I don't know why everyone in the comment section has doubt in the secondary circuit which is absolutely correct but no one saw the actual mistake in the primary circuit i.e what sir were trying to teach in the secondary circuit, it seems that he himself forgot to apply that in the primary circuit. The mistake is that sir took the voltage across primary coil to be 1000V which he used to calculate the emf induced across the secondary coil and got 200V. But the voltage across the primary coil is not 1000V, it is 990V because 1000V is across the source and the current in the circuit is 10A. So there will be a potential drop of (10x1=10V) across the internal resistance of the primary coil and hence (1000-10=990V) across the primary coil. So now if we use [ Ep/Es=Np/Ns ], we will get [Es=990x1/5=198V] . So the actual potential difference across the secondary coil is 198V rather than 200V. Hope that i have explained it correctly. Feel free to ask if there is something which i haven't explained completely/correctly. 🙏
Arey Ep is given, it is fixed, it is the voltage to be changed, coming to the question, if you draw the circuit of the primary side then it will be simple, with ac source and a resistor (coil) so of course the job of the ac source is to keep the voltage across the ends of the resistor 1000v so it is fixed, Ep=1000V, and hence the power = VI, this 10000 W is the power supplied by the ac source to the transformer, now when this power goes in, first it suffers cu loss in the primary circuit, then iron loss, then another copper loss in the secondary circuit, which is in total 1000W, Now In, Es the problem is that there is a load, and the load is the one supposed to dissipate the final power and give the required voltage , i.e if you are having output power as 9000 W then this 9000 W will be dissipated across the load only,( 14:35, maybe replay this part) so to get the required voltage across the load, two resistors are involved (cu one and load one, whereas in the primary circuit, only cu one was present), so as both these resistors are dissipating power, the total power disspated will obviously be 9000+200 =9200. So ig sir is right… this was my theory from what I have understood after replaying the part 100 times😂
52:15 part 8 , load ke pass power 9000w gaya woh bhi 200w loss hone ke bad iska matlab woh circuit ke pass total 9200w power tha..jiss se 200 w resistance udane ke bad baki bacha power load ke pass gaya.
@@Jjk308 Yes, 700 watts of iron loss plus 100 watts of copper loss in the primary coil, however this current also causes 200 watt copper loss in the secondary coil resulting in a final output power of 9000 watts. The output power is not the power delivered to the secondary coil on the iron core, it is the final power sent out of the transformer.
Sir at 53:00, secondary circuit ko Poutput mila 9000w agar isme se 200w ka copper loss hua h toh 200w minus hona chahiye so net power of secondary becomes 8800w.
no bro, the efficiency given tells us the power received by the load, not the entire seccondary circuit... that power is 9000, power in secondary circuit is 9200, but that 200 is lost in copper and only 9000 received by load
Hlo.... primary ke coils ka inductance bhi to hoga aur resistance vese h hi ..to LR ckt ho gya ....ise E= iR na lgke E=iZ lgna chahiye....but uski information question me nhi h
Shouldnt the power loss due to resistance of circuit be counted along with the total power loss of the circuit(including load)..then why did we take total power in that circuit to be 9200 Watts?( I was thinking the net power loss in the circuit was 9000Watts which includes the resistance)
I too didnt understand why sir hadnt included that. Think of it this way Input source provides 10000watts, out of which 800watt goes to copper loss(primary circuit) nd iron loss, therefore 9200 watts of power goes to the secondary circuit which will be equal to EsIs.
Sir in part 5 at 46:43 , the resistance of the coil is given so why we can't use P(primary)=I²R(primary) , to get I as 100A ? why not possible? or anyone else ?
BECAUSE you dont know the exact resistance of primary circuit you only know the resistance in the coils and not in the iron region, so we have to use formula independent of resistance.
Ye result hmesha valid h .... derive krte vkt sir ne kbhi nhi bola ki it is only for ideal transformer ....only jisme current ata h ...vo ideal k liye valid h
sir transformer wale Q8 me to 200W uss 9000W ka hi part hai na Kyuki total Power loss = PCu + PFe to Phir correct answer 45A hi aa raha hai Aur sir agar ye correct hai to phir same concept primary ke liye bhi lagta na
@@krrishkumar6812 The induced EMF in the secondary coil is 200V. The power delivered to the secondary circuit is 9200W. Now, the current of 46A due to the induced EMF would result in a 200W copper loss in the secondary coil and the final power delivered to the load is 9kW. The misconception here is due to the incorrect idea that the potential difference across the load is equal to the potential difference induced in the secondary coil. There is a small potential drop due to the resistance of the coil. The actual potential difference across the load in this case would be 9000/46 = 195.65V, and the potential drop due to the coil would be the cause for the 4.35V deficit.
