# Stuck on caculate primary impedance question

Discussion in 'Homework Help' started by Sam2002tii, Feb 8, 2014.

1. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
Using the formula Zp = (Np/Ns)^2 x Zs

One primary winding of a transformer with 100 turns

Two secondary windings

One of the secondary windings has 50 turns with 10ohms of RL and the second has 25 turns with 25ohms of RL

Calculate the primary impedance Zp in the primary winding.

100/50 sqrd x 10ohms of Zs which is RL gives 40 ohms for the zp of one of the secondary windings.

100/25 sqrd x 25ohms of Zs which is RL of the other secondary winding gives 400 ohms of Zp

so I have 400 zp and 40zp for secondary winding 1 and 2 respectively.
The answer in the back of the book is 36.36ohms Zp for the primary winding.

Not sure how they ended up with that number.

Any help is greatly appreciated,

Thanks, Sam

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,970
616
400*40/(400+40)=36.36

3. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
Ok thanks, I see where they got the 36.36ohms.

Why do you need to take the product of each secondary winding Zp and divide it by the sum of each secondary Zp?

I added 400 + 40 and assumed that you could just add the two Zp secondary values to get the primary. Obviously it was not correct.

4. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
Just figured it out. Resistance in parallel formula, product over sum for two parallel branches.