In both cases, there will be no continuous output at the secondary side. A sudden voltage (and in turn, current) spike can be recorded in the secondary side due to initial current inrush and nothing else later. We use this kind of testing on low rating transformers, for determining the polarity. It' s the change in flux that produces the voltage in secondary, and dc source provides a constant flux. For ideal transformer, as there is zero winding resistance (inductive reactance is zero anyway, for dc), there will be huge current through the winding, due to no heat loss the winding will remain intact(ideally). But, in case of a practical transformer, the winding will burn if a sustained uncontrolled source is applied; the source current must be controlled to ensure continuous working...
Speaking of first case since transformer taken is ideal the inductance value will be ideal so resistance will not be there so in case of pure inductor it tries to absorb the energy from the source and in infinite time it will act as short circuit during transient voltage in primary will be exponentially decreasing and we know differentiation of exponential is only exponential so we will get a voltage on secondary which will be exponential and ultimately it will become zero because when the inductor becomes short circuit the voltage in primary is zero so in secondary the voltage will be zero coming to second case it is a practical transformer so resistance will be there along with inductor now this time the voltage will not be 0 in steady state there will be some fine amount of voltage developed across the resistor in primary so in the secondary also And this time also the curve will be exponentially decaying all the discussion about is based on assumption that winding is capable of holding infinite amount of current
@@dineshkolotiya631 during the transient state the current through the primary is exponentially increasing, so there is change in flux so there will be a voltage on secondary...but during steady state current is sc current so no change in flux so no induced emf on the secondary winding
@@dineshkolotiya631 pretty simple.. ideal transformer has zero resistance only inductance so the primary current would keep growing to infinity and never stop. But it's rate of change would steadily decay which means the induced voltage across the secondary would start off at infinity and then steadily decay towards zero (but would never become completely zero). Practical transformer has a winding resistance so the current will grow from zero to a max value of V/R in time approximately 7*L/R seconds and then stop growing. Secondary voltage will start from infinity and decay to zero in the same time.
When we give DC supply it has constant magnitude at all time so there no change in flux also the winding of transformer is static no there is no change of flux in secondary flux is present in core but not in changes with respect to time so no flux linkage no EMF induced in secondary coil
At 9:20 sir in question 4 if see the current first enters the dot and then when it passed the dot, it appears that the current is leaving (which is the case we took into consideration) so how to know whether current is leaving or entering?
@@the_achiever_girlroy6697 inductor behaves as a short circuit for dc voltage so if it is an ideal transformer and dc voltage is applied, an infinite amount of current flows in the primary winding.
1)in first case in ideal transformer core loss will be zero and if no loss then there will be no resistance and if we are applying constant voltage source then current continously willl increase linearly it means mmf or flux vary linearly due to which an emf induced in the secondry winding the magnitude of induced emf will be constant 2)in second case in practical transformer transformer will have some resistance due to which that will be the case of RL charging circuit where after 4 times of time constant almost it will achieve its maximum value and got short cicuited after long time no emf will induced in secondry winding..
According to me the value of E2=0 because it induced due to mutal induction . And for a mutal induction there should be varying magnetic flux but we supply a dc current which induces only static magnetic flux so at starting when the value of our dc source raise from zero to rated the vale of E2 is not equal to zero but once it come to rated value E2 become zero
First case: as there is no resistance... Inductor will absorb energy from the source nd in infinite time it will be fully charged... Current across it will increase linearly with time nd at steady state it will behave as a short circuit.. Nd voltage e1 at steady state will be v1 nd corresponding e2 is related to e1 by turns ratio... Second case: as resistance is included.. Current will rise exponentially nd at steady state it will also behave like a short circuit but it will dissipate it's energy through resistor.. Nd current will exponentially decrease... Nd e1 will vary Or increase exponentially nd e2 will behave corresponding (as related to e1 by turns ratio)
In additive polarity, flux adds up ,so lenzs law is not followed in secondary. Then how does it work ....or what is phaser diagram in this case. Please make video with phasor diagram sir
Case 1: Ideal case permeability the transformer is infinite so the transient voltage induced in the transformer is very very high it may cause insulation breakdown Case 2:Due to low resistance of the winding it allows the more and more current winding continuously losses in the transformer is more, winding gets damaged.
