Question

The balancing length for a cell is $560cm$ in a potentiometer experiment. When an external resistance of $10\Omega$ is connected in parallel to the cell, the balancing length changes by $60cm$. If the internal resistance of the cell is $\frac{N}{{\displaystyle 10}}\Omega$, the value of N is

Answer 1

Step 1: Given data

Balancing length for cell$\left(L\right)$=$560cm$

Connected external resistance$\left(R\right)$=$10\Omega$

Change in balancing length after connecting external resistance=$60cm$

The internal resistance of cell $r=\frac{N}{{\displaystyle 10}}\Omega$

Step 2: Compute the EMF

Consider the following figure that shows the case $\left(1\right)$ i.e when the cell is connected.

Suppose the emf of the cell is $\epsilon$ and the internal resistance is $r$. And the potential gradient of potentiometer wire be $x$.

We know that $\begin{array}{rcl}potentialgradient\left(x\right)& =& \frac{EMF\left(\epsilon \right)}{Lengthofwire\left(L\right)}\end{array}$

$\begin{array}{rcl}x& =& \frac{\epsilon }{560}\\ \epsilon & =& 560x----\left(1\right)\end{array}$

Step3: Compute internal resistance

Now consider the following modified figure that shows the case $\left(2\right)$ i.e when the external resistor is connected.

Now the length $L=500cm$

From the figure, it is clear that $V_{AB}=V_{DF}=V_{MN}$. as they are connected in parallel connection So,

$\begin{array}{rcl}potentialgradient\left(x\right)& =& \frac{V}{Length}\\ x& =& \frac{V}{500}\\ V& =& 500x----\left(2\right)\end{array}$

It is understood that $\epsilon =I\left(R+r\right)$

where $\epsilon =EMF,I=current,R=Resis\tan ceinthecircuit,r=Internalresis\tan ce$

Substitute the known values in the formula $I=\frac{\epsilon }{\left(R+r\right)}$,

$I=\frac{\epsilon }{10+r}----\left(3\right)$

According to Ohm's law $V=IR$. So,

Substitute the values from equation $\left(2\right)$ and equation $\left(3\right)$ in $V=IR$,

$500x=\frac{\epsilon }{10+r}\times 10----\left(4\right)$

Substitute the value of emf from equation $\left(1\right)$ to equation $\left(4\right)$,

$$\begin{gathered} 500x=\frac{560x}{10+r}\times 10 \\ r+10=11·2 \\ r=1·2 \end{gathered}$$

Step 4: Determine the value of N

It is given that $r=\frac{N}{10}$,

$$\begin{gathered} 1·2=\frac{N}{10} \\ N=12 \end{gathered}$$

Hence, the value of $N$ is $12.$