This bridge provide us the most suitable method for comparing the two values of capacitor if we neglect dielectric losses in the bridge circuit. The circuit of De Sauty's bridge is shown below.

Battery is applied between terminals marked as 1 and 4. The arm 1-2 consists of capacitor c_{1} (whose value is unknown) which carries electric current i_{1} as shown, arm 2 - 4 consists of pure resistor (here pure resistor means we assuming it non inductive in nature), arm 3 - 4 also consists of pure resistor and arm 4 - 1 consists of standard capacitor whose value is already known to us.

Let us derive the expression for capacitor c_{1} in terms of standard capacitor and resistors.

At balance condition we have,

It implies that the value of capacitor is given by the expression

In order to obtain the balance point we must adjust the values of either r_{3} or r_{4} without disturbing any other element of the bridge. This is the most efficient method of comparing the two values of capacitor if all the dielectric losses are neglected from the circuit.

Now let us draw and study the phasor diagram of this bridge. Phasor diagram of De Sauty bridge is shown below:

Let us mark the voltage drop across unknown capacitor as e_{1}, voltage drop across the resistor r_{3} be e_{3}, voltage drop across arm 3 - 4 be e_{4} and voltage drop across arm 4 - 1 be e_{2}. At balance condition the electric current flows through 2 - 4 path will be zero and also voltage drops e_{1} and e_{3} be equal to voltage drops e_{2} and e_{4} respectively. In order to draw the phasor diagram we have taken e_{3} (or e_{4}) reference axis, e_{1} and e_{2} are shown at right angle to e_{1} (or e_{2}). Why they are at right angle to each other? Answer to this question is very simple as capacitor is connected there, therefore phase difference angle obtained is 90°.

Now instead of some advantages like bridge is quite simple and provides easy calculations, there are some disadvantages of this bridge because this bridge give inaccurate results for imperfect capacitor (here imperfect means capacitors which not free from dielectric losses). Hence we can use this bridge only for comparing perfect capacitors.

Here we interested in modify the De Sauty's bridge, we want to have such a kind of bridge that will gives us accurate results for imperfect capacitors also. This modification is done by Grover. The modified circuit diagram is shown below:

Here Grover has introduced electrical resistances r

_{1}and r

_{2}as shown in above on arms 1 - 2 and 4 - 1 respectively, in order to include the dielectric losses. Also he has connected resistances R

_{1}and R

_{2}respectively in the arms 1 - 2 and 4 - 1. Let us derive the expression capacitor c

_{1}whose value is unknown to us. Again we connected standard capacitor on the same arm 1 - 4 as we have done in De Sauty's bridge. At balance point on equating the voltage drops we have:

On solving above equation we get:

This the required equation.

By making the phasor diagram we can calculate dissipation factor. Phasor diagram for the above circuit is shown below

Let us mark δ_{1} and δ_{2} be phase angles of the capacitors c_{1} and c_{2} capacitors respectively. From the phasor diagram we have tan(δ_{1}) = dissipation factor = ωc_{1}r_{1} and similarly we have tan(δ_{2}) = ωc_{2}r_{2}.

From equation (1) we have

on multiplying ω both sides we have

Therefore the final expression for the dissipation factor is written as

Hence if dissipation factor for one capacitor is known. However this method is gives quite inaccurate results for dissipation factor.