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How to Download a PDF of De Sauty Bridge Experiment

De Sauty Bridge is a type of AC bridge that is used to compare the capacitances of two capacitors or to measure the capacitance of an unknown capacitor. It is based on the principle of Wheatstone bridge, where the ratios of impedances are equal when the bridge is balanced. De Sauty Bridge consists of two resistors and two capacitors connected in series in four arms of the bridge, as shown in the figure below.

De Sauty Bridge experiment is a common practical course for B.Sc. Physics students, who need to determine the value of an unknown capacitor by using a known capacitor and two resistance boxes. The experiment also involves adjusting the frequency of an audio frequency oscillator and observing the current through a milliampere meter or a headphone. The experiment can be performed by following the steps given in the PDF file below.

If you are looking for a PDF file of De Sauty Bridge experiment, you can download it from one of the following sources:

B.Sc. Physics Laboratory Manual - MSUNIV: This PDF file contains the aim, formula, principle, procedure, precautions, result and table for De Sauty Bridge experiment, along with other practical courses for B.Sc. Physics students. The PDF file is available for free download from the official website of Manonmaniam Sundaranar University.[^1^]

De Sauty Bridge PDF - Scribd: This PDF file contains a brief description, formula, circuit diagram, procedure, precautions, result and table for De Sauty Bridge experiment. The PDF file is available for free download from Scribd, a popular online platform for reading and publishing documents.[^2^]

De Sauty Bridge Electrical4U: This web page contains a detailed explanation, formula, circuit diagram, derivation and example for De Sauty Bridge experiment. The web page also provides a link to download a PDF file of De Sauty Bridge experiment at the end of the article. The web page is part of Electrical4U, a website that provides free online tutorials on electrical engineering topics.[^3^]

By downloading a PDF file of De Sauty Bridge experiment from any of these sources, you can learn how to perform the experiment and measure the capacitance of an unknown capacitor by using De Sauty Bridge.

De Sauty Bridge is based on the concept of resonance, which occurs when the frequency of the applied AC voltage matches the natural frequency of the circuit. At resonance, the impedance of the circuit is minimum and the current is maximum. The resonant frequency of a series LCR circuit is given by:

$$f_r = \\frac{1}{2\\pi\\sqrt{LC}}$$

where L is the inductance, C is the capacitance and fr is the resonant frequency.

In De Sauty Bridge, the inductance is replaced by a capacitor, so the resonant frequency becomes:

$$f_r = \\frac{1}{2\\pi\\sqrt{C_1C_2}}$$

where C1 and C2 are the capacitances of the two capacitors in the bridge.

When the bridge is balanced, the voltage across the two capacitors is equal and the current through them is zero. This implies that the capacitive reactance of the two capacitors is equal, which can be written as:

$$X_C = \\frac{1}{2\\pi fC}$$

where XC is the capacitive reactance, f is the frequency and C is the capacitance.

Therefore, we can write:

$$\\frac{1}{2\\pi fC_1} = \\frac{1}{2\\pi fC_2}$$

or

$$C_1 = C_2$$

This means that the capacitance of the unknown capacitor C2 is equal to the capacitance of the known capacitor C1.

If we introduce a resistance R in series with each capacitor, then the impedance of each arm becomes:

$$Z = R + jX_C = R + j\\frac{1}{2\\pi fC}$$

where Z is the impedance and j is the imaginary unit.

The condition for balance now becomes:

$$R_1 + j\\frac{1}{2\\pi fC_1} = R_2 + j\\frac{1}{2\\pi fC_2}$$

This equation can be simplified by multiplying both sides by 2πfC1C2, which gives:

$$(R_1C_2 + R_2C_1) + j(C_2 - C_1) = 0$$

This equation can be satisfied if both the real and imaginary parts are zero, which implies that:

$$R_1C_2 = R_2C_1$$

and

$$C_2 = C_1$$

The first equation gives us another way to find the capacitance of the unknown capacitor C2, by using the resistances R1 and R2. The second equation confirms that the capacitance of C2 is equal to C1. aa16f39245