For more problems on Magnetically Coupled Circuits, the downloadable google drive link is given in the description. While solving if you have any questions, do let me know here.
In the first problem,according to single phase notation we need to consider 50V as maximum value but in the problem 50V is considered as rms value.Hence analysis is wrong.(In single phase circuits if the voltages are specified in polar form,by default it must be maximum value.If it is rms value it need to specified Source :NPTEL Videos)
At 6:29, the value of jX has been calculated and it is j7 ohm. Now, if we take j5+j8+2jXm, then, in that case, you will never get jX as j7 and it will be more than j13. It is only possible when 2jXm is subtracted from the (j5+j8). I hope, it will clear your doubt.
Yes, you are correct Raj. Rightly observed. It should have been framed like power is dissipated rather than absorbed. In resistor power only gets dissipated in form of heat. My Mistake.
No, once you consider that power as dissipated power in the same example then procedure is alright. I mean consider the power in the example as dissipated power rather than absorbed power.
In the question, it is given that the inductance between 1 am 2 is 4H, when 3 and 4 terminals are open. And it is 3H when 3 and 4 are short-circuited. Please check the question once again, you will get it.
@@ALLABOUTELECTRONICS Hello :) I agree with Chandana Priya. Why we used a value of 4H as Lp in the second problem instead of 3H? Value of 4H is present when secondary side (terminals 3-4) is opened. Equations (1) and (2) are for the case when secondary side (terminals 3-4) is short circuited and in that case Lp is 3H. Still you used value of 4H as Lp.
voltage V2 is the voltage between terminal 3 and 4. When 3 and 4 is short-circuited then terminal 3 and 4 will be at the same potential. Or in other words V2 =0. I hope it will clear your doubt.
As I mentioned at 8:50, when terminal 3-4 open-circuited, no current is flowing through the secondary coil. That means there won't be any coupling on the primary side due to secondary coil (as no current is flowing through Ls). So, whatever inductance measured at the primary side is entirely a self-inductance Lp. That's why Lp = 4H I hope it will clear your doubt.
Because, it's the inductance seen at the primary side when secondary is short circuited. He equations are written when secondary is short circuited. I hope it will clear your doubt. If you still have any doubt please let me know here.
@@ALLABOUTELECTRONICS as you said these equations are regarding to the second condition Then why you kept lp=4 As they mentioned in the question when it (3,4 terminals short circuited)lp=3 Why you take lp=4 At 11:20
@@RupeshChandu1 Lp is the self-inductance on the primary side. Using the first condition, when the secondary side is open, we found the self-inductance Lp on the primary side and using the second condition, we found mutual inductance. As, you are aware for the two mutually coupled coils, the voltage across any coils is due to self-inductance and the mutual inductance. At 11:32, exactly that is written. I hope it will clear your confusion.
I think, you didn't understand the question properly. The thing is, the mutual inductance can be additive or subtractive. Since here the dot convention is not given, so from the given circuit parameters we need to find out, whether it is additive or subtractive. So, first from the given power, the total reactance X was calculated. And from that, the mutual inductance equation which satisfies this reactance is selected. Please watch that part again, you will get it. If you still have any doubt then let me know here.
CORRECTION:
Example 1: Instead of absorbed power, please consider it as the dissipated power in the circuit.
For more problems on Magnetically Coupled Circuits, the downloadable google drive link is given in the description.
While solving if you have any questions, do let me know here.
ans 1 is 2.5044 H
ans 2 is -20j ohm
am I right ?
Sir can u made video on problem of coupled ckt like
to find volt or power
Developed across any elements
ans 1 - 2.5044H
ans 2 - 9 ohm
ans 3 - 9.8 H
correct?
can someone explain the 2nd ques of pdf
A brilliant and very clear cut explanation! Thanks!
Took my time to unpack it and I learned the lesson well. Thanks again!
In the first problem,according to single phase notation we need to consider 50V as maximum value but in the problem 50V is considered as rms value.Hence analysis is wrong.(In single phase circuits if the voltages are specified in polar form,by default it must be maximum value.If it is rms value it need to specified Source :NPTEL Videos)
I really like this channel, video was really helpful, thank you
i love your videos thanks for posting
Can u pls explain why 8-M^2/2= 3?
sir how to you take equation j5+j8-2jxm ...
and why we cant take quation j5+j8+2jxm....
in example no. 1
At 6:29, the value of jX has been calculated and it is j7 ohm. Now, if we take j5+j8+2jXm, then, in that case, you will never get jX as j7 and it will be more than j13. It is only possible when 2jXm is subtracted from the (j5+j8). I hope, it will clear your doubt.
ALL ABOUT ELECTRONICS
ya its clear
how can u take Imax in power dissipated equation it should be Irms
You say power always absorbed only across resistive elements.
My question is reactive element why can not absorb power?
Thanks 🙏🙏🙏🙏🙏🙏🙏
sir doubt power absorption takes place in inductors and capacitors but u said power is absorbed in resistor how.
Yes, you are correct Raj. Rightly observed.
It should have been framed like power is dissipated rather than absorbed.
In resistor power only gets dissipated in form of heat.
My Mistake.
thanks for ur reply sir then the procedure which u followed i think wrong sir.
No, once you consider that power as dissipated power in the same example then procedure is alright.
I mean consider the power in the example as dissipated power rather than absorbed power.
Why did we consider the value of lp is 4 in 2nd prblm it's probably to be 3 right
In the question, it is given that the inductance between 1 am 2 is 4H, when 3 and 4 terminals are open. And it is 3H when 3 and 4 are short-circuited. Please check the question once again, you will get it.
@@ALLABOUTELECTRONICS Hello :) I agree with Chandana Priya. Why we used a value of 4H as Lp in the second problem instead of 3H? Value of 4H is present when secondary side (terminals 3-4) is opened. Equations (1) and (2) are for the case when secondary side (terminals 3-4) is short circuited and in that case Lp is 3H. Still you used value of 4H as Lp.
How the value of v2 =0 . when we short the second circuit sir..
voltage V2 is the voltage between terminal 3 and 4. When 3 and 4 is short-circuited then terminal 3 and 4 will be at the same potential. Or in other words V2 =0.
I hope it will clear your doubt.
sir can u please make a video on ideal and real transformers?
Sir very helpfull videos
can you please pleas make a detailed vido on telephone hybrid circuit
How can we determine the coupled equation after getting jx value
Would you please provide the exact timing in the video where you are referring?
Please make a video on transient and steady state conditions
I have already made videos on it. Please check the network analysis playlist.
Thank you sir
Sir why you consider only lp=4
Why lp=3
Please reply sir
As I mentioned at 8:50, when terminal 3-4 open-circuited, no current is flowing through the secondary coil. That means there won't be any coupling on the primary side due to secondary coil (as no current is flowing through Ls).
So, whatever inductance measured at the primary side is entirely a self-inductance Lp.
That's why Lp = 4H
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS
Why (8-m*m)2=3?
Why can't be 4
At 12:06
Because, it's the inductance seen at the primary side when secondary is short circuited.
He equations are written when secondary is short circuited.
I hope it will clear your doubt.
If you still have any doubt please let me know here.
@@ALLABOUTELECTRONICS as you said these equations are regarding to the second condition
Then why you kept lp=4
As they mentioned in the question when it (3,4 terminals short circuited)lp=3
Why you take lp=4
At 11:20
@@RupeshChandu1 Lp is the self-inductance on the primary side. Using the first condition, when the secondary side is open, we found the self-inductance Lp on the primary side and using the second condition, we found mutual inductance.
As, you are aware for the two mutually coupled coils, the voltage across any coils is due to self-inductance and the mutual inductance.
At 11:32, exactly that is written. I hope it will clear your confusion.
why we didn't wrote V1= Lp di1/dt + M di2/dt ????
Your equation is in time domain. He is solving in frequency domain. In frequency domain d/dt becomes jw where w is the frequency in rad/sec
why lp is taken as 4 but it is given 3 in 2nd prblm?
It is 4H when terminals 3 and 4 are open and Lp is 3H when 3 and 4 are short-circuited. Please check the question once again, you will get it.
In the first problem why you take j5+j8-2jXm instead of j5+j8+2jXm
I think, you didn't understand the question properly. The thing is, the mutual inductance can be additive or subtractive. Since here the dot convention is not given, so from the given circuit parameters we need to find out, whether it is additive or subtractive. So, first from the given power, the total reactance X was calculated. And from that, the mutual inductance equation which satisfies this reactance is selected. Please watch that part again, you will get it. If you still have any doubt then let me know here.
how is 8-sq(m)/2 =3
s same doubt
i love all about electronics
nice.
can an english version be made the accent is very annoying
It's indian accent whether you like it or not