@@existential_turtle yes, your explanation is perfect.And I don't know why everyone in the comment section has doubt in the secondary circuit which is absolutely correct but no one saw the actual mistake in the primary circuit i.e what sir were trying to teach in the secondary circuit, it seems that he himself forgot to apply that in the primary circuit. The mistake is that sir took the voltage across primary coil to be 1000V which he used to calculate the emf induced across the secondary coil and got 200V. But the voltage across the primary coil is not 1000V, it is 990V because 1000V is across the source and the current in the circuit is 10A. So there will be a potential drop of (10x1=10V) across the internal resistance of the primary coil and hence (1000-10=990V) across the primary coil. So now if we use [ Ep/Es=Np/Ns ], we will get [Es=990x1/5=198V] . So the actual potential difference across the secondary coil is 198V rather than 200V. Hope that i have explained it correctly. Feel free to ask if there is something which i haven't explained completely/correctly. 🙏
@@monkey-ob9fbThe fundamental difference between the primary and the secondary circuit is that in the primary circuit emf is supplied while in the secondary circuit emf is induced across the coil.So,the coil in the primary circuit is a load and the coil in the secondary circuit is a source.
Sir last question mein primary circuit k power ko 10000 liya gya hai aur secondary circuit ko 9200. Sir secondary wale se ham Fe loss hata diye hain par primary wale se ni hataye usko poora 10000 rakha gya hai. Sir aisa kyun? Primary wale ka power toh 9300 watt hona chaahiye na ?
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8:32 transformer
12:38 reasons
21:00
39:17 feel aa chuka bhai
49:00 53:26
58:19 flux leakage
1:04:12 eddy current
1:09:18
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I don't know why everyone in the comment section has doubt in the secondary circuit which is absolutely correct but no one saw the actual mistake in the primary circuit i.e what sir were trying to teach in the secondary circuit, it seems that he himself forgot to apply that in the primary circuit. The mistake is that sir took the voltage across primary coil to be 1000V which he used to calculate the emf induced across the secondary coil and got 200V. But the voltage across the primary coil is not 1000V, it is 990V because 1000V is across the source and the current in the circuit is 10A. So there will be a potential drop of (10x1=10V) across the internal resistance of the primary coil and hence (1000-10=990V) across the primary coil. So now if we use [ Ep/Es=Np/Ns ], we will get [Es=990x1/5=198V] . So the actual potential difference across the secondary coil is 198V rather than 200V.
Hope that i have explained it correctly. Feel free to ask if there is something which i haven't explained completely/correctly. 🙏
Please correct me if I am wrong
hi there
there is no error, because Ep is given in question, not calculated.
Bhai toh Pin-Pout = 1000 kahan aa rha usse
Arey Ep is given, it is fixed, it is the voltage to be changed, coming to the question, if you draw the circuit of the primary side then it will be simple, with ac source and a resistor (coil) so of course the job of the ac source is to keep the voltage across the ends of the resistor 1000v so it is fixed, Ep=1000V, and hence the power = VI, this 10000 W is the power supplied by the ac source to the transformer, now when this power goes in, first it suffers cu loss in the primary circuit, then iron loss, then another copper loss in the secondary circuit, which is in total 1000W, Now In, Es the problem is that there is a load, and the load is the one supposed to dissipate the final power and give the required voltage , i.e if you are having output power as 9000 W then this 9000 W will be dissipated across the load only,( 14:35, maybe replay this part) so to get the required voltage across the load, two resistors are involved (cu one and load one, whereas in the primary circuit, only cu one was present), so as both these resistors are dissipating power, the total power disspated will obviously be 9000+200 =9200. So ig sir is right… this was my theory from what I have understood after replaying the part 100 times😂
52:15 part 8 , load ke pass power 9000w gaya woh bhi 200w loss hone ke bad iska matlab woh circuit ke pass total 9200w power tha..jiss se 200 w resistance udane ke bad baki bacha power load ke pass gaya.
