# Ideal Transformer Question

Discussion in 'Homework Help' started by xz4chx, Sep 27, 2013.

1. ### xz4chx Thread Starter Member

Sep 17, 2012
71
1
So for part A, I have an ideal transformer connected to a couple resistors.

*All voltages and currents I am assuming are rms in the problem because i don't want to type RMS for each one.*

a) So I did a source transformation to make the current source into a voltage source.
So $V_{th}= 240<0 [mV]$
Since I did a source transformation:
$R_{th}=5+15=20 [k\Omega]$

So i know in order to get the max amount of power from the load I have to make the source see a 20 [kΩ] through the number of turns in the coils. I am not 100% on how to do that (Find N2)?

b) That is just a voltage divider so I get $P= V^2/R$
$P= (120)^2/20 = 720 [mW]$

c) The voltage before the inductors and load is 120 V so if i put V2 across N2 then I have $V_1+V_2=120 [V]$, also the voltage per turn is the same so but with dot convention they are opposites
$V_1/N1=-V_2/N2
V_2=-(N2/N1)*V_1$

Putting that back into first equation

$V_1-(N2/N1)V_1 = 120$ (Solve for V1 which will give me V2)

d) Putting the circuit back to its original form before source transformation, I need to find the voltage across the source
KVL gives me $V_{cs} = 120V + (240/40000)*5000 = 150 [V]$
(240 [V] /40000 [Ω]) [<-- current before transformation back]
So $P_{cs} = 150 * 10^-3 = 2.4 [W]$
$% = (.72/2.4)*(100)= 30%$
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So I just need to know how to find N2. I have been given a formula sheet which does not seem to help and we have not gone over this in class yet but my homework is due next tuesday and I am interested on how to solve for this.

Thanks for the help and time
Zach

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The winding voltages must be directly proportional to their individual number of turns as you have already indicated. What is the voltage V2? If you know a certain number of turns gives you voltage V1 (for which you have developed an equation) then a commensurate number of turns will give voltage V2.