Hello every one...Hope you fine.

I am studying Transformer circuit analysis...I have faced a simple circuit where the text says that it found the equivalent resistance of a parallel inductor and a resistor but I don't know how!!!

If anyone can explain this situation and give the interpretation of a general one that would be great.
The problem is attached.

Thanx.

 

Here is a simple derivation.

The equivalent resistance is the resistance that the primary 'sees'.
In other words it is the resistance we would have to place in the primary circuit to draw the same power as the 9 ohms in the secondary.

Since this is a step up voltage transformer it steps the current down in the inverse ratio ie the secondary to primary current ratio is 2 : 3.

Consider any resistance Rs driven by the secondary and let Ip be the primary current.
Then the secondry current is 2Ip/3 . If Rp is the equivalent resistance as seen by the primary

Then equating power

Primary power = Secondary power


\(I_p^2{R_p} = I_s^2{R_s} = {\left( {\frac{{2{I_p}}}{3}} \right)^2}{R_s}\)

\(I_p^2{R_p} = I_p^2{\left( {\frac{2}{3}} \right)^2}{R_s}\)

\({R_p} = {\left( {\frac{2}{3}} \right)^2}{R_s}\)

Edit I note you are using complex quantities so

A more complete formula for reflected impedance is given here

http://www.insula.com.au/physics/1221/L17.html

 

Now i understand the formula ie power in S = power in P ....But still don't know why Is=(2/3)*Ip....where did this come from....based on what???

 

It is one of the basic properties of a transformer.

If n is the turns ratio (secondary/primary) = T2 / T1

Then the ratio of the then the voltage in the secondary = n times the voltage in the primary Vs = nVp or Vs/Vp = n

The current is given by the reciprocal so the current in secondary = 1/n times the current in the primary.

This is proved in the first two section of the link I gave you for complex quantities.

You should learn this it is very important.

 

Do these properties apply only for only Ideal transformer huh?

 

Yes and no.

The output voltage of an opencircuit (no load) transformer will be pretty close to ideal. Of course there is no load current.

As we take more and more load current the output voltage drops away from ideal.
But we can use a more sophisticated model of the transformer to account for this.

We are also able to design and manufacture transformers that are over 90% efficient in their operating range (not all are as good as this, especially cheap ones) so we are operating quite close to ideal conditions.

Incidentally I'm sorry I mixed up the current and voltage turns ratio in the original. I have put this right so that later viewers will not be confused.
So posts 2 and 4 now say the same thing.

 

Really thankful for your help