Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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 ofthe battery is:(A) $\dfrac{30 E}{100}$(B) $\dfrac{30 E}{100.5}$(C) $\dfrac{30 R}{(100-0.5)}$(D) $\dfrac{30(E-0.5 i)}{100}$

Last updated date: 20th Jun 2024
Total views: 54.9k
Views today: 0.54k
Verified
54.9k+ views
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.

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.