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# A potentiometer wire has a length $10m$ and resistance $20\Omega$. Its terminals are connected to a battery of EMF $4V$ and internal resistance $5\Omega$. What are the distances at which the null points are obtained when two cells of EMF $1.5V$ and $1.31V$ are connected so as to (a) assist or (b) oppose each other.

Last updated date: 13th Jun 2024
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Hint: We will use the formula of current in terms of voltage and resistance first to get current. Then we will calculate potential drop by using voltage and resistance. After this we should calculate potential gradients. After finding these quantities we will use the formula of EMF in terms of balancing length and potential gradient.
Formula Used:
We will use the following formula to solve the given problem:-
$l=\dfrac{E}{\kappa }$

Total current ( $i$ ) will be given as the ratio of voltage ( $V$ ) by total resistance ( $R$ ), as follows:-
$i=\dfrac{V}{R}$ ……………. $(i)$
Putting values in $(i)$ we get
$i=\dfrac{4}{20+5}$
$\Rightarrow i=\dfrac{4}{25}$……………. $(ii)$
Now we will calculate voltage drop $({{V}_{d}})$ which is equal to the product of current and resistance $(R)$. We will use the following formula:-
${{V}_{d}}=i\times R$ …………….. $(iii)$
Putting values we get
${{V}_{d}}=\dfrac{4}{25}\times 20$
${{V}_{d}}=\dfrac{16}{5}V$ ……………. $(iv)$
Now potential gradient $\left( \kappa \right)$ will be given as follows:-
$\kappa =\dfrac{{{V}_{d}}}{L}$ ………….. $(v)$ Where $L=10m$ is the length of wire.
Putting values in $(v)$, we get
$\kappa =\dfrac{16}{5\times 10}$
$\Rightarrow \kappa =\dfrac{16}{50}$ ………….. $(vi)$
Now, the EMF of the cells is given as the product of potential gradient and balancing length. For the case when there is assist, EMF is given as follows:-
${{E}_{1}}=\kappa \times {{l}_{1}}$ ……………. $(vii)$ Where balancing length for assist is given as ${{l}_{1}}$ .
According to the question ${{E}_{1}}=1.5+1.3=2.8V$, putting this in $(vii)$ we get
$2.8=\dfrac{16}{50}\times {{l}_{1}}$
$\Rightarrow {{l}_{1}}=\dfrac{2.8\times 50}{16}$
Therefore, ${{l}_{1}}=8.75m$
Hence, balancing distance in the case of assist will be equal to $8.75m$ .
Now, for the case of opposing each other, EMF of the cells, ${{E}_{2}}=1.5-1.3=0.2V$
Hence, ${{E}_{2}}=\kappa \times {{l}_{2}}$ ………….. $(viii)$
Putting the values of the respective parameters in $(viii)$ we have
$0.2=\dfrac{16}{50}\times {{l}_{2}}$. Where ${{l}_{2}}$ is the balancing distance for opposition.
$\Rightarrow {{l}_{2}}=\dfrac{0.2\times 50}{16}$
Therefore, ${{l}_{2}}=0.625m$
Hence, balancing distance in case of opposition will be equal to $0.625m$ .

Note:
We should remember the fact that the total EMF of a cell is the sum of individual EMFs of individual cells in the case of assist and it is subtracted when they oppose each other. Relations should be used carefully without any mistake. It should be noted that EMF is a terminal potential when no current flows but not be considered as force.