Electrical Engineering - Transformers - Discussion

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

Why transformer can not work on DC?

Assuming the transformer to be ideal,as losses in primary and secondary are equal.

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

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

I1/I2=SQRT(320/80).

I1/I2=2.

So the answer is 2.

Transformer works on the principle of Faradays Laws of Electro Magnetic Induction; with mutual Induction concept.

By Ohms Law ;current I = V/Z.
V =supply voltage- bak EMF.

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

We know that incase of DC supply,frequency f is zero;no induced EMF because there is no change of flux with DC(Induced EMF e= N dq/dt;).

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

Here, numerator increases(back EMF is Zero) and denominator decreases(as X=0 and R is very less);consequently I increases to a dangerous level leading to burning of winding's if it is not properly fused.

Hence DC supply should not be connected to Transformer ; in-case connected a mechanism should be made to switch ON and OFF frequently(like SCR).

We know in ideal Transformer v2/v1 = i1/i2 = k.

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

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

After calculation result is 2.

What is converter transformer?

What is the turns ratio the ratio of transformer needed to match 160 ohm load?

Turn ratio = square root of load impedance upon source impedance.

Copper losses are current square*resistance losses. Or winding losses. {I*I*R}.

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 ?