Potentiometer

In a potentiometer experiment when terminals of the cell are connected at distance of $52 \mathrm{cm}$ on the wire, then no current flows through it. When $5 \mathrm{\Omega }$ shunt resistance is connected across the cell the balancing length is $40 \mathrm{cm}.$ The internal resistance of the cell (in $\mathrm{\Omega }$) is $\frac{n}{2}$ what is the value of $n$?

A wire $AB$ (of length $1 \mathrm{m}$, area of cross-section $\pi {\mathrm{m}}^2$) is used in the potentiometer experiment to calculate the emf and internal resistance $\left(r\right)$ of the battery. The emf and internal resistance of the driving battery are $15 \mathrm{V}$ and $3 \mathrm{\Omega }$ respectively. The resistivity of wire $AB$ varies as $\rho ={\rho }_{0}x,$ where $x$ is distance from $A$ in meters and ${\rho }_0=24\pi \mathrm{\Omega }$. The distance of null point from $A$ is obtained at $\sqrt{\frac{2}{3}} \mathrm{m}$ when switch $S$ is open. Find the value of $E$.

A $10 \mathrm{m}$ long potentiometer wire has a resistance of $20 \mathrm{ohm}$. It is connected in series with a battery of emf $3 \mathrm{V}$ and a resistance of $10 \mathrm{\Omega }$. The internal resistance of cell is negligible. If the length can be read accurately up to $1 \mathrm{mm}$, the potentiometer can read voltage with an accuracy of $\frac{k}{10}$ (in millivolt). Then the value of $k$ is

A wire $AB$ (of length $1 \mathrm{m},$ area of cross-section $\pi {\mathrm{m}}^2$) is used in potentiometer experiment to calculate emf and internal resistance $\left(r\right)$ of battery. The emf and internal resistance of driving battery are $15 \mathrm{V}$ and $3 \mathrm{\Omega }$, respectively. The resistivity of wire $AB$ varies as $\rho ={\rho }_{0}x,$ where $x$ is distance from $A$ in meters and ${\rho }_0=24\pi \mathrm{\Omega }-\mathrm{m}.$ The distance of null point from $A$ is obtained at $\sqrt{\frac{2}{3}} \mathrm{m}$ when switch $S$ is open. Find the value of $E.$

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

The balancing length for a cell is $560 \mathrm{cm}$ in a potentiometer experiment. When an external resistance of $10 \mathrm{\Omega }$ is connected in parallel to the cell, the balancing length changes by $60 \mathrm{cm}.$ If the internal resistance of the ceil is $\frac{\mathrm{N}}{10}\Omega ,$ where $N$ is an integer then value of $N$ 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 :-

In the given diagram length of potentiometer wire $AC$ is $10 \mathrm{m}$ and resistance is $R\mathit{.}$ Internal resistance of cell $E$ is negligible. Experimental cell $V$ is balanced at balancing length of $\ell =5 \mathrm{m}.$ Find new balancing length in centimetre is $\frac{k}{4}$ if wire $AC$ is replaced by another wire of length $20 \mathrm{m}$ and resistance $4\mathrm{R}$. Find $k$.

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

Why is a copper wire not used in potentiometer?

How will the position of balance point in the potentiometer will be affected if, (I) current in the potentiometer be increased.

Name the device used for measuring the internal resistance of a secondary cell.

Why is a potentiometer preferred over a voltmeter in measuring the emf of a cell?

What is meant by the sensitivity of a potentiometer? How can it be increased?