Actually in case of DC supply there will be no rate of change of flux so that's why induced emf at primary side will be zero so that at secondary side also it will become zero and the nature of the current is steady state current.
if there is a switch in the primary then on closing the switch the the voltage value across the primary winding will be equal to V1 but because at time t=0 ,i=0 so it will act as open circuit but after time t=0+ the current starts to rise, due to change in current we get voltage at primary winding as Vp=Ldi/dt where L is the inductance of the primary winding, but as time passes the current flowing in the primary winding increases and at time t=infinity(steady state) it is equal to maximum thus change in current equals to zero(di/dt=0),thus the voltage across the primary winding becomes zero,thus making the winding(inductor) short at steady state due to short circuit no mmf will be produced and if no mmf then no flux andif o flux no induction process will take place and so this voltage will be induced in the secondary too since permeability is infinity due to ideal transformer so voltage across the secondary will be of decreasing function and finally become zero at steady state. In the second case it is simple RL circuit, same concept can be applied over here as above but the difference is that the current which will flow is be decided by the winding resistance and the voltage in the primary will decrease exponentially and so in the secondary due to coupling process ,and due to winding resistance power loss will be here and at steady state again the voltage across the winding of primary and secondary will be equal to zero but there will be a voltage drop across the winding resistance.
basic is transformer always works when there is time-varying input dc signal will not change wrt time...e=n(d pfi/dt) = 0 no , pfi=constant...so primary winding saturates.
In case of second case in practical transformer we can say that the induced emf at the secondary side will be zero because its a DC supply but here some winding resistance it will be burn out
1. The emf wont get induced. e=d(phi)/dt for DC d(phi)/dt=0 so e=0(Induced emf is zero 2. The Primary of the transformer gets damaged the reason is I=V/Z; Z=R+jXL XL=2*pi*f*L if f=0 (In case of DC) XL=0 Z=R so I =V/R R
Due to constant flux rate of change of flux will be zero then due to zero rate of change in flux there will not be any induced EMF in the secondary windings. And there will not be any burning of coils because in ideal transformer the winding resistance is considered to be zero.
Case 1 when dc supply is given winding acts as short ckt .......to satisfy kvl v become 0..... Case 2 due to some resistance.... I2r losses heat will produce and ckt burnts
In first case the current would be infinite because resistance is zero and due to dc no XL will role it's part and hence current is infinite so e1 is very small or we can say tends to zero and e2 is infinite in ideal case and it is all happen for a short time until flux is not going to be constant
@Genique Education aap machine mein jis topic karwa rahein hain usme questions ban rahe hain. sir mera har subject accha hain appllication part . par machine ka concept janthe hue bhi concept application mein aur question karne mein dikkath ho raha hain . mein chahtha hu aap mere problem ko solve kare . thank you .
i think that since the voltage applied is dc so there is no change in flux thus in both of the cases the voltage on the secondary side being induced is zero and yeah since second one id unit impulse so there will be some voltage on secondary side for some time but it will decay eventually
Sir,as Dc supply has constant magnitude at all time so there won't be any change in flux with respect to time so no emf will be induced in the secondary of coil .
Sir emf induced is directly proportional to frequency therefore induced emf should be zero, may be I am missing something but i will try to think about it in other way Also bcz now coil will act like a pure inductor and for dc after a certain time it will become short circuited As permeability is finite that means there is requirement of some current For ideal case the reluctance can be negligible so there may can lead to induction of emf e2 by the current which will flow during transient time bcz as the poles are aligned and constant flux will induced but without variation in flux or coil motion how emf will get induced
E2 = -[N2 Φm e^(-t/T)] / T Where, e = 2.7182 T = time constant of whole transformer circuit. N2 = number of turns in secondary For large value of time (t) e^(-t/T) --> 0 So, E2 = 0 ( for t > 4T )
case-1 when xmer is ideal, for dc supply- current is linearly increasing in nature (Ldi/dt is constant),i.e. flux is also increasing so in secondary winding induced voltage is constant and depend V1*N2/N1. case-2 in practical case, after some time xmer goes into saturation and there is no change in current i.e. flux, so induced voltage is zero.