Nice explanation bro..
@@Pram-ly-r thanks bro
But if total power of circuit is 9200 then that means total power loss is 800 Watts right
@@Jjk3089000 watt power has to reached to out appliance hence 1000w loss
@@Jjk308 Yes, 700 watts of iron loss plus 100 watts of copper loss in the primary coil, however this current also causes 200 watt copper loss in the secondary coil resulting in a final output power of 9000 watts. The output power is not the power delivered to the secondary coil on the iron core, it is the final power sent out of the transformer.
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great content and explanation sir...skipping aakash classes to attend your classes
Sir at 53:00 in 8 part if we say Esis = 9200 watts then won't it mean that total power loss = 800 watts as total power loss = Epip - Esis ?
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@@shirshgyan6778 bhai bords bbi ho jayega na isi se??
36:00 52:00 Mera bhi 45 A aaya tha😶🤦🏻♀️
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U a dropper or in 12th?
Damn you are 1 month ahead of me
@@kashyaptandel5212 are you guys dropper
@@kashyaptandel5212 are u in 12
@@ranbirkaurarora9979 no I am in 12th
Completed AC 4/11/22.
Done 17 jan 2023 2:58am tysm sir❤
At 52:15 in qsn 8,can anyone explain in little detail,why power become 9200,how 200 increase in it!
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Thanks Sir. Completed 27 September 2023. JEE 2024
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Completed this chapter on 27 Nov 2023
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16 May 2022
hi brother, using your physics notes of abj sir, thanks.
@@lamelauncelot1385 from where?
Completed ✅29-8-24
Sir at 53:00, secondary circuit ko Poutput mila 9000w agar isme se 200w ka copper loss hua h toh 200w minus hona chahiye so net power of secondary becomes 8800w.
no bro, the efficiency given tells us the power received by the load, not the entire seccondary circuit... that power is 9000, power in secondary circuit is 9200, but that 200 is lost in copper and only 9000 received by load
Sir at 56:00 , can't we use E= IR to find the current in copper, if we use it we get different ans , why is it so? Kindly guide 🙏
How did you find the value of R
@@Human-og7oj its given in question 1 ohm
Hlo.... primary ke coils ka inductance bhi to hoga aur resistance vese h hi ..to LR ckt ho gya ....ise E= iR na lgke E=iZ lgna chahiye....but uski information question me nhi h
Is liye hme majburan P= ei use krna pda
Thanks sir (:
49:35 and mai pehle se (I)s = 45A nikal karr baitha tha...🗿💀
Ps = Es.is se naaaa
Shouldnt the power loss due to resistance of circuit be counted along with the total power loss of the circuit(including load)..then why did we take total power in that circuit to be 9200 Watts?( I was thinking the net power loss in the circuit was 9000Watts which includes the resistance)
I too didnt understand why sir hadnt included that.
Think of it this way
Input source provides 10000watts, out of which 800watt goes to copper loss(primary circuit) nd iron loss, therefore 9200 watts of power goes to the secondary circuit which will be equal to EsIs.
Completed 18th July 2022 😊
thank you so much sir
Sir to wo 1st wire me power loss kyun nahi liya CU wire ke chalte??
Sir when are you going to start the next chapter?
advance lvl lectures with such low views😥
badhiya hai na bhai
Sir in part 5 at 46:43 , the resistance of the coil is given so why we can't use P(primary)=I²R(primary) , to get I as 100A ?
why not possible?
or anyone else ?
bhai sir ne 44:00 pe input power ko 10000 W le liya tha
BECAUSE you dont know the exact resistance of primary circuit you only know the resistance in the coils and not in the iron region, so we have to use formula independent of resistance.
Same problem bro
Bro it just simple as power given in trans is total power supplied which is not fully consumed by resistor
Further you have to use 100watt instead of 10000watt as 100 watt is consumed by resistor
at 42:33 how can sir use that result if transformer is not ideal?
Ye result hmesha valid h .... derive krte vkt sir ne kbhi nhi bola ki it is only for ideal transformer ....only jisme current ata h ...vo ideal k liye valid h
Es/Ep = Ns/Np is always valid
But Es/Ep = Ns/Np = Ip/Is is valid only if transformer is ideal
(See derivation again)
lekin power dissipated to V2/R bhi hota hai, to 33000V??
