Potentiometer

The emf of a cell is $2 \mathrm{V}$ and its internal resistance is $2\mathrm{ohm}$. A resistance of $8\mathrm{ohm}$ is joined to battery in parallel. This is connected in secondary circuit of potentiometer. If $1 \mathrm{V}$ standard cell balances for $100 \mathrm{cm}$ of potentiometer wire, the balance point of above cell is

A potentiometer can measure emf of a cell because :

A cell of $2\mathrm{V},\mathrm{ }1\mathrm{\Omega }$ is balanced at $1.9\mathrm{ }\mathrm{m}$. Then what is balanced length for ideal cell of $2\mathrm{V}$ ?

The potential gradient of potentiometer is $0.2 \mathrm{volt} {\mathrm{m}}^{-1}$. $A$ current of $0.1 \mathrm{amp}$ is flowing through $a$ coil of $2 \mathrm{Ohm}$ resistance. The balancing length in meters for the$\mathrm{p}.\mathrm{d}.$ at the ends of this coil will be

The length of a wire of a potentiometer is $100 \mathrm{cm},$ and the emf of its standard cell is $E \mathrm{volt}$. It is employed to measure the emf of a battery whose internal resistance is $0.5 \mathrm{Ohm}$. If the balance point is

Resistivity of potentiometer wire is ${10}^{-7}\text{Ωm}$ and its area of cross-section is ${10}^{-6}{ \mathrm{m}}^2.$ When a current $i=0.1 \mathrm{A}$ flows through the wire, its potential gradient is :

When a cell is balanced on potentiometer wire, then balancing length is $125 \mathrm{cm}.$ If resistance of $2 \mathrm{ohm}$ is connected across the ends of cell, then balancing length is $100 \mathrm{cm},$ then internal resistance of cell is

A cell is balanced at $100 \mathrm{cm}$ of a potentiometer wire when the total length of the wire is $400 \mathrm{cm}.$ If the length of the potentiometer wire is increased by $100 \mathrm{cm},$ then the new balancing length for the cell will be (Assume pd across potentiometer wire is constant)

If the specific resistance of a potentiometer wire is ${10}^{-7} \mathrm{\Omega } \mathrm{m}$, the current flowing through it is $0.1 \mathrm{A}$ and the cross-sectional area of the wire is ${10}^{-6} {\mathrm{m}}^2$ then, the potential gradient will be

A standard cell of $1.08 \mathrm{V}$ is balanced by the PD across $90\mathrm{ }\mathrm{c}\mathrm{m}$ of a meter long wire supplied by a cell of emf $2 \mathrm{V}$ through a series resistor of resistance $2\mathrm{ }\mathrm{\Omega }$. If the internal resistance of the cell in the primary circuit is zero then the resistance per unit length of potentiometer wire is

In a potentiometer the balancing with a cell is at length of $\mathrm{ }220\mathrm{ }\mathrm{c}\mathrm{m}$. On shunting the cell with a resistance of $3\mathrm{ }\mathrm{\Omega }$ balance length becomes$\mathrm{ }130\mathrm{ }\mathrm{c}\mathrm{m}$. Then the value of internal resistance of this cell is

The temperature coefficient of resistance of a potentiometer wire should be,

In a potentiometer experiment, the emf of a cell is balanced at $L$ length. Another cell of same emf $\mathrm{\epsilon }$ is connected parallel to it, then, new balancing length will be,

Figure shows a potentiometer. Length of the potentiometer wire $AB$ is $100 \mathrm{cm}$ and its resistance is $100 \mathrm{\Omega }$. EMF of the battery $E$ is $2 \mathrm{V}$. A resistance $R$ of $50 \mathrm{\Omega }$ draws current from the potentiometer. What is the voltage across $R$ when the sliding contact $C$ is at the mid-point of $AB$?

A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves:

In a potentiometer experiment, the balancing with a cell is at a length of $240 \mathrm{cm}$. On shunting the cell with a resistance of $2 \mathrm{\Omega }$ the balancing length becomes $120 \mathrm{cm}$. The internal resistance of the cell is :-

The potentiometer wire $10 \mathrm{m}$ long and $20 \mathrm{\Omega }$ resistance is connected to a $3 \mathrm{V}$ emf battery and a $10 \mathrm{\Omega }$ resistance. The value of potential gradient in $\mathrm{V} {\mathrm{m}}^{-1}$ of the wire will be

A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves

Potential gradient $\mathrm{K}$ remains constant until: 

Null point in galvanometer is obtained when a cell of emf $\epsilon$ and internal resistance $\mathrm{r}$ is connected across the length of $22 \mathrm{cm}$ wire of potentiometer. Now a resistance of $10\Omega$ is connected across the terminals of the cell (by closing the key $\mathrm{k}$ ) and null point is obtained against the length of $20 \mathrm{cm}$. Then the internal resistance rof the cell is: