With the help of a schematic diagram discuss the principle , construction of a potentiometer. Describe any two application of a potentiometer.
Answers
Answer:
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Explanation:
The Potentiometer is an electric instrument used to measure the EMF (electromotive force) of a given cell, the internal resistance of a cell. And also it is used to compare the EMFs of different cells. It can also use as a variable resistor in most of the applications. These potentiometers are used in huge quantities in the manufacture of electronics equipment that provides a way of adjusting electronic circuits so that the correct outputs are obtained. Although their most obvious use must be for volume controls on radios and other electronic equipment used for audio.Construction and Working Principle
The potentiometer consists of a long resistive wire L made up of magnum or with constantan and a battery of known EMF V. This voltage is called as driver cell voltage. Connect the two ends of the resistive wire L to the battery terminals as shown below; let us assume this is a primary circuit arrangement. One terminal of another cell (whose EMF E is to be measured) is at one end of the primary circuit and another end of the cell terminal is connected to any point on the resistive wire through a galvanometer G. Now let us assume this arrangement is a secondary circuit.
Construction of Potentiometer
The basic working principle of this is based on the fact that the fall of the potential across any portion of the wire is directly proportional to the length of the wire, provided wire has a uniform cross-sectional area and the constant current flowing through it.“When there is no potential difference between any two nodes there is electric current will flow”.
Now the potentiometer wire is actually a wire with high resistivity (ῥ) with uniform cross-sectional area A. Thus, throughout the wire, it has uniform resistance. Now this potentiometer terminal connected to the cell of high EMF V (neglecting its internal resistance) called driver cell or the voltage source. Let the current through the potentiometer is I and R is the total resistance of the potentiometer.
Then by Ohms law V=IR
