ac bridge offset voltage

[Carlos100pre]

Can anybody explain this formula thanks in advance, I need to find the maximum bridge offset voltage amplitud, the only information I have is all the capacitors = 880pF, E = 10vrms, f = 10kHz, what is j? isn't an imaginary number how can I use it?

\[\Delta\]E= E[(R3 - j/\[\omega\]Cx)/ (R1 + R3 - j/\[\omega\]C3) - (R4 - j/\[\omega\]Cx)/R2 + R4 - j/(\[\omega\]Cx)]

 

[chuckey]

j is a term to indicate that the quantity has a direction that is at right angles to a quantity you can measure. It comes from J squared = -1 . if you have a sine wave and you multiply it by -1, its the same amplitude, but when the sinewave was + its now -, so j X the sinewave means the sinewave is at 90 degrees to the original. There are special rules for the j operator which you will have to learn, such as dividing by a j term, what you have to do is to multiply, top and bottom by -j, so the j on the bottom line becomes - j X j or 1, so it can be ignored.
In general, the voltage at the half way point on one leg is, Vin X (Zbottom/(Z top + Zbottom)), this occurs with the other leg, call it Vin X (Zbottom2/(Ztop2 + Zbottom2)). The voltage out is then the difference between the two, which is what you have.
Frank

 

[Carlos100pre]

j is a term to indicate that the quantity has a direction that is at right angles to a quantity you can measure. It comes from J squared = -1 . if you have a sine wave and you multiply it by -1, its the same amplitude, but when the sinewave was + its now -, so j X the sinewave means the sinewave is at 90 degrees to the original. There are special rules for the j operator which you will have to learn, such as dividing by a j term, what you have to do is to multiply, top and bottom by -j, so the j on the bottom line becomes - j X j or 1, so it can be ignored.
In general, the voltage at the half way point on one leg is, Vin X (Zbottom/(Z top + Zbottom)), this occurs with the other leg, call it Vin X (Zbottom2/(Ztop2 + Zbottom2)). The voltage out is then the difference between the two, which is what you have.
Frank

Frank thanks for the explanation, one more question what about if i need to find the value of the resistance that i need to use, how can I find the sesistance values(R1, R2)

 

[chuckey]

Without a circuit I don't know where these components are. From what I said, the ratio of the resistors is the same in each arm. :-
Vin ----------------------
........... |................... |............... R3/R1+R3 = R4/R2+R4
.......... R1................. R2
............|.................... |.............. or R1/R3 =R2/R4
........... ---- Vo---------
............ |.................... |
............ R3................. R4
.............|.....................|
......------------------------
If there are some capacitors involved, they cause a phase shift in the Vo from that side, so to balance it up, the phase shift must be the same on the other side.
Hope the text circuit works LOL It does not!!!. It edits out spaces even when they occur during a line so I filled in the spaces with dots>.......
Frank