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written 4 months ago by |

Answer: This method of measurement is particularly suited for the measurement of inductance having high Q values.Hay bridge differs from Maxwells bridge by having a resistance in series with a capacitorinstead of being parallel.For large phase angles, $R_1$ needs to be low,therefore the bridge is more convenient for measuring high Q values. The schematic diagram for Hay bridge is shown below :

When the bridge is balanced,$Z_1Z_x=Z_2Z_3$

Here, $Z_1=R_1-\dfrac{j}{\omega C_1}$

$Z_2=R_2$

$Z_3=R_3$

$Z_x=R_x+j\omega L_x$

Putiing all the above values ,we get

$R_1R_x+j \omega L_xR_1-\dfrac{jR_x}{\omega C_1}+\dfrac{L_x}{C_1}=R_2R_3$

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

Equating real and imaginary parts, we get

$R_1R_x+\dfrac{L_x}{C_1}=R_2R_3$.....................(1)

$\omega L _xR_1-\dfrac{R_x}{\omega C_1}=0$

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

Putting this in equation 1 , we get

$R_x[R_1+\dfrac{1}{\omega^2R_1C_1^2}]=R_2R_3$

$R_x=\dfrac{\omega^2C_1^2R_1R_2R_3}{1+\omega^2R_1^2C_1^2}$

Putting the above value in (2)

$\therefore L_x=\dfrac{C_1R_2R_3}{1+\omega^2R_1^2C_1^2}$............................(3)

From the above formula , we can calculate unknown inductance.Since , $\omega$ appears in the expression for $L_x$ , the bridge is frequency sensitive.

Quality factor (Q) for capacitor is given as

$Q=\dfrac{1}{\omega R_1C_1}$

Putting this in equation (3)

$\therefore L_x=\dfrac{C_1R_2R_3}{1+\dfrac{1}{Q^2}}$

If (Q>10) , the equation becomes $L_x=C_1R_2R_3$ the equation obtained is similar to Maxwells bridge.