A potentiometer circuit is set up as shown. The potential gradien… - Vidyadip

MCQ

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is $k$ $volt/cm$ and the ammeter, present in the circuit, reads $1.0\,\, A$ when two way key is switched off. The balance points, when the key between the terminals $(i)$ $1$ and $2$ $(ii)$ $1$ and $3,$ is plugged in, are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes, of the resistors $R$ and $X,$ in $ohms$, are then, equal, respectively, to

A
$k(l_2 -l_1)\,Ω , kl_2\,Ω$
✓
$kl_1 \,Ω , k(l_2 - l_1)\,Ω$
C
$\;{\rm{k}}\left( {l_2 - l_1} \right)\,\Omega {\rm{\;}},{\rm{\;\;k}}l_1{\rm{\;}}\,\Omega {\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}$
D
$kl_1 \,Ω , kl_2\, Ω$

✓

Answer

Correct option: B.

$kl_1 \,Ω , k(l_2 - l_1)\,Ω$

b
When the two way key is switched off, then The current flowing in the resistors $R$ and $X$ is

$I=1\, \mathrm{A}$ .......$(i)$

When the key between the terminals $1$ and $2$ is plugged in, then

Potential difference across $R=I R=k l_{1}$ ......$(ii)$

where $k$ is the potential gradient across the potentiometer wire

When the key between the terminals $1$ and $3$ is plugged in, then

Potential difference across $(R+X)=I(R+X)=k l_{2}$ ....$(iii)$

From equation $(ii),$ we get

$R=\frac{k l_{1}}{I}=\frac{k l_{1}}{1}=k l_{1} \Omega$ .......$(iv)$

From equation $(iii),$ we get

$R + X = \frac{{k{l_2}}}{I} = \frac{{k{l_2}}}{1} = k{l_2}\,\Omega \quad {\rm{ (Using }}({\rm{i}}))$

$X = k{l_2} - R$

$ = k{l_2} - k{l_1}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,{\rm{(Using}}\left( {iv} \right){\rm{)}}$

$=k\left(l_{2}-l_{1}\right) \,\Omega$

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