Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\, V$ and a negligible internal resistance. The potentiometer wire itself is $4\, m$ long. When the resistance $R$, connected across the given cell, has values of (i) infinity (ii) $9.5\, \Omega $ the balancing lengths on the potentiometer wire are found to be $3 \,m$ and $2.85\, m$, respectively. The value of internal resistance of the cell is

• A. $0.25\, \Omega$
• B. $0.95\, \Omega$
• C. $0.5\, \Omega$
• D. $0.75\, \Omega$

Answer 1

The Correct Option is C

The internal resistance of the cell is
$r = \bigg(\frac{l_1}{l_2}-1\bigg) R$
Here, $l_1=3\, m,t_2=2.85\, m, R=9.5\, \Omega$
$\therefore \, \, \, \, r = \bigg(\frac{3}{2.85}-1\bigg) (9.5\, \Omega)= \frac{0.15}{2.85} \times 9.5\, \Omega=0.5\, \Omega $

Answer 2

First, we need to understand the working method of a potentiometer. Let's follow the given figure. This figure is a simple setup of a potentiometer when it is used for measuring the internal resistance of a cell.

A potentiometer has a long wire with a known resistance (R), and its total length is L (AB). It is attached to a cell and a potential difference is generated across the wire.

J is a moveable point for finding the balancing length. When 0 current in the galvanometer, the circuit is balanced.

When the circuit is balanced, the voltage (V) through the resistance (R) is directly proportional to the balancing length (l) i.e. 

V∝l

Therefore, V = kl, where k is a constant.

When the circuit is balanced, 

V=E−ir and V=iR, where i is the current in the circuit PQRS and r is the internal resistance of the cell (E).

It means that E - ir = iR.

⇒E=i(r+R) ⇒ i=E(r+R)

Therefore,

⇒V=E−ir=E−\((\frac{E}{(r+R)})\)r

⇒V=E(1−\(\frac{r}{(r+R)}\))

⇒V=\(\frac{ER}{(r+R)}\)

⇒V=\(\frac{\frac{ER}{R}}{\frac{r+R}{R}}\)

= \(\frac{E}{\frac{r}{R}+1}\)

When a resistance of infinity, let the voltage across the resistance be 

V1 and the balancing length be l1

Therefore, V₁=kl₁ …. (i).

And 

V₁=\(\frac{E}{\frac{r}{R}+1}\)

But, if 

R→∞ then 

r/R→0

⇒V₁=\(\frac{E}{0+1}\)=E…. (ii).

Let the voltage across the shunted resistance of 9.5Ω be V₂ and the balancing length be l₂

Therefore, V₂=kl₂…..(iii).

And 

V₂=\(\frac{9.5E}{r+9.5}\) ….. (iv).

Divide equations (i) and (iii).

⇒

\(\frac{V_1}{V_2}\) =\(\frac{I_1}{I_2}\)

Here, l₁= 3 cm and l₂ = 2.85cm.

This gives us that

\(\frac{V_1}{V_2}\)=\(\frac{3}{2.85}\)

Now, divide equations (ii) and (iv).

\(\frac{V_1}{V_2}\)=\(\frac{E}{\frac{9.5E}{r+9.5}}\) =\(\frac{r+9.5}{9.5}\)

But \(\frac{V_1}{V_2}\) =\(\frac{3}{2.85}\)

Therefore,

\(\frac{r+9.5}{9.5}\) = \(\frac{3}{2.85}\)

⇒2.85(r+9.5)=3(9.5)

⇒2.85r+27.075=28.5

⇒r=\(\frac{(28.5-27.075)}{2.85}\) = \(\frac{1.425}{2.85}\) = 0.5Ω

Therefore, the internal resistance of that cell is 0.5Ω

Option (C)  is correct.

Answer 3

Here,

 The formula for the Internal resistance, r = (\(\frac{E-V}{V}\))R = (\(\frac{I_1 - I_2}{I_2}\))R

=(\(\frac{3-2.85}{2.85}\))(9.5)Ω = 0.5Ω

So, Option (c) is the correct answer.