# De Sauty bridge with RC parallels

The bridge is described in this way:

Whit the equilibrium condition, when the current is zero between A and B, I'll have:

$$Z_x*R_2=Z_c*R_1$$

derived by Wheatstone Bridge's relation.

Evaluating the parallel impedances ( $R_x\; ||\; 1/(j\omega C_x)$ ), I obtain:

$$R_x/(1+j\omega R_x*C_x) * R_2 = R_c/(1+j\omega R_c*C_c) * R_1$$

And so:

$$R_x*R_2*(1+j\omega R_c*C_c)= R_c*R_1*(1+j\omega R_x*C_x)$$

Finally, It's obtained these results:

$$R_x=(R_c*R_1)/R_2$$ $$C_x=(R_2*C_2)/R_1$$

Solutions of the De Sauty Bridge. But, how are obtained this results starting from the previous equation?

If we had not parallel scheme, as impedance between $C_x$ and $R_x$, the result should be obvious for $C_x$ starting by the equilibrium condition; but, as it is, I cannot figure out which steps are made.

• I hope you can see LaTeX...
– FdT
Commented May 14, 2013 at 13:15
• Not sure what you mean to reference by 'starting from the previous equation'. Could you add numbers to the formula's and clarify the question a bit? Commented May 14, 2013 at 18:50
• Hi jippie, I don't know what's happened, but there was already an answer to this question. To calculate $$R_x$$ and $$C_x$$ it need to divide the impedence in real and imaginary part, separately. Not both as I made. Probably is the correct way.
– FdT
Commented May 14, 2013 at 19:10

$$Z_x=Z_c\frac{R_1}{R_2}$$ Now it's convenient to invert this equation:
$$\frac{1}{Z_x}=\frac{1}{Z_c}\frac{R_2}{R_1}$$
Using $1/Z_x=j\omega C_x+1/R_x$ and $1/Z_c=j\omega C_c+1/R_c$ we obtain
$$j\omega C_x+1/R_x = (j\omega C_c+1/R_c)\frac{R_2}{R_1}$$ Comparing real and imaginary parts we get
$$C_x=C_c\frac{R_2}{R_1}\text{ and }R_x=R_c\frac{R_1}{R_2}$$