Thank you sir
sir par electric poles me jo wires use hote hai woh toh aluminium ke bne hote hai na copper ke nhi??
52:40 pe humne , Es/Ep = ip/ is use kyu nhi kiya 🙄
since it is not an ideal transformer as efficiency is 90%.
sir transformer wale Q8 me to 200W uss 9000W ka hi part hai na
Kyuki total Power loss = PCu + PFe
to Phir correct answer 45A hi aa raha hai
Aur sir agar ye correct hai to phir same concept primary ke liye bhi lagta na
Vhi to mai bhi revise krte samay confuse ho gya
@@krrishkumar6812 The induced EMF in the secondary coil is 200V. The power delivered to the secondary circuit is 9200W. Now, the current of 46A due to the induced EMF would result in a 200W copper loss in the secondary coil and the final power delivered to the load is 9kW. The misconception here is due to the incorrect idea that the potential difference across the load is equal to the potential difference induced in the secondary coil. There is a small potential drop due to the resistance of the coil.
The actual potential difference across the load in this case would be 9000/46 = 195.65V, and the potential drop due to the coil would be the cause for the 4.35V deficit.
@@existential_turtle yes, your explanation is perfect.And I don't know why everyone in the comment section has doubt in the secondary circuit which is absolutely correct but no one saw the actual mistake in the primary circuit i.e what sir were trying to teach in the secondary circuit, it seems that he himself forgot to apply that in the primary circuit. The mistake is that sir took the voltage across primary coil to be 1000V which he used to calculate the emf induced across the secondary coil and got 200V. But the voltage across the primary coil is not 1000V, it is 990V because 1000V is across the source and the current in the circuit is 10A. So there will be a potential drop of (10x1=10V) across the internal resistance of the primary coil and hence (1000-10=990V) across the primary coil. So now if we use [ Ep/Es=Np/Ns ], we will get [Es=990x1/5=198V] . So the actual potential difference across the secondary coil is 198V rather than 200V.
Hope that i have explained it correctly. Feel free to ask if there is something which i haven't explained completely/correctly. 🙏
The potential across the coil is 1000 V. The resistance of 1 ohm is of the coil itself.
@@monkey-ob9fbThe fundamental difference between the primary and the secondary circuit is that in the primary circuit emf is supplied while in the secondary circuit emf is induced across the coil.So,the coil in the primary circuit is a load and the coil in the secondary circuit is a source.
thanks sir
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Completed
22 Feb , 2023
JEE 2023
Today alternating current complete
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sir ac completed now
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Iske bad EMW hoga n sir
8:25 12:48 20:31 27:08 31:30 38:05 39:00
JEE ADV. 2025 ----> AC DONE ON 5 OCT 2024 AT 1.00 AM.
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Kya sir Prabal batch ke sabhi lecture free available honge
I have a doubt that by cal calculating current by p = vi and v/r equal to I is different ??
Same doubt
Yup same doubt
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Sir is this chap completed now
58:26
Sir sawtooth curve wala kuch problem kab karaoge
Sir jee main ke liye reso ki package ka Kaun sa exercise Karein??
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EX 1 AND EX 3 PARTN 2 BASSSS...
Sir last question mein primary circuit k power ko 10000 liya gya hai aur secondary circuit ko 9200. Sir secondary wale se ham Fe loss hata diye hain par primary wale se ni hataye usko poora 10000 rakha gya hai. Sir aisa kyun? Primary wale ka power toh 9300 watt hona chaahiye na ?
i know i am late but bhai ye toh source hai na secondary mein jake power loss hoga na
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Sir plz start thermo
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Dt. 7 Aug 2023
at 47.05 why Ip=Ep/(resistance)p
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Completed on 16/07/2023 at 4:00 pm makiiiiinnnnnnnnnnnnnnnn histtttttttttooooooooorrrrrrrrrrrrrryyyyyyyyyy
Sir modern physics kb start karenge?
42:00 3 D
Completed Alternating Current
15/feb/2023
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Sir please correct the alternating current playlist
One vedio is out of the track..
Are haa vo bhakti geet...
Which one bro tell me pls
Please tell which video is missing?
@@rajeshkumarsinghgangaprasa5374 kansa video alternating current ki track ki bahar hai
Sir chapter ki limitations hill jayega
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Is this chapter completed ???
39:32
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09/09/2022 to 16/09/2022
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