Electrical Engineering - Transformers - Discussion
Discussion Forum : Transformers - General Questions (Q.No. 2)
2.
The turns ratio required to match an 80 
source to a 320 load is
source to a 320 load isDiscussion:
101 comments Page 2 of 11.
Siri said:
1 decade ago
Cu losses does not depends on voltage, iron losses depends on voltage and Cu losses depends on current moreover power factor of t/f depends on load hence rated in kva where as in motors power factor depends only on construction hence rated in kw.
Sarath said:
1 decade ago
The impedance or resistance wrt both primary and secondary must be same which implies Z01 = Z02.
Where Z01 = Z1+(Z2/k^2).
Z02 = Z2+(Z1*k^2).
Solving the above equations I got '1' which isn't the answer.
Is it the correct approach ?
Where Z01 = Z1+(Z2/k^2).
Z02 = Z2+(Z1*k^2).
Solving the above equations I got '1' which isn't the answer.
Is it the correct approach ?
Khanindra Kalita said:
4 years ago
Source to a 320 ohm, mean primary side resistance is 320, and secondary side resistance is 80 ohm.
Now ,
We know that,
R1 = R2 * (N1/N2)^2,
320 = 80 * (N1/N2)^2,
(N1/N2)^2 = 320/80 = 4,
N1/N2 = √4.
= 2.
Now ,
We know that,
R1 = R2 * (N1/N2)^2,
320 = 80 * (N1/N2)^2,
(N1/N2)^2 = 320/80 = 4,
N1/N2 = √4.
= 2.
(7)
Saranya said:
1 decade ago
Single phase transformer 50hz 200 kva 11kv/230v test results: no load test: input = 1600watts short circuit test : 2600 watts calculate all day efficiency duty cycle 160kw 0.8 pf 8 hr 100kw unity pf 6hr remaining hrs no load.
Sanchayan Singha Roy said:
5 years ago
Let, the impedance of the secondary side be Zs and that of primary is Zp.
So, the formula is, Zs = Zp* (N1/N2)^2.
now, we can say here is, sq. root(Zs/Zp)= N1/N2.
Therefore, N1/N2= √(320/80).
And Turns ratio = 2.
So, the formula is, Zs = Zp* (N1/N2)^2.
now, we can say here is, sq. root(Zs/Zp)= N1/N2.
Therefore, N1/N2= √(320/80).
And Turns ratio = 2.
(2)
Venkatesh S said:
1 decade ago
In question we understood that we convert the 320 ohm resistance in secondary as 80 ohm in the primary side (means source side). r'2 is 80, r2 is 320. Turns ratio is K.
r'2= (r2/K^2).
So that k=sqrt of (320/80).
K=2.
r'2= (r2/K^2).
So that k=sqrt of (320/80).
K=2.
Krishnakumar said:
1 decade ago
During current loss test (short circuit) we had applied rated current of the transformer. Transformer reaches its rated current at some voltage. It is very low current loss only depend on voltage. So it is negligible.
GAUTAM said:
10 years ago
Clearly given in question that:
Primary or source resistance R1 = 80 ohm.
Secondary resistance R2 = 320.
We know that:
Transformation ratio k = N1/N2 = V1/V2 = R2/R1 = I2/I1.
SO K = 320/80 = 4 is correct answer.
Primary or source resistance R1 = 80 ohm.
Secondary resistance R2 = 320.
We know that:
Transformation ratio k = N1/N2 = V1/V2 = R2/R1 = I2/I1.
SO K = 320/80 = 4 is correct answer.
Mallikarjun Bhirade said:
8 years ago
Explanation:- Given Data, R1=80 Ohms, R2=320 Ohms, We have, R2/R1=K^2
Where, K=Transformation Ratio
Now, K^2=320/80=4 => K=√(4)=2
But, Turns Ratio = 1/K, Now, Turns Ratio = 1/2 =0.5 only!
Where, K=Transformation Ratio
Now, K^2=320/80=4 => K=√(4)=2
But, Turns Ratio = 1/K, Now, Turns Ratio = 1/2 =0.5 only!
Anirudhya said:
1 decade ago
v1 I1=v2 I2
v1*v1/r1=v2*v2/r2
sq of (v1/v2)=r2/r1
sq of (1/k)=r2/r1=4
1/k=2------this is turns ratio
k=1/2-----this is transformation ratio
bcoz--n2/n1==transformation ratio
and
n1/n2==turns ratio
v1*v1/r1=v2*v2/r2
sq of (v1/v2)=r2/r1
sq of (1/k)=r2/r1=4
1/k=2------this is turns ratio
k=1/2-----this is transformation ratio
bcoz--n2/n1==transformation ratio
and
n1/n2==turns ratio
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