Discussion page for Q.439 electrical engineering questions and answers (2023)

The turns ratio required to match 80Source to a 320burden is

=2

Turns ratio = sq rt (n2/n1) = 2

r'2= (r2/K^2).
So k = square of (320/80).
K=2.

Therefore turns ratio (k) = square root of (R2/R1)
= square root of (320/80)
= 2.

According to Ohm's law; Current I = V/Z.
V = supply voltage - back EMF.

Impedance Z =R+jX.
Reactance X =2 Pi f L.

We know that in the case of a DC supply, the frequency f is zero; no induced EMF, since there is no flux change at DC (Induced EMF e= N dq/dt;).

Current I =(supply voltage-back EMF)/R=j+ X.

Here the numerator increases (back EMF is zero) and the denominator decreases (since X = 0 and R is very small); consequently, I rises to a dangerous level, leading to burning of the windings if not properly fused.

Therefore, the DC supply should not be connected to the transformer; If connected, a mechanism for frequent switching on and off (such as SCR) should be provided.

(I1)^2/R1=(I2)^2/R2.

(I1/I2)^2=R2/R1.

I1/I2=SQRT(320/80).

I1/I2=2.

So the answer is 2.

v2/v1 = (v1/r1)*(r2/v2).

(v2/v1)^2 = r2/r1.

After calculation, the result is 2.

Where Z01 = Z1+(Z2/k^2).
Z02 = Z2+(Z1*k^2).

Solving the above equations I got '1' which is not the answer.

Is it the right approach?

= v1/r1*r2/v2.

(v2/v1)^2 = r2/r1.

(k)^2 = r2/r1.

k^2 = 320/80.

k^2 = 4 .

k = 2.

Lastwiderstand (R2) = 320 Ohm.

R1 = R2/K^2.
80 = 320/K^2.
K^2 = 320/80 = 4.
K = quadrat(4) = 2.

In fact, we know that whenever primary is referred to secondary,
r1'=kSquare*r1.

So here,

ksquare*r1 = 320, which means that

kQuadrat = 320/r1=320/80=4.

i.e. k² = 4 then k=2.

Here k = transformation ratio.

Then the answer is 4.

80*k^2 = 320.
k^2 = 4.
k = 2.

When going from secondary to primary, the secondary impedance must be multiplied by 1/k^2. Match it to the primary impedance to match. The same result is obtained.

[v2/v1 = k,i2*r2/i1*r1 = k, R(hv)/R(lv) = k^2; k = 2; r2/r1 = k^2].

(k)^2 = r2/r1.
k^2 = 320/80.
k^2 = 4 .
k = 2.

I(p) N(p) = I(s) N(p).

E(P)N(P) / R(P) = E(S)N(S)/R(S)-2.

From 1 and 2.

Sq (N(S)/N(P)) = R(s)/R(p).

Turns ratio = E1/E2 = V1/V2 = N1/N2 = 1/k.

Primary or source resistance R1 = 80 Ohm.

Secondary resistance R2 = 320.

We know that:

Transmission ratio k = N1/N2 = V1/V2 = R2/R1 = I2/I1.

SO K=320/80=4 is the correct answer.

n^2 = 320/80 = 4.

n = 2.

Gear ratio (n) = N1/N2 = V1/V2 = I2/I1.
The same transformer ratio is (K) = 1/n.

Bedeutet V2/V1 = N2/N1 = I1/I2 = K.

The question is turns ratio?
V1*I1=V2*I2.
(v1)^2/R1= (V2)^2/R2.
Also (V1/V2)^2=R1/R2.

Gear ratio n = (N1/N2)^2=R1/R2=80/320=1/4.

If we do sqrt it will come =1/2.

What do you think of the above solution?

The correct law is: vs/sp = NS/NP = IP/IS.

Please revise again.

Because the turns ratio is Np/Ns, not Ns/Np.

It should be necessary to measure the current and voltage of the transformer from the secondary. Then you can know the KVA rating of the transformer!

Np/Ns=Zp/Zs.
Wo,
Np=number of turns in the primary winding.
Ns=number of turns in the secondary side.
Zp=primary impedance or (source).
Zs = secondary impedance or (load).
Floor; Which Z = 80 Ohm.
Last Z = 320 Ohm.
Zp/Zs=80/320.
Turns Ratio = 1:2 Answer = 2.
The 1:2 turns ratio matches the 80 ohm source impedance to a 320 ohm load.

R1 = 80 Ohm
R2 = 320 Ohm
transformer ratio,
K = N1/N2 = V1/V2=R2/R1=I2/I1.
K=R2/R1.
= 320/80=4.

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Life, K^2=320/80=4 => K=√(4)=2
But turns ratio = 1/K, now turns ratio = 1/2 = only 0.5!

So the formula is Zs = Zp* (N1/N2)^2.
now here we can say square root (Zs/Zp)= N1/N2.
Hence N1/N2= √(320/80).
And turns ratio = 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.

So the answer should be 1/2.

V1^2/R1 = V2^2/R2.

V2^2 / V1^2 = R2/R1.
(k)^2 = r2/r1.
k^2 = 320/80.
k^2 = 4 .
k = 2.