Hay's bridge is the electrical circuit used for the measurement of self-inductance. It is an alternating current bridge similar to Maxwell's bridge with small modifications.

The main difference between Hay's bridge and Maxwell's bridge is that Hay's bridge employs a resistance in series with the standard capacitor, whereas Maxwell's bridge uses a resistor in parallel with the standard capacitor.

## Construction of Hay’s Bridge :

The Hay's bridge is a modified form of Maxwell's inductance-capacitance bridge. It measures inductance by comparing it with a standard variable capacitance. The circuit diagram of Hay's bridge is shown below.

It consists of an inductor with an inductance L_{1} and internal resistance R_{1} in arm AB and non-inductive standard resistances R_{2} and R_{3} in arms AD and BC respectively, and a known variable standard capacitance C_{4} in series with known non-inductive variable standard resistance R_{4} in arm CD.

## Operation and Theory of Hay’s Bridge :

The bridge can be balanced by adjusting the values of R_{4} and C_{4}. From the above figure,

#### Under balanced condition, we have,
Equating real and imaginary terms on both sides, we get,
Substituting equation 2 in 1, we get,
Substituting equation 3 in 2, we get,
Now, the quality factor of an inductor is given by,
Substituting equation 5 in 3, we have,
For high Q coils i.e., Q > 10, 1/Q^{2} is almost negligible. Hence the above equation reduces to,*L*_{1} = R_{2} R_{3} C_{4}

*L*

_{1}= R_{2}R_{3}C_{4}From the above equations, we can say that for high Q coils the expression for L_{1} is free from the frequency term. For low Q coils, 1/Q^{2} cannot be neglected and hence to find L_{1}, the frequency of source is to be accurately known. Therefore, the bridge suits only for the measurements of inductance of high Q coils.

## Phasor Diagram of Hay’s Bridge :

The below shows the phasor diagram of the bridge under balanced conditions. By taking inductor current I_{1} as the reference phasor. It is the current of arm AB, then the voltage drop across R_{1} will I_{1} R_{1} which will be in-phase with I_{1}. Similarly, the voltage drop across L_{1} will be I_{1} ωL_{1} which leads the current I_{1} with 90°. Now the total voltage drop V_{1} of arm AB will be the sum of voltage drops across R_{1} and L_{1}.

When the bridge is balanced, B and D will be at the same potential and there will be a null-deflection i.e., V_{1} = V_{2} and V_{3} = V_{4}, also I_{1} = I_{3} and I_{2} = I_{4}. Therefore, the phasor V_{2} lies along with V_{1} with equal magnitude, and the voltage drop I_{2} R_{2} (in AD arm) and current I_{2} will be in-phase with V_{2}. Also, when the bridge is balanced I_{4} lies along with I_{2} and I_{3} with I_{1}. Similarly, the voltage drop along the BC arm will be I_{3} R_{3} and will be in-phase with I_{3}. Thus I_{3} R_{3} lies along phasor I_{3} which is nothing but V_{3}, and V_{4} will be equal to V_{3} under balanced condition.

Now the voltage drop in arm CD is the sum of voltage drops across capacitor and resistor i.e., I_{4} R_{4} + I_{4}/ωC_{4}. Due to capacitance, the drop I_{4}/ωC_{4} lags behind the I_{4} by 90° and I_{4} R_{4} lies along with I_{4}. Therefore, the resultant of drop I_{4} R_{4} and I_{4}/ωC_{4} will be V_{4} (also equal to V_{3}). The resultant of V_{1} and V_{3} will be the V_{s} since V_{s} will be equal to (V_{1} + V_{3}) or (V_{2} + V_{4}).

### Advantages of Hay's Bridge :

- The expression obtained for the Q-factor of the coil using Hay's bridge is not a complicated one.
- From the above expression, it can be seen that the resistance R
_{4} is inversely proportional to the Q-factor. Lower the resistance higher the Q-factor. Thus for high Q coils, the value of resistance R_{4} should be quite small. Hence, the bridge requires the resistance of low value. - Hay's bridge is suitable for coils whose quality factor is greater than 10 (Q > 10). Also, it gives a simple expression for unknown inductance for high Q coils.

_{4}is inversely proportional to the Q-factor. Lower the resistance higher the Q-factor. Thus for high Q coils, the value of resistance R_{4}should be quite small. Hence, the bridge requires the resistance of low value.### Disadvantages of Hay's Bridge :

The major drawback of Hay's bridge is that it cannot be used for measuring coils having a Q-factor less than 10. We have,

For lower values of Q(<10) the term (1/Q)^{2} in the above expression cannot be neglected. Hence this bridge cannot be used for coils that have a Q-factor less than 10. For such coils, Maxwell's bridge can be used.