Solved Examples
In a De Sauty Bridge arm BC contains a resistor of value 1400Ω, arm AD contains a capacitor of 0.15µF and arm CD contains a resistor of 1800Ω.If the bridge is in a balanced state, find the unknown capacitance in arm AB.
C2=0.15µF
R3=1400Ω
R4=1800Ω
At balanced condition:
C1=C2·(R4/R3)
=(0.15 x 10-6 x 1800)/1400
= 0.192µF
C1=0.192µF
The balanced condition of a De Sauty bridge is obtained at the following values. R3=2KΩ, R4=1500Ω and the standard capacitance C2=2µF. Find the unknown capacitance C1
C2=2µF
R3=2000Ω
R4=1500Ω
At balance condition:
C1=C2·(R4/R3)
= 2 x 10-6 x (1500/2000 )
C1 =1.5 x 10 -6F
De Sauty Bridge
De Sauty Bridge is an AC bridge used to find the unknown capacitance in a circuit. French engineer Paul de Sauty invented it. De Sauty Bridge compares the unknown capacitance with another standard or known capacitor. It can also be used to compare the capacitors used in a circuit. De Sauty Bridge has a high degree of accuracy in measuring unknown capacitance over a wide range of capacitances. It works on the principle of null deflection. It is the simplest way to compare two pure(non-inductive) capacitors. The circuit design of the De Sauty Bridge is very simple and straightforward and the calculations are also simple. However, it can only measure capacitors that are free from dielectric loss. To overcome this disadvantage Grover modified the De Sauty Bridge by adding extra resistors.
Table of Content
- Defintion
- Working Principle
- Components
- Construction
- Formula
- Phasor Diagram
- Modified De Sauty Bridge
- Advantages and Disadvantages
- Application