De Sauty bridge with RC parallels

Question

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.

Answer 1

Your first equation is of course correct:

$$Z_x=Z_c\frac{R_1}{R_2}$$ 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

$$C_x=C_c\frac{R_2}{R_1}\text{ and }R_x=R_c\frac{R_1}{R_2}$$