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Derive an expression for inductance measurement using Hay bridge
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Hay Bridge is used to measure high value inductances.

The circuit diagram is shown below.

When bridge is balance,

$\begin{gather*} Z_1 \ Z_x = Z_2 \ Z_3 \\[2ex] Z_2=R_2 \\ \\[2ex] Z_3=R_3 \\ \\[2ex] Z_x=R_{x\ }+j\omega{}L_x \\ \end{gather*} $

From equation 1,

${(R}_1-\dfrac{j}{\omega{}C_1})\left(R_x+j\omega{}L_x\right)=R_2R_3$

$R_1R_x+j\omega{}L_xR_1\frac{-jR_x}{\omega{}C_1}+\frac{L_x}{C_1}= R_3 \ R_2$

$(R_1R_x+\dfrac{L_x}{C_1})+j\left(\omega{}L_xR_1-\dfrac{R_x}{\omega{}C_1}\right)=R_2R_3 $

Equating real and imaginary terms,

$R_1R_x+\dfrac{L_x}{C_1}=R_2R_3$

$\omega{}L_xR_1-\ \dfrac{R_x}{\omega{}C_1}= 0$

$\therefore L_x = \dfrac{R_x}{\omega{}{^2}R_1C_1}$

$\\\begin{gather*} R_1\ R_x+\dfrac{R_x}{\omega{}{^2}R_1C{^2}}=R_2\ R_3 \\ \\[2ex] R_x\ \left\lfloor{}R_1+\ \dfrac{1}{\omega{}{^2}R_1C_1{^2}}\right\rfloor{}=R_2\ R_3 \\ \\[2ex] R_x\ \times{}\dfrac{\omega{}{^2}\ R_1{^2}C_1{^2}+1}{\omega{}{^2}R_1C_1{^2}}=R_2\ R_3 \\ \\[2ex] \therefore{}\ R_x=\dfrac{\omega{}{^2}C_1{^2}R_1R_2R_3}{1+\omega{}{^2}R_1{^2}C_1{^2}} \\ \\[2ex] \therefore{}L_x=\dfrac{\omega{}{^2}C_2{^2}R_1R_2R_3}{\left(1+\omega{}{^2}R_1{^2}C_1{^2}\right)\ \left(\omega{}{^2}R_1C_1\right)} \\ \\[2ex] \therefore{}L_x=\dfrac{C_1R_2R_3}{\left(1+\omega{}{^2}R_1{^2}C_1{^2}\right)} \\\end{gather*}$

A w appears in the expression for Lx, this bridge is frequency sensitive.

Quality Factor (Q) for capacitance is

$\\\begin{gather*}Q=\dfrac{1}{\omega{}R_1C_1} \\ \\[2ex] Or\ \ \omega{}R_1C_1=1/Q \\ \\[2ex] \therefore{}L_x=\dfrac{C_1R_2R_3}{1+1/Q{^2}} \\\end{gather*} $

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