How to get this expression from the circuit

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SUMMARY

The discussion focuses on deriving the output voltage expression \( V_o \) from a circuit where \( R1 = R2 \). The user initially applied the voltage divider rule and attempted to calculate \( V_o \) using midpoints of the branches. The correct expression is confirmed to be \( V_o = V_s \left(\frac{1}{2} - \frac{R}{R + \frac{1}{jwC}}\right) \). The key takeaway is the importance of proper algebraic manipulation and parentheses in achieving the correct formula.

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nhrock3
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it is given that R1=R2

http://i43.tinypic.com/b8359w.jpg
i used voltage divider
i tried to calculate it by taking v1 as the midle poit of the left brench
and v2 as the middle point of the of the other branch

so Vout=V_s/2 -Vs[r/(r+1/jwc )=v_s/2(1-rjwc/2(rjwc+1))

i can't get the expression in the photo
 
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nhrock3 said:
it is given that R1=R2

http://i43.tinypic.com/b8359w.jpg
i used voltage divider
i tried to calculate it by taking v1 as the midle poit of the left brench
and v2 as the middle point of the of the other branch

so Vout=V_s/2 -Vs[r/(r+1/jwc )=v_s/2(1-rjwc/2(rjwc+1))

i can't get the expression in the photo

I get the expression in the photo, so it must just be an algebra problem in your work. Be careful with your parenthesis when writing it out like you did above... the last expression is missing some parenthesis that make a difference.

In Latex:

[tex]V_o = V_s (\frac{1}{2} - \frac{R}{R + \frac{1}{jwC}})[/tex]

[tex]V_o = V_s (\frac{1}{2} - \frac{R}{\frac{jwRC}{jwC} + \frac{1}{jwC}})[/tex]

... and so on ...
 
Last edited:

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