Yes sir I have watch these 2 Lectures that you have mentioned. In kvl you have said that if there is Rise up in voltage then use + sign. similarly in this example voltage across 1ohm resistor is rise up and even if you have taken '-I'. Please clear my doubt
Hello All. I think there is a potential increase across 1 Ohm Resistor and hence while writing KVL equation, there would be '+' sign. Per the last lecture, "Rise in Voltage = '+' Sign" and "Drop in Voltage = Negative Sign". Current direction is same as Sir has indicated i.e. Clockwise. My KVL equation starting from point 'Va' is : -6V + 1 Ohm x I - 2 Ohm x I + 3V = 0 and hence I = -3A Apply KVL in upper loop using the same logic: Va - 6V + 1 Ohm x (-3 Amp) = Vb and hence Va - Vb = 9V Can someone pl point out my mistake or it may be an oversight by Sir. Thanks in advance
Thank you so much for this tutorial. I have a question, what if you actually had connections at A and B. Would they still be simple node? I'm trying to understand the safe context of the assumption that the node at A and B are simple nodes.
sir,,,thank q for uploading videos in regular basis....sir i have a one question regarding d completion of network syllabus..sir could u please tell me...in how many days u will complete d syllabus...? ..it will be better if u complete dis subject as soon as possible because this subject have an application in other subjects...
Nice one sir. I have one doubt,while solving the lower part why there is +2i, why not it is -2i,and also how to know that there is raising the potential.
Sir it Is last question. please explain me why you taken voltage across R1 as '-I' Not '+I'. similarly for voltage across R2 explain the reason. this will definately clear my doubt. Thank you in advance.
+Neso Academy Thank you sir i have saw the lecture in which you have given the homework. from thet lecture I have get answer for my question. Now my doubt is clear Thank you sir.
How can we say that this small circuit is independent circuit and is not a part of any other circuit ? It is not given in circuit so, a and b nodes can be principal nodes also!....?
But when i solve the this question using source transformation i get Vab=4.95 but I also have a resistance valued 0.66 in series with it. So is it correct or not????
Sir, while solving 2 equations of kvl for finding v(a)-v(b) , you have used the conventions as current entering into 6v source =-6v and at the same time u have also used -3v for current leaving the 3v source. How is this possible...? You cant change conventions at once in a particular problem. We may follow any one convention but must stick to it strictly for that problem as a whole.
The convention is not changed, Initially to calculate the current in the circuit I applied the KVL in clock-wise direction (right to left in bottom branch) and while doing so we got +3V. After we calculated the current, to find out the potential difference between A and B points, We traveled from A to B in the bottom branch (left to right), due to this we have -3V. The convention is same, the direction is changed.
Again sorry sir But the thing is that it is about direction of current with respect to polarity of voltage source. Which are contaradictory in the two equations used.
I don't think anything is contradictory here, we've not changed the direction of current and it is the same. We have only applied KVL in the direction opposite to that of the obtained current's direction. I hope you have seen the previous lectures on KVL.
#nesoacademy Sir, whenever current passes through a constant resistor does it only create voltage drop?? Which means we don't need to be careful about whether it connected to +ve or - ve terminal of voltage source??
Sir please explain me How you take potential drop across 1 ohm resistor is negavite? Because 1 ohm resistor is connected to negative terminal of voltage source so _ to + there potential is raise but why should you take this negative sing please explain me please please
Its all about sign convention...if you feel take +ve sign after crossing resistor but you have to apply the same sign with all the resistors in the direction of current...final answer would come same, use same sign convention in all equation..
@@richamishra6160 Same here. I did the same equation as yours. Sir's initial assumption of current flowing clockwise in that circuit was wrong, that's why he got -1 calculating I. Our assumption was right that's why we've got just 1 amp (The sign just represents the direction of flow)
Sir, Thank you very much for this great free series on electrical circuits.
Sir u r dedication levels r peaks....😘😘
Yes sir I have watch these 2 Lectures that you have mentioned.
In kvl you have said that if there is Rise up in voltage then use + sign.
similarly in this example voltage across 1ohm resistor is rise up and even if you have taken '-I'.
Please clear my doubt
You are best tutor on You tube 👍
Sir while writing 1st equation after 6v we move from -ve to +ve that is potential rise.Therefore we get +1i amp right...
No sir. We are moving from +ve to -ve drop potential
You are the best teacher.
It took me few minutes, but i tried myself and the answer was right 😁😁
Sir plz give atleast one lecture over graph theory and steady state theorm
Great lecture✨
Nice explanation!
Amazing sir.....keep on going....
Hello All. I think there is a potential increase across 1 Ohm Resistor and hence while writing KVL equation, there would be '+' sign. Per the last lecture, "Rise in Voltage = '+' Sign" and "Drop in Voltage = Negative Sign". Current direction is same as Sir has indicated i.e. Clockwise. My KVL equation starting from point 'Va' is :
-6V + 1 Ohm x I - 2 Ohm x I + 3V = 0 and hence I = -3A
Apply KVL in upper loop using the same logic:
Va - 6V + 1 Ohm x (-3 Amp) = Vb and hence Va - Vb = 9V
Can someone pl point out my mistake or it may be an oversight by Sir.
Thanks in advance
sir there is voltage drop polarity in 2ohm resistor as you said 8:09 but y not on 1 ohm resistor
Preparing for jee advance is now enjoying 😍😍👍
If the A terminal is common (0 volt) then the voltage from A to B terminal is -5 volt.
What can you say about this please...I got -5 volt as well
Sir it's very helpful.
