Question

In a potentiometer experiment for the determination of the internal resistance of a cell, when an external resistance of R is connected parallel to the cell, the balancing length decreases by 10%. The internal resistance of the cell is

• A. R/9
• B. R/7
• C. R/5
• D. R/11

Hint:

The potentiometer works on the principle that potential difference is proportional to length (\(V \propto l\)). Therefore, \( \frac{\epsilon}{V} = \frac{l_1}{l_2} \). Combining this with \( \epsilon = I(R+r) \) and \( V = IR \), we get \( \frac{R+r}{R} = \frac{l_1}{l_2} \), which leads directly to the formula \( r = R(\frac{l_1}{l_2} - 1) \).

Answer 1

The Correct Option is A

The formula for the internal resistance (r) of a cell using a potentiometer is:
\( r = R \left( \frac{l_1}{l_2} - 1 \right) \), where R is the external resistance.
\( l_1 \) is the balancing length when the cell is in an open circuit (measuring the EMF, \(\epsilon\)).
\( l_2 \) is the balancing length when the external resistance R is connected across the cell (measuring the terminal voltage, V).
We are told that the balancing length decreases by 10% when R is connected.
This means \( l_2 \) is 10% less than \( l_1 \).
So, \( l_2 = l_1 - 0.10 \times l_1 = 0.9 \times l_1 \).
The ratio \( \frac{l_1}{l_2} \) is therefore \( \frac{l_1}{0.9 l_1} = \frac{1}{0.9} = \frac{10}{9} \).
Now, substitute this ratio back into