the 9ohms bcus that one is also part of the other part of the resistor.
Sir for t
Before the switch is opened at t = 0, the capacitor would have been filled with charge and therefore, no currentl will pass through the capacitor and therefore no current will flow through the 1 ohm resistor.
Sir could you please please explain your statement @30.secs why we don't have current through the capacitor before t=0-.
With the switch open current cannot flow through the circuit. (You must have a closed loop)
@@MichelvanBiezen sir how can we say the capacitor is charged full at t=0 or it is just representing that
sir, if we happen to have much bigger resistor on our 1 ohm, does it make any difference? or regardless how big the resistance it will have no function?
Changing resistors will always change the function of the circtuit. The best way to become familiar with that is to change the value a few times (like 2, 5, 10, 20....) and see how it changes the currents in the circuit and the potential across the load each time.
at 1.40, why do we not consider the 1ohm resistance in equivalent circuit? when the switch is closed, isn't the current flow into the 1ohm resistance as well? im so confused here
Before the switch opens, there is no current flow through the branch with the capacitor, so there is no current flowing through the 1 ohm resistor (and it can be removed from the equivalent circuit.
Thank you sir, you work is excellent
You are most welcome
Excellent!!
Thank you. 🙂
if i have to determine the voltage across a condenser can i use ( Vc * E^(-t/rc)) or (Vc * (1- E^(-t/rc) ? sorry i am a bit confuse about those equations!..
why wouldnt the 7.5volts be shared between the 1 ohm and the capacitor??
At t = 0, (when the switch first opens), the capacitor is filled with charge and there is no current flowing through 1 ohm resistor and all the voltage is across the capacitor. A moment later when current begins to flow through the resistor, the voltage will be shared.
Sir, I've seen this equation : V(t)=V(open circuit)+ [V(t=0) - V(open circuit)]e^-t/tow. Why doesn't V(open circuit) apply in your videos ? Or this equation does not exist ?
This is super late, but it's because that is for step response. Videos in the mid 30s use that equation
Excuse me sir, I wonder why the 1 ohm at the circuit (t>0) isn't included in calculations? Thank you sir. Love from Indonesia.
Before the switch is opened (at t < 0) the capacitor is fully charged and there is no current flowing through the 1 ohm resistor. Welcome to the channel!
So, even if the capacitor starts to discharge, the 1 ohm will keep being excluded, sir?
Once a current flows through the 1 ohm resistor, you can no longer ignore it.
@@MichelvanBiezen ok, thank you, Sir, for your time and teaching. It's helpful✨
Hello sir. I would like to know do you have any book related to these lessons?
No, we don't have any books.
@@MichelvanBiezen sad to hear it. I am a beginner Electric Engineering student. My teachers are so weak, so I do not get they explanations. May you introduce some books for me??
The book writted by Sadiku is pretty good. It was used by one of our sons when he studied electrical engineering.
@@MichelvanBiezen thanks for your quick response. Sir may you provide me with your email? Cuz I will share some materials and you suggest me your playlist related to them😊
I thought the equation for a steady state RC circuit was V0+V(inf)-v(0)e^(-t/T) not just v(0)e^(-t/T)??
so i see that you drew the t=inf circuit at 4:43 but whats the voltage of the capacitor there
Don't forget to replace t with infinity which causes e^-t to go to zero.
1 ohms and 9 ohs are in series?
Sir, if the voltage accross the 9 ohm resistors is equal to 7.5 volt, why the V of 9 ohm equal to V capacitor?
isnt the correct equatiion is (V 9ohm = V 1ohm + Vc)
i understand in the first statement that the circuit is in steady state , so the current accross the voltage is equal to zero, but i couldnt understant why you put V9 ohm = Vc, should the correct equation is V9ohm = v 1ohm + Vc?
The voltage across the 9 ohm resistor must be equal to the voltage across the capacitor AND the voltage across the 1 ohm resistor added together.
sorry , why is the total resistance of circuit 10 ohms?
At the moment the switch opens (at t = 0), the capacitor will discharge through both the 1 ohm and 9 ohm resistors. Since they are then in series, they add up to 10 ohms.
@@MichelvanBiezen thank u
why do we short the 1 ohm resistor
Before the switch opens, there is no current flowing through the 1 ohm resistor, so it acts like it is not there.
@@MichelvanBiezen why? i am still lost. thanks for your reply
@@preciousa.4384 You can let the 1 Ohm resistor on the circuit, but remember that for DC circuits, the capacitor behaves as an open circuit, and thus the current on that mesh is zero. If the current is zero, there won't be a voltage drop on the 1 Ohm resistor (V = iR = 0 * 1 = 0) and therefore, the voltage on the capacitor terminals will be the one measure on the 9 Ohm resistor, which corresponds to the voltage given by the voltage division as explained in the video, 10V(9/(9+3)). Hope this makes sense now.
thats cap, how is v*r=v? 2:40
Which "v" are you asking about?
LMAO
boss
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