Thank you so much for this tutorial. I have a question, what if you actually had connections at A and B. Would they still be simple node? I'm trying to understand the safe context of the assumption that the node at A and B are simple nodes.
sir,,,thank q for uploading videos in regular basis....sir i have a one question regarding d completion of network syllabus..sir could u please tell me...in how many days u will complete d syllabus...? ..it will be better if u complete dis subject as soon as possible because this subject have an application in other subjects...
Sir.. I m preparing for gate... So please complete in less days at least in 90days... Thanks in advance..
Sourav Pati bro netwotk is not that hard subject
I = 3A is correct for Question 1
Thank you sir
Nice one sir.
I have one doubt,while solving the lower part why there is +2i, why not it is -2i,and also how to know that there is raising the potential.
Because we move in opposite direction of the assumed current.. that's why +2i
@@Avinashkumar-gi8dv 👌
*Sir your all videos are very good but you can't make video of Electrical Machine*
Why currents from 2 batteries are not taken sir?
Sir it Is last question.
please explain me why you taken voltage across R1 as '-I' Not '+I'.
similarly for voltage across R2 explain the reason.
this will definately clear my doubt.
Thank you in advance.
+Neso Academy Thank you sir i have saw the lecture in which you have given the homework.
from thet lecture I have get answer for my question. Now my doubt is clear
Thank you sir.
@@akashjagtap2784 can u pls explain me the answwr of your question
wow !!
please explain the Va and Vb thing, didn't understood
Sorry voltage is rise up so it should be +IR1
Resistor always dissipates energy so - IR1 is correct
sr magnetic circuits are not uploaded yet...plzz upload
why does current division does not take place at node Va ? Should'nt they divide at Va and join again at Vb?
How can we say that this small circuit is independent circuit and is not a part of any other circuit ?
It is not given in circuit so, a and b nodes can be principal nodes also!....?
Bro what if a battery is connected to the vab so that the current divides know?
How do you decide the direction of current? that current will flow from Va to 6V source at first and not to 3V source at first??
Sir pls help i used the second convention that is i take rise in potential as -ve then how to solve the last vab through that path,it gave me -5
stay on one convention...check + and -
9volt.
By using the rules explained by sir, just solve in your own way.
But when i solve the this question using source transformation i get Vab=4.95 but I also have a resistance valued 0.66 in series with it. So is it correct or not????
Is it possible to take the loop from vb to vb and then proceed to do va-vb or vice versa??
Good
-6+3 should have been -3 but +3 was taken further. Why?
sir please explain how va = vb?
Sir Why are you not indicate the polarities across resistors.
sir if we will take it clockwise then its coming -5v
can anybody derive the current passes through 1 ohm and 2 ohm resistors?
Sir, while solving 2 equations of kvl for finding v(a)-v(b) , you have used the conventions as current entering into 6v source =-6v and at the same time u have also used -3v for current leaving the 3v source.
How is this possible...?
You cant change conventions at once in a particular problem.
We may follow any one convention but must stick to it strictly for that problem as a whole.
I have not taken -3V, see carefully it is +3V. The same convention is used to for the equation.
Thanks sir
But sorry once again
You may observe video at 8:04 time duration
Please answer my query as am dependent on you for my studies.
The convention is not changed, Initially to calculate the current in the circuit I applied the KVL in clock-wise direction (right to left in bottom branch) and while doing so we got +3V. After we calculated the current, to find out the potential difference between A and B points, We traveled from A to B in the bottom branch (left to right), due to this we have -3V. The convention is same, the direction is changed.
Again sorry sir
But the thing is that it is about direction of current with respect to polarity of voltage source.
Which are contaradictory in the two equations used.
I don't think anything is contradictory here, we've not changed the direction of current and it is the same. We have only applied KVL in the direction opposite to that of the obtained current's direction. I hope you have seen the previous lectures on KVL.
could you please write KVL in a path from Va to Vb path using other convention please??
6+I+2I-3=0 => I=-1A ; -Vab+6-1=0 => Vab=5v =>Va-Vb=5v
#nesoacademy In 1ohm resistor current entering into -ve terminal and leaving +ve terminal..isn't that rise in potential??
#nesoacademy Sir, whenever current passes through a constant resistor does it only create voltage drop?? Which means we don't need to be careful about whether it connected to +ve or - ve terminal of voltage source??
Sir i don't understand, while applying kvl in a loop why you take (-IR1). Because voltage is dropped you have to take it +IR1?
please clear my doubt
It's mistake I think sir I have just taken that clip BCz if that ir is -ve them V must be positive
Can’t understand 😢
Why I - 1 added to 6v I thought its current not a voltage
What signs for va,vb
Isme Vb ko include nhi kiya gya ????q
Sir please explain me
How you take potential drop across 1 ohm resistor is negavite?
Because 1 ohm resistor is connected to negative terminal of voltage source so _ to + there potential is raise but why should you take this negative sing please explain me please please
I don't know English properly but I can understand so forgive me any mistakes to my comment
Its all about sign convention...if you feel take +ve sign after crossing resistor but you have to apply the same sign with all the resistors in the direction of current...final answer would come same, use same sign convention in all equation..
Sir, by taking 2nd convention i.e. Rise in potential = -ve & Drop in potential= +ve...I'm getting i= 1A
Sir please reply as i always follow this convention only n get correct answers...but here i'm getting different answer
Changing the direction of current (Anticlockwise) -6+3+2i+1i=0
-3+3i=0
i=1A
(+2i+1i ; current alwaus flows from +ve to -ve or HP to LP)
@@richamishra6160
Same here. I did the same equation as yours. Sir's initial assumption of current flowing clockwise in that circuit was wrong, that's why he got -1 calculating I. Our assumption was right that's why we've got just 1 amp
(The sign just represents the direction of flow)
@@Kaspesky same here I am also getting current as 1A. By this the potential is coming out to be 0.