Question

To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at \(1\)m when cell is shunted by \(3\Omega\) resistance and at \(1.5\)m length when cell is shunted by \(6\Omega\) resistance. The internal resistance of the cell is

• A. \(2\Omega\)
• B. \(4\Omega\)
• C. \(6\Omega\)
• D. \(8\Omega\)

Hint:

The balance length is proportional to the terminal voltage E R over R plus r; take the ratio for the two shunt resistances.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
When the cell is shunted by \(R\), its terminal voltage is \(V=\frac{ER}{R+r}\), and the balance length is \(l=kV\).

Step 2: Take the ratio
\(\frac{l_{1}}{l_{2}}=\frac{3/(3+r)}{6/(6+r)}=\frac{1}{1.5}=\frac{2}{3}\)

Step 3: Calculation
\(\frac{3(6+r)}{6(3+r)}=\frac{2}{3}\Rightarrow\frac{6+r}{2(3+r)}=\frac{2}{3}\)
\(18+3r=12+4r\Rightarrow r=6\,\Omega\)

Step 4: Compare the options
This is option (C).

Final Answer:
The internal resistance of the cell is 6 ohm. \[ \boxed{\text{(C) }6\,\Omega} \]