Potentiometer - Principle And Applications - Practice Questions & MCQ

Potentiometer

• The potentiometer is a device which does not draw current from the given circuit and still measures the potential difference.
• It is a device used to measure e.m.f of a given cell and to compare e.m.f's of cells
• It is also used to measure the internal resistance of a given cell

Potentiometer consists of wire of length 5 to 10 meters arranged on a wooden block as parallel strips of wires with 1-meter length each and end of wires are joined by thick coppers. The wire has a uniform cross-section and is made up of the same material. A driver circuit that contains a rheostat, key, and a voltage source with internal resistance r. The driver circuit sends a constant current (I) through the wire.

Potential across the wire  AB having length L  is given as V=IR, Where R is the resistance of the wire AB

Since the driver circuit sends a constant current (l) through the wire so $V \quad \alpha R$ Using $R=\frac{\rho L}{A}$ we can say that $R \quad \alpha \quad L$ since area and resistivity are constant. Therefore we get V is proportional to length. l.e $V \propto L$

The secondary circuit contains cell/resistors whose potential is to be measured. Whose one end is connected to a galvanometer and another end of the galvanometer is connected to a jockey which is moved along the wire to obtain a point where there is no current through the galvanometer. So that potential of the secondary circuit is proportional to the length at which there is no current through the galvanometer. This is how the potential of a circuit is measured using the potentiometer.

For the above figure
$l_1$ is the balancing length obtained when cell with emf $E_1$ is included in the secondary circuit. That is key is at position $1 . l_2$ is the balancing length obtained when cell with emf $E_2$ is included in the secondary circuit. That is key is at position 2.

So since $E \propto l$ we get

$$
\frac{E_1}{E_2}=\frac{l_1}{l_2}
$$

With the help of the above ratio, we can compare the emf of these cells.

Determine the internal resistance of a cell

Note-The cell in the secondary circuit has emf E and internal resistance r

Here $l_1$ is the balancing length obtained when key $\mathrm{K}^{\prime}$ is open that is we include only the cell in the secondary circuit. So corresponding potentials of wire of balancing length $l_1$ is E . And we know that $E \propto  l_1 \ldots$. (1)

Similarly $l_2$ is the balancing length obtained when key $\mathrm{K}^{\prime}$ is closed that is both cell and $R^{\prime}$ is connected in the secondary circuit.So corresponding potentials of wire of balancing length $l_2$ is $V$.

$
\begin{aligned}
& \text { And we know that } V \propto l_2 \ldots(2) \\
& \\
& \qquad \begin{array}{l}
I R^{\prime} \propto l_2 \\
\text { or we can say that } \\
\Rightarrow \frac{E}{r+R^{\prime}} * R^{\prime} \propto l_2 \ldots
\end{array}
\end{aligned}
$

So taking ratio of equation (1) to equation (2)

$
\begin{aligned}
& \text { we get } \frac{E}{V}=\frac{l_1}{l_2} \\
& \frac{E}{\frac{E R^{\prime}}{r+R^{\prime}}}=\frac{l_1}{l_2} \\
& \Rightarrow \frac{r+R^{\prime}}{R^{\prime}}=\frac{l_1}{l_2}
\end{aligned}
$
 

 Then the internal resistance is given by

$\begin{aligned} & r=\left(\frac{l_1-l_2}{l_2}\right) R^{\prime} \\ & r=\left(\frac{E}{V}-1\right) R^{\prime}\end{aligned}$

Comparison of resistances

The balance point is at a length $\mathrm{I}_1 \mathrm{~cm}$ from A when jockey $J$ is plugged in between $Y$ and $X$, while the balance point is at a length $I_2 \mathrm{~cm}$ from $A$ when jockey $J$ is plugged in between $Y$ and $Z$.

Then we get a ratio of resistances as

$$
\frac{R_2}{R_1}=\frac{l_2-l_1}{l_1}
$$

With the help of this ratio, we can compare these resistances.