Electrical Engineering - Transformers - Discussion

Discussion Forum : Transformers - General Questions (Q.No. 16)
16.
A certain amplifier has 600 internal resistance looking from its output. In order to provide maximum power to a 4 speaker, what turns ratio must be used in the coupling transformer?
8
0.8
0.08
80
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 2 of 2.

Rakesh alliwar said:   9 years ago
Yes, 0.08 is the correct answer.

E.sravani said:   1 decade ago
Power = i^2r.

Input power = i1^2r1 = i1^2(600).

Output power = i2^2r2 = i2^2(4).

P(in) = P(out).

i1^2(600) = i2^2(4).

K^2 = (4/600) {(i1/i2)^2 = k^2}.

K = 0.08.

Ashim said:   1 decade ago
Turns ratio K = N2/N1= sqrt (R2/R1) = sqrt (4/600) = 0.08.

Love said:   1 decade ago
But it is asked to determine turn ratio = 1/K.

Akash.bhure said:   1 decade ago
Internal resistance(R1)=600 ohms.

Output speaker(R2)=4 ohms.

R1 = R2/K^2.
600 = 4/K^2.
K^2 = 4/600 = 0.0066.
K = sqrt(0.0066) = 0.081.

Manish said:   1 decade ago
V2/I2 = 4.

V1/I1 = 600.

V2xI1/V1xI2 = 4/600.

N2 = 4/600N1 = 0.08.

Siri said:   1 decade ago
@Saritha.
Transformation ratio is unit less quantity.
Not measured in ohms.

Saritha said:   1 decade ago
Here,

Load resistance(destination)=4 ohms
Source resistance=600 ohms

k=(r2/r1)^0.5
=(4/600)^0.5
=0.0816
k=0.0816 ohms

ShaijuTS said:   1 decade ago
Pow((Ns/Np),2) = Zs/Zp
ie. pow(k,2) = 4/600
k=sqrt(4/600)
k=0.08


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