Owen's bridge is an ac circuit similar to Maxwell and Hay's bridge used for the measurement of self-inductance over a wide range of values. The principle of operation is similar to Maxwell and Hay's bridge i.e., balancing the loads on its four arms and comparing the unknown value with the known value. Here the unknown inductance to be determined is compared with the standard capacitor. Hence, it gives the value of self-inductance in terms of capacitance.

## Construction of Owen’s Bridge :

The circuit diagram of Owen's bridge for the measurement of inductance in terms of standard capacitance is shown in the figure below. It consists of an inductor in arm AB with inductance L_{1} and internal resistance R_{1} which is to be measured. A standard variable resistance and capacitance of R_{2} and C_{2} are placed in arm AD.

The arms BC and CD consist of standard resistance R_{4} and standard capacitance C_{4} respectively. A null indicator or detector is connected between junctions B and D, to balance the arms of the bridge.

## Operation and Theory of Owen’s Bridge :

When the detector shows null deflection, it means the bridge is balanced. Since at balanced condition, the junctions B and D are at the same potential and no current flows through the detector. This can be done by varying the standard variable resistor R_{2} and capacitor C_{2} in arm AD.

#### From the figure above, Under balanced condition, we have, Equating the real and imaginary terms on both sides, we get, Hence, the unknown inductor in arm AB is compared with the known capacitor in arm CD, and the value of inductance and internal resistance of the inductor are determined in terms of capacitance.

## Phasor Diagram of Owen’s Bridge :

Let us draw the phasor diagram for Owen's bridge by using the relation between voltage drops and currents in each arm. Firstly, it should be noted that, across a resistance, the voltage and current through it are in the same phase.

Whereas in the case of the inductor and capacitor, the voltage leads the current by 90° and lags the current by 90° respectively. The phasor diagram of the bridge under balance conditions is shown below.

Taking current I_{1} as the reference phasor, as it is flowing in the arm AB. The voltage drop across R_{1} will be I_{1} R_{1} which lies in phase with I_{1}. The voltage drop across the inductor will be I_{1} jωL_{1} which leads the current I_{1} by 90°. Thus, the total voltage drop V_{1} across arm AB will be the resultant of the sum of the drops across R_{1} and L_{1}.

We know that when the bridge is balanced the junctions B and D will be at the same potential and no current flows through the null indicator, thus I_{1} = I_{3} and I_{2} = I_{4}, also the drops V_{1} = V_{2} and V_{3} = V_{4}. Hence, phasor V_{2} = V_{1}. The drop V_{2} in arm AD is the sum of drops in resistor R_{2} and capacitor C_{2} i.e., I_{2} R_{2} + I_{2}/jωC_{2}.

Now the voltage drop V_{3} in arm BC i.e., in resistor R_{3}, is I_{3} R_{3}. Since I_{3} will be equal to I_{1} in the balance condition, thus the drop I_{3} R_{3} will lie along with I_{1}. Also, V_{3} = V_{4} under balance condition, hence the drop V_{4} (i.e., I_{4}/jωC_{4}) will be same as V_{3}.

The supply voltage V is the sum of drops V_{1} + V_{3} or V_{2} + V_{4}. Therefore, the resultant of phasor V_{1} and V_{3} or V_{2} and V_{4} gives the supply voltage V.

### Advantages of Owen's Bridge :

- The equation obtained is independent of frequency. Hence, slight variations in frequency do not affect the balance.
- The two balance equations are independent of each other, as R
_{2} and C_{2} are chosen as variable elements. - A wide range of measurements of inductance can be done.

_{2}and C_{2}are chosen as variable elements.