Sir, now i have understood dot convention. Thank-you so much.
In both cases, there will be no continuous output at the secondary side. A sudden voltage (and in turn, current) spike can be recorded in the secondary side due to initial current inrush and nothing else later. We use this kind of testing on low rating transformers, for determining the polarity.
It' s the change in flux that produces the voltage in secondary, and dc source provides a constant flux.
For ideal transformer, as there is zero winding resistance (inductive reactance is zero anyway, for dc), there will be huge current through the winding, due to no heat loss the winding will remain intact(ideally). But, in case of a practical transformer, the winding will burn if a sustained uncontrolled source is applied; the source current must be controlled to ensure continuous working...
excellent
Speaking of first case since transformer taken is ideal the inductance value will be ideal so resistance will not be there so in case of pure inductor it tries to absorb the energy from the source and in infinite time it will act as short circuit during transient voltage in primary will be exponentially decreasing and we know differentiation of exponential is only exponential so we will get a voltage on secondary which will be exponential and ultimately it will become zero because when the inductor becomes short circuit the voltage in primary is zero so in secondary the voltage will be zero
coming to second case it is a practical transformer so resistance will be there along with inductor now this time the voltage will not be 0 in steady state there will be some fine amount of voltage developed across the resistor in primary so in the secondary also
And this time also the curve will be exponentially decaying all the discussion about is based on assumption that winding is capable of holding infinite amount of current
I want to answer this in OLA(one line approach)..
@@dineshkolotiya631 during the transient state the current through the primary is exponentially increasing, so there is change in flux so there will be a voltage on secondary...but during steady state current is sc current so no change in flux so no induced emf on the secondary winding
@@dineshkolotiya631 pretty simple.. ideal transformer has zero resistance only inductance so the primary current would keep growing to infinity and never stop. But it's rate of change would steadily decay which means the induced voltage across the secondary would start off at infinity and then steadily decay towards zero (but would never become completely zero).
Practical transformer has a winding resistance so the current will grow from zero to a max value of V/R in time approximately 7*L/R seconds and then stop growing. Secondary voltage will start from infinity and decay to zero in the same time.
Thanku sir....
Very nice way of teaching sir....
Thanks sir
When we give DC supply it has constant magnitude at all time so there no change in flux also the winding of transformer is static no there is no change of flux in secondary flux is present in core but not in changes with respect to time so no flux linkage no EMF induced in secondary coil
Super teaching .very good and short than nptel
At 9:20 sir in question 4 if see the current first enters the dot and then when it passed the dot, it appears that the current is leaving (which is the case we took into consideration) so how to know whether current is leaving or entering?
Case 1:- No flux is there so E2=0( NO burning due to no winding resistence )
Case 2:- No flux is there so E2=0( But burn due to winding resistence )
How it will be burn? Since E2=0 so no current in secondary means no resistance loss
@@the_achiever_girlroy6697 primary winding will burn due to dc voltage at practical
@@the_achiever_girlroy6697 inductor behaves as a short circuit for dc voltage so if it is an ideal transformer and dc voltage is applied, an infinite amount of current flows in the primary winding.
Nice lecture
1)in first case in ideal transformer core loss will be zero and if no loss then there will be no resistance and if we are applying constant voltage source then current continously willl increase linearly it means mmf or flux vary linearly due to which an emf induced in the secondry winding the magnitude of induced emf will be constant
2)in second case in practical transformer transformer will have some resistance due to which that will be the case of RL charging circuit where after 4 times of time constant almost it will achieve its maximum value and got short cicuited after long time no emf will induced in secondry winding..
emf will never be induced bcs it dc. Brother!!!!!
