Why is a potentiometer preferred over a voltmeter to measure emf of a cell?

A potentiometer being a null device, does not draw current from the balance point at the balance point. Therefore, potentiometer measures the actual emf of the cell. Whereas, a voltmeter always draws current from the cell and measures the terminal voltage of the cell instead of the actual emf of the cell .

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Are the paths of electrons straight lines between successive collisions (with positive ions of the metal) in the (i) absence of electric field? (ii) presence of electric field? Establish a relation between drift velocity ‘v_{d}’ of an electron in a conductor of cross-section ‘A’ carrying current ‘I’ and concentration ‘n’ f free electrons per unit volume of conductor.

i) In the absence of electric field the path of electrons are straight lines between successive collisions.

ii) When an electric field is set up from positive to negative charge, the electrons are accelerated towards the negative charge. Therefore, in the presence of electric field the electrons follow a curved path in between successive collisions.

Consider a conductor of length'l' and of uniform area of cross section 'A'.

Therefore,

Volume of conductor, V = Al

If, n is the number density of electrons, then

Total number of free electrons in the conductor = Aln

If, e is the total charge on each electron then,

Total charge on electrons in the conductor, q = Alne

Let, a constant potential difference 'V' be applied across the ends of a conductor. Then,

Electric field set up across the conductor, E = V/l

Because of this field, the free electrons present in the conductor begin to move with a drift velocity, v_{d} towards the left hand side.

Therefore,

Time taken by the free electrons to cross the conductor, t = *I /V _{d} *Hence, current, I = $\frac{\mathit{q}}{\mathit{t}}\mathit{=}\frac{\mathit{A}\mathit{l}\mathit{n}\mathit{e}}{\mathit{l}\mathit{/}{\mathit{v}}_{\mathit{d}}}$

$\Rightarrow $ $\overline{)\mathrm{I}=\mathrm{Anevd}}$

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Obtain an expression for the potential gradient ‘k’ of potentiometer whose wire of length l has a resistance r. The driving cells has an emf E, is connected in series with an external resistance R.

Given,

Length of the potentiometer wire = l

Internal resistance of the wire = r

Emf of the driving cell =E

External resistance of the cell = R

Current drawn by the wire, I= $\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$

Now, using Ohm's law V =IR

i.e, V =$\frac{\mathrm{ER}}{\mathrm{R}+\mathrm{r}}$

Now, potential gradient of the wire is given by

k = V/l

=$\frac{\mathrm{ER}}{(\mathrm{R}+\mathrm{r})\mathrm{l}}$

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Define the term electrical resistivity of a material. How is it related to its electrical conductivity? Of the factors, length area of cross-section, nature of material and temperature, which one controls the resistivity value of a conductor?

Resistivity is defined as the measure of resistance to electrical conduction for a given size of material.

For a particular material at specified temperature, resistivity is the electrical resistance per unit length and per unit of cross sectional area.

Resistivity is the inverse of conductivity.

The resistivity of a material depends upon the material and temperature of the material and is independent of the length and area of cross-section.

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Let, the drift velocities of the wires be v_{d }and v_{d}' respectively.

Area of cross- section of two wires be A and A'.

Then,

A : A' = 1:2

This implies,

A' = 2A

i) When the two wires are connected in series, the same amount of current flows through both the wires.

Implies,

n A v_{d} e = n A' v_{d}' e

where,

n is the number of electrons per unit volume and,

e is the charge on electron.

Therefore,

$\frac{{\mathrm{v}}_{\mathrm{d}}}{{{\mathrm{v}}_{\mathrm{d}}}^{\text{'}}}=\frac{A\text{'}}{A}=\frac{2A}{A}=2:1$

ii) When the two wires are connected in parallel, the potential difference across the two wires would be the same.

Then,

$\mathrm{n}\mathrm{A}{\mathrm{V}}_{\mathrm{d}}\mathrm{e}\mathrm{\rho}\frac{\mathrm{l}}{\mathrm{A}}=\mathrm{n}\mathrm{A}\text{'}{\mathrm{v}}_{\mathrm{d}}\text{'}\mathrm{e}\mathrm{\rho}\frac{\mathrm{l}}{\mathrm{A}\text{'}}$

where, $\mathrm{\rho}$ is the resistivity of the two wires.

Therefore,

$\frac{{\mathrm{v}}_{\mathrm{d}}}{{{\mathrm{v}}_{\mathrm{d}}}^{\mathrm{i}}}=1$

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