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The length of a wire of a potentiometer is $100 \mathrm{cm}$, and the emf of its standard cell is $E$ volt. It is employed to measure the emf of a battery whose internal resistance is
$0.5 \Omega. $ If the balance point is obtained at $l=30 \mathrm{cm}$ from the positive end, the emf of
the battery is:
(A) $\dfrac{30 E}{100}$
(B) $\dfrac{30 E}{100.5}$
(C) $\dfrac{30 R}{(100-0.5)}$
(D) $\dfrac{30(E-0.5 i)}{100}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: It is known to us that a potentiometer is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. If only two terminals are used, one end and the wiper, it acts as a variable resistor or rheostat. 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 be used as a variable resistor in most of the applications. Potentiometers are commonly used to control electrical devices such as volume controls on audio equipment. Potentiometers operated by a mechanism can be used as position transducers, for example, in a joystick.

Complete step by step answer
It is known to us that the electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field. A potential is a scalar field that describes the potential energy per unit of some quantity due to a vector field. It is closely related to potential energy. The higher potential is the point of higher concentration of charges and lower potential is the point with a lesser concentration of charges. Hence considered the point which is more positive as a higher potential point.
We can also say that the electric potential is the electric potential energy of a test charge divided by its charge for every location in space. Because it's derived from energy, it's a scalar field. These two fields are related. The electric field and electric potential are related by displacement.
Let us consider that $\mathrm{V}$ be the potential across the balance point and one end of wire.
Hence according to the principle of potentiometer $\mathrm{V} \propto \mathrm{l}$
Also, if a cell of emf $\mathrm{E}$ is employed in the circuit between the ends of potentiometer wire of length $\mathrm{L}$ then $\mathrm{E} \propto \mathrm{L}$
Therefore, $\dfrac{\mathrm{V}}{\mathrm{E}}=\dfrac{\mathrm{l}}{\mathrm{L}}$
$\mathrm{V}=\dfrac{1}{\mathrm{L}} \mathrm{E}=\dfrac{30}{100} \mathrm{E}=\dfrac{30 \mathrm{E}}{100}$

So, the correct answer is option A.

Note: We know that the potentiometer consists of L which is a long resistive wire and a battery of known EMF V whose voltage is known as driver cell voltage. One end of the primary circuit is connected to the cell whose EMF E is to be measured and the other end is connected to galvanometer G. A potentiometer is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. If only two terminals are used, one end and the wiper, it acts as a variable resistor or rheostat. The potentiometer works on the principle that when a constant current flows through a wire of uniform cross-sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points.