Can a Bridge Voltage Divider be Used to Calculate Voltage Gain?

[Pablo3]

Good morning,I'm french and I need help for this exercice.
It's a exercice it is an exercise on the voltage gain,and on the first scheme there are the correction of my teatcher,but I but I was wondering if we can't calculate the voltage gain with a bridge voltage divider like I did.(It's in french but only the formula are important).
The second scheme,it's the same thing(the method of my teacher ,and bridge voltage divider),but is it good?

 

[BvU]

First diagram: yes. One more step and you can see that the $V_s\over V_e$ ratios are identical !

Second diagram: I don't see what your professeur did ?

But you want to check your $1\over Z_{eq}$ !

 

[Pablo3]

Hello,thank you for your help :),so my teacher did not do the second exercise,but I tried to use his methode.
His method is in red on this diagram(or scheme ):
Is-it a question?
No I don't want to check my 1/Zeq,this calculation is right I think but I have not always trusted me.
I must speak better in english to better understand you and write better.
But yes professeur=professor in english :).

 

[BvU]

Method of your professeur does not give 1: numerator and denominator are different.

Time to point out that method of professeur and your method are not different: $$ {V_s\over V_e} = {Z' I_e \over \left ( Z + Z' \right ) I_e} = {Z' \over Z + Z' } $$

 

[BvU]

No I don't want to check my 1/Zeq, this calculation is right

Of course not$${1\over 2} = {1\over 3} + 0.16667 \Rightarrow 2 = 3 + 6 \ \ \ \ {\rm ?} $$

 

[Pablo3]

Of course not$${1\over 2} = {1\over 3} + 0.16667 \Rightarrow 2 = 3 + 6 \ \ \ \ {\rm ?} $$

Yes yes,I have forgot one thing,and to answer your question no that does not mean it,but yes effectively the methods are similar.
Thank you very much !
I wish you a good day :).

 

[BvU]

Avec plaisir. You're welcome and I'll be glad to look at your result for the (R//C) / (R+C + R//C) case...

 

[Pablo3]

Yes yes,I have forgot one thing,and to answer your question no that does not mean it,but yes effectively the methods are similar.
Thank you very much !
I wish you a good day :).

Normally equal to Zeq :

 

[BvU]

So far, so good !