Electrical Engineering - Transformers - Discussion
Discussion Forum : Transformers - General Questions (Q.No. 4)
4.
If a transformer has 50 turns in the primary winding and 10 turns in the secondary winding, what is the reflective resistance if the secondary load resistance is 250 
?
?250
25
6,250
62,500
Discussion:
46 comments Page 1 of 5.
Bhagyashree said:
1 decade ago
R2=250
k=n1/n2=50/10
R1=R2*k^2
=250*(2500/100)
R1=6250ohm
k=n1/n2=50/10
R1=R2*k^2
=250*(2500/100)
R1=6250ohm
HJHGJK said:
1 decade ago
K=N2/N1
N1=50
N2=10
K=(10/50)=1/5
SECONDARY TO PRIMARY R/K^2
SO 250/(1/5)^2=250*25=6250 OHMS
N1=50
N2=10
K=(10/50)=1/5
SECONDARY TO PRIMARY R/K^2
SO 250/(1/5)^2=250*25=6250 OHMS
Prakash. D. said:
1 decade ago
Turns ratio K= n1/n2= 50/10
K=5,
w.k.t..., R1/R2=K^2
R1=250*25
R1=6250 ohms.
K=5,
w.k.t..., R1/R2=K^2
R1=250*25
R1=6250 ohms.
Chintan said:
1 decade ago
N1=50 , N2=10
Turn ratio K=N2/N1=10/50
K=1/5=0.2
now,R2/R1=K^2
R1=R2/K^2
=250/0.2^2
R1=6250 ohms.
Turn ratio K=N2/N1=10/50
K=1/5=0.2
now,R2/R1=K^2
R1=R2/K^2
=250/0.2^2
R1=6250 ohms.
Siddu said:
1 decade ago
Turns ratio will proportional to square of load then square of resistance in sec/pri 2n square.
Pinnapareddy .sandhya lakshmi said:
1 decade ago
r1=r2/k^2
k=n2/n1=10/50=1/5
r2=250
r1=250/(1/5)^2=6250
r1=6250
k=n2/n1=10/50=1/5
r2=250
r1=250/(1/5)^2=6250
r1=6250
Divij said:
1 decade ago
r1=(n1/n2)^2*r2
WHERE n1=50 , n2=10,r2=250
Ans: 6250
WHERE n1=50 , n2=10,r2=250
Ans: 6250
Sanjeev Kr. Rajoria said:
1 decade ago
Transformation ratio a = N1/N2
so
a=50/10=5
now
load resistance referred to primary= (a^2)*(load resistance)
so
the reflected load = 5^2*250
= 25*250
= 6250 ohm
So the correct ans is (c)= 6250 ohm
so
a=50/10=5
now
load resistance referred to primary= (a^2)*(load resistance)
so
the reflected load = 5^2*250
= 25*250
= 6250 ohm
So the correct ans is (c)= 6250 ohm
Kuldeep Bisht said:
1 decade ago
Primary resistence R'=R2/(K)2
here k=N2/N1=10/50=0.2
hence R'=250/0.2*0.2=6250 Ohm
here k=N2/N1=10/50=0.2
hence R'=250/0.2*0.2=6250 Ohm
Amit Ranjan said:
1 decade ago
Eqv.resistance referred to primary=r1+r2/(k*k)
Here r1 is not given
k=10/50=1/5
Hence ans = 250/(1/25) = 6250 ohm.
Here r1 is not given
k=10/50=1/5
Hence ans = 250/(1/25) = 6250 ohm.
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