According to me the value of E2=0 because it induced due to mutal induction . And for a mutal induction there should be varying magnetic flux but we supply a dc current which induces only static magnetic flux so at starting when the value of our dc source raise from zero to rated the vale of E2 is not equal to zero but once it come to rated value E2 become zero
First case: as there is no resistance... Inductor will absorb energy from the source nd in infinite time it will be fully charged... Current across it will increase linearly with time nd at steady state it will behave as a short circuit.. Nd voltage e1 at steady state will be v1 nd corresponding e2 is related to e1 by turns ratio...
Second case: as resistance is included.. Current will rise exponentially nd at steady state it will also behave like a short circuit but it will dissipate it's energy through resistor.. Nd current will exponentially decrease... Nd e1 will vary Or increase exponentially nd e2 will behave corresponding (as related to e1 by turns ratio)
I am agreee with your both reasons that you have given (e2=e1/a)....
Awsome video sir
thank you sir !! and thats really great that you are about to move in questions format because thats what the need of time !very nice teaching
In additive polarity, flux adds up ,so lenzs law is not followed in secondary. Then how does it work ....or what is phaser diagram in this case. Please make video with phasor diagram sir
Case 1: Ideal case permeability the transformer is infinite so the transient voltage induced in the transformer is very very high it may cause insulation breakdown
Case 2:Due to low resistance of the winding it allows the more and more current winding continuously losses in the transformer is more, winding gets damaged.
Thanks a lot sir
Sir, what do you mean by the steady state when dc current is applied to the transformer. When will it reach the steady state as you mentioned?
Sir, full course
sir,In my opinion due to dc source a constant flux is produced in primary winding but flux is not a time varying due to this E2=0
Actually in case of DC supply there will be no rate of change of flux so that's why induced emf at primary side will be zero so that at secondary side also it will become zero and the nature of the current is steady state current.
Sir when v give dc supply there is no rate of change of current so there is no flux change and hence e2=0V
if there is a switch in the primary then on closing the switch the the voltage value across the primary winding will be equal to V1 but because at time t=0 ,i=0 so it will act as open circuit but after time t=0+ the current starts to rise, due to change in current we get voltage at primary winding as Vp=Ldi/dt where L is the inductance of the primary winding, but as time passes the current flowing in the primary winding increases and at time t=infinity(steady state) it is equal to maximum thus change in current equals to zero(di/dt=0),thus the voltage across the primary winding becomes zero,thus making the winding(inductor) short at steady state due to short circuit no mmf will be produced and if no mmf then no flux andif o flux no induction process will take place and so this voltage will be induced in the secondary too since permeability is infinity due to ideal transformer so voltage across the secondary will be of decreasing function and finally become zero at steady state.
In the second case it is simple RL circuit, same concept can be applied over here as above but the difference is that the current which will flow is be decided by the winding resistance and the voltage in the primary will decrease exponentially and so in the secondary due to coupling process ,and due to winding resistance power loss will be here and at steady state again the voltage across the winding of primary and secondary will be equal to zero but there will be a voltage drop across the winding resistance.
Sir discuss about sense of winding also...
basic is transformer always works when there is time-varying input dc signal will not change wrt time...e=n(d pfi/dt) = 0 no , pfi=constant...so primary winding saturates.
In case of second case in practical transformer we can say that the induced emf at the secondary side will be zero because its a DC supply but here some winding resistance it will be burn out
1. The emf wont get induced.
e=d(phi)/dt
for DC d(phi)/dt=0
so e=0(Induced emf is zero
2. The Primary of the transformer gets damaged
the reason is
I=V/Z;
Z=R+jXL
XL=2*pi*f*L
if f=0 (In case of DC)
XL=0
Z=R
so
I =V/R
R
Due to constant flux rate of change of flux will be zero then due to zero rate of change in flux there will not be any induced EMF in the secondary windings.
And there will not be any burning of coils because in ideal transformer the winding resistance is considered to be zero.
