Question

A battery of emf $E_0=12$ $\text{V}$ is connected across a $4$ $\text{m}$ long uniform wire having resistance $4\Omega /\text{m}$. Two cells of emf $E_1=2\text{V}$ and $E_2=4\text{V}$ having internal resistance $2\Omega$ and $6\Omega$ respectively are connected as shown in the figure. If galvanometer shows no deflection at the point $\text{N}$, the distance of point $\text{N}$ from the point $\text{A}$ is equal to

• A. $\frac{5}{3}\text{m}$
• B. $\frac{25}{24}\text{m}$
• C. $\frac{3}{2}\text{m}$
• D. $\frac{24}{25}$ $\text{m}$

Answer 1

The correct option is B $\frac{25}{24}\text{m}$

Under the balanced condition, the galvanometer will show zero deflection and no current will flow through it.
The equivalent emf of two batteries $E_1$ and $E_2$ is:

$E=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$

$\Rightarrow E=\frac{\left(\frac{2}{2}\right)+\left(\frac{4}{6}\right)}{\left(\frac{1}{2}\right)+\left(\frac{1}{6}\right)}$

or, $E=\frac{\left(\frac{10}{6}\right)}{\left(\frac{4}{6}\right)}=2.5\text{volt}$

Now, the potential difference across $\text{AN}$ is equal to the equivalent emf of two cells connected in parallel.

$\Rightarrow V_{AN}=E$

or, $I_{AN}\times \left(4l_{AN}\right)=E$

In the potentiometer circuit,

$I=\frac{E_0}{R_{AB}}=\frac{12}{4+(4\times 4)}$

or, $I_{AN}=\frac{12}{20}=\frac{3}{5}$

Now,

$I_{AN}\times l_{AN}\times 4=2.5$

$\Rightarrow \frac{3}{5}\times L_{AN}=\frac{5}{2\times 4}$

$\therefore L_{AN}=\frac{25}{24}\text{m}$

Hence, option (b) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In problems involving multiple cell connected in}\\ \text{parallel in potentiometer circuit, the key concept}\\ \text{is to equate the potential difference of balanced}\\ \text{length of wire with parallel combination of cell}\end{array}$