Discussion :: Transformers - General Questions (Q.No.4)
|Jeet said: (Mar 30, 2011)|
|Because I is inversely proportional to no. of turns.|
|Velan said: (Aug 21, 2011)|
|Es/Ep = Ns/Np = Ip/Is
V increase I decrease
N(no of turns) increase V increase
|Sukumar said: (Feb 24, 2012)|
|I think this effect the primary current in the form of increasing.
Am I right?
|Avinash said: (Apr 8, 2012)|
|In this case the ratio Ip/Is will increase but it will not affect the primary current as no.of turns in the primary is same.|
|Ramya Wijesundera said: (Apr 26, 2012)|
|The number of turns are more resistance is more so the current is less. Further, when number of turns are more generated e.m.f. is more.
Therefore, according to the equation P=VI, secondary I (current) should be less than primary current because power is not increasing.
|Vijay Kumar said: (May 25, 2012)|
|We know that in a transformer,
Ns/Np = Ip/Is.
From the above relation, The relation btwn no.of turns and current is inversely propotional to each other.
|Purnima said: (Aug 17, 2012)|
|If no of turn is increase then the resistance is also increases thats why the current is decreases.|
|Ravi Kumar said: (Feb 28, 2013)|
|Because I is inversely proportional to no.of turns.|
|Veera said: (Mar 8, 2014)|
|The work of resistance is its oppose the current. So only resistance is increasing means current is decreasing.|
|Rudresh.R said: (Jan 23, 2015)|
|Krishna Chilakabattini said: (Mar 30, 2015)|
|V1/V2 = I2/I1 = N1/N2.|
|Malik said: (Sep 18, 2015)|
|If we give 220 on primary 100 turns and we take from secondary 110 and what should be no of turns of secondary winding.|
|Buba said: (Mar 4, 2017)|
|Answer incorrect, without load increasing turns of secondary doesn't change any current.|
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