1st one zero voltage and second one the winding is burn out
Case1:
L circuit with time constant =L/R, R=0. Therefore, Time constant -> infinite.
Case2.
Transformer will get heat up.
Voltage at secondary side linearly increase ,V=L(di/dt) there is only inductance so current increase upto infinity.
There no voltage signal across secondary
E2=0
Case 1 when dc supply is given winding acts as short ckt .......to satisfy kvl v become 0.....
Case 2 due to some resistance.... I2r losses heat will produce and ckt burnts
E2 = v1/a. Ideal case
In first case the current would be infinite because resistance is zero and due to dc no XL will role it's part and hence current is infinite so e1 is very small or we can say tends to zero and e2 is infinite in ideal case and it is all happen for a short time until flux is not going to be constant
In practical case same happen as ac but now i1 is very large and e2
is also very large for a small time
@Genique Education aap machine mein jis topic karwa rahein hain usme questions ban rahe hain. sir mera har subject accha hain appllication part . par machine ka concept janthe hue bhi concept application mein aur question karne mein dikkath ho raha hain . mein chahtha hu aap mere problem ko solve kare . thank you .
From today you will get three video on problem solving
i think that since the voltage applied is dc so there is no change in flux thus in both of the cases the voltage on the secondary side being induced is zero and yeah since second one id unit impulse so there will be some voltage on secondary side for some time but it will decay eventually
initially we get the vtg due to sudden chng in dc signal....after that it goes to steady state
Sir,as Dc supply has constant magnitude at all time so there won't be any change in flux with respect to time so no emf will be induced in the secondary of coil .
Sir emf induced is directly proportional to frequency therefore induced emf should be zero, may be I am missing something but i will try to think about it in other way
Also bcz now coil will act like a pure inductor and for dc after a certain time it will become short circuited
As permeability is finite that means there is requirement of some current
For ideal case the reluctance can be negligible so there may can lead to induction of emf e2 by the current which will flow during transient time bcz as the poles are aligned and constant flux will induced but without variation in flux or coil motion how emf will get induced
Won't the core be saturated in DC as frequency is zero
In the ideal x-mer no resistance is there so the voltage V1 is equal to V2
Emf is not induced in secondary E2=0 because flux is not time varying in case 2
Sir this case me ye apply nahi ho raha he ..plz..help me
I think DC supply is not able to produce flux in the core that's why the induced voltage in secondary is zero E2 =0
In the first case E2 will be 0
In the Second case E2 also will be 0,provided there is heat loss.
Only alternating current able to produce flux in the core
Transformer to hogaya ....sir....jab rotating machine padhayenge ..to thora electromechanical energy conversion principle ko bhi padhaiyega
E2 = -[N2 Φm e^(-t/T)] / T
Where,
e = 2.7182
T = time constant of whole transformer circuit.
N2 = number of turns in secondary
For large value of time (t)
e^(-t/T) --> 0
So,
E2 = 0 ( for t > 4T )
in each video you ask a Q and conclude plz...
Practical transformer mein toh E2 will be exponentially decreasing...while in ideal transformer there will be no E2
case 1 . flux zero so E2=0
CASE2.SOME amount of E2 will be.
E2 will not induced bcoz not flux
Emfs in both cases would be zero
case-1 when xmer is ideal, for dc supply- current is linearly increasing in nature (Ldi/dt is constant),i.e. flux is also increasing so in secondary winding induced voltage is constant and depend V1*N2/N1.
case-2 in practical case, after some time xmer goes into saturation and there is no change in current i.e. flux, so induced voltage is zero.
Ideal transformer ki permeability the aap finite kaise maan sakte hai .Please give me solution
Baka
Both cases emf is zero
But it dc no flux
E2=0
Resultant flux will be very high
Heating of core will occur
E2 will be zero
0
E2=0 & primery coil will burn due to high current.
Superb sir...
In dc there is no change In flux so voltage will not enduce into the secondary so e2 will be 0