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 $\varepsilon_{1}$ and $\varepsilon_{2}$
connected in the manner shown are obtained at a distance of $120 \ cm $ and $300 \ cm$ from the end $A$.
Find $(i) \varepsilon_{1} / \varepsilon_{2}$ and $(ii)$ position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased?
Using Kirchoffs rules determine the value of unknown resistance $R$ in the circuit so that no current flows through 4 $\Omega$ resistance. Also find the potential difference between $A$ and $D.$

Q: 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 $\varepsilon_{1}$ and $\varepsilon_{2}$ connected in the manner shown are obtained at a distance of $120 \ cm $ and $300 \ cm$ from the end $A$. Find $(i) \varepsilon_{1} / \varepsilon_{2}$ and $(ii)$ position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased? Using Kirchoffs rules determine the value of unknown resistance $R$ in the circuit so that no current flows through 4 $\Omega$ resistance. Also find the potential difference between $A$ and $D.$

1. $\varepsilon_{1} + \varepsilon_{2} = 300k \ (k$ is potential gradient in $\text{volt/cm)}$ $\varepsilon_{1} - \varepsilon_{2} = 120 \text{k}$ $\Rightarrow\frac{\varepsilon_{1}}{\varepsilon_{2}} = 7 / 3$ 2. $\varepsilon_{1} + \varepsilon_{2} = 300\text{k}$ $\therefore\varepsilon_{1} + \frac{3}{7}\varepsilon_{1} = 300 \text{k}$ $\Rightarrow\varepsilon_{1} = 210 \text{k}$ Therefore, balancing length for cell $\varepsilon_{1}$ is $210\ cm.$ 3. By decreasing potential gradient. $[$Or through Increasing length, reducing potential drop across wire, increasing resistance put in series with the main cell etc.$]$ Alternate Answer Applying Kirchhoff's Voltage rule for loop $\text{ABEFA}$ $-9 + 6 + 4 x 0 +2 I = 0$ $2 I – 3 = 0$ $\text{I} =\frac{3}{2}\text{A} = 1.5\text{A}$ For loop $\text{BCDEB}$ $3 + I R + 4 x 0– 6 = 0$ $\therefore\text{IR} = 3 $ Substituting the value of current $I,$ $\frac{3}{2}\times\text{R} = 3 $ $\therefore\text{R} = 2\Omega$ Potential difference between $A D$ Through path $\text{ABCD}$ $+9V – 3V – IR = V_{AD}$ $ + 9 - 3 - \frac{3}{2}\times2 = \text{V}_{AD}$ $\Rightarrow\text{V}_{AD} = 3\text{V}$ Alternate Answer through path $\text{AFD}$ $\frac{3}{2}\times2 =\text{V}_{AD}$ $\Rightarrow\text{V}_{AD} = 3 \text{V}.$

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 $\varepsilon_{1}$ and $\varepsilon_{2}$
connected in the manner shown are obtained at a distance of $120 \ cm $ and $300 \ cm$ from the end $A$.
Find $(i) \varepsilon_{1} / \varepsilon_{2}$ and $(ii)$ position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased?

Using Kirchoffs rules determine the value of unknown resistance $R$ in the circuit so that no current flows through 4 $\Omega$ resistance. Also find the potential difference between $A$ and $D.$

CBSE DELHI - SET 1 2012

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Solution

$\varepsilon_{1} + \varepsilon_{2} = 300k \ (k$ is potential gradient in $\text{volt/cm)}$

$\varepsilon_{1} - \varepsilon_{2} = 120 \text{k}$
$\Rightarrow\frac{\varepsilon_{1}}{\varepsilon_{2}} = 7 / 3$

$\varepsilon_{1} + \varepsilon_{2} = 300\text{k}$

$\therefore\varepsilon_{1} + \frac{3}{7}\varepsilon_{1} = 300 \text{k}$
$\Rightarrow\varepsilon_{1} = 210 \text{k}$
Therefore, balancing length for cell $\varepsilon_{1}$ is $210\ cm.$

By decreasing potential gradient.

$[$Or through Increasing length, reducing potential drop across wire, increasing resistance put in series with the main cell etc.$]$
Alternate Answer
Applying Kirchhoff's Voltage rule
for loop $\text{ABEFA}$
$-9 + 6 + 4 x 0 +2 I = 0$
$2 I – 3 = 0$
$\text{I} =\frac{3}{2}\text{A} = 1.5\text{A}$
For loop $\text{BCDEB}$
$3 + I R + 4 x 0– 6 = 0$
$\therefore\text{IR} = 3 $
Substituting the value of current $I,$
$\frac{3}{2}\times\text{R} = 3 $
$\therefore\text{R} = 2\Omega$
Potential difference between $A D$
Through path $\text{ABCD}$
$+9V – 3V – IR = V_{AD}$
$ + 9 - 3 - \frac{3}{2}\times2 = \text{V}_{AD}$
$\Rightarrow\text{V}_{AD} = 3\text{V}$
Alternate Answer
through path $\text{AFD}$
$\frac{3}{2}\times2 =\text{V}_{AD}$
$\Rightarrow\text{V}_{AD} = 3 \text{V}.$

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