Question

In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ${\epsilon }_1$ and ${\epsilon }_2$ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end Find(i)${\epsilon }_1/{\epsilon }_2$ and (ii) position of null point for the cell ${\epsilon }_1$. How is the sensitivity of a potentiometer increased?
OR
Using kirchoff's rules determine the value of unknown resistance R is the circuit so that no current flows through $4\Omega$ resistance. Also find the potential difference between A and D.

Answer 1

Given: a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ${\epsilon }_1$ and ${\epsilon }_2$ connected in the manner as shown are obtained at a distance of $120cm$ and $300cm$ from the end

To find (i) $\frac{{\epsilon }_1}{{\epsilon }_2}$ (ii) position of null point for the cell ${\epsilon }_1$. How is the sensitivity of a potentiometer increased

Solution:

(i) Apply Kirchhoff's law in loop ACFGA(refer fig(i):-

$\varphi \left(120\right)={\epsilon }_1-{\epsilon }_2$

$\varphi$= potential drop per length

Or, ${\epsilon }_1={\epsilon }_2+\varphi \left(120\right).....\left(i\right)$

Loop AEHIA:-

$\varphi \left(300\right)={\epsilon }_2+{\epsilon }_1$

Or,${\epsilon }_2+({\epsilon }_2+\varphi (120\left)\right)=\varphi \left(300\right)$( By substituting value ${\epsilon }_1$ from equation (i))

Or, $2{\epsilon }_2=(300-120)\varphi$

Or, ${\epsilon }_2=90\varphi ....\left(ii\right)$

Thus, ${\epsilon }_1=90\varphi +120\varphi$

${\epsilon }_1=210\varphi ...\left(iii\right)$

Hence $\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{210}{90}=\frac{7}{3}$

(ii) As we know,

$\epsilon =\varphi l$

Thus from equation (ii) and (iii).

Null point for cell ${\epsilon }_2$is $90cm$

And for cell ${\epsilon }_1$, itis $210cm$

Sensitivity of the potentiometer can be increased by:

(a) Increasing the length of the potentiometer wire

(b) Decreasing the resistance in the primary circuit

Refer fig(ii) for second pat of the question.

Apply Kirchhoff's law in loop ABCFA:-

$I+I+4I_1=9-6⟹2I+4I_1=\mathrm{3....}\left(i\right)$

As there is no current flowing through the $4\Omega$ reisistance.

$I_1=0$

Or, $2I=3⟹I=1.5A$

Thus the current through resistance $R$ is $1.5A$

As there is no current through branch CF, thus equivalent circuit will be, (refer fig(iii))

By alpplying Kirchhof's looplaw we get,

$1.5+1.5+R\left(1.5\right)=9-3⟹R=2\Omega$

Potential difference between A and D = $(9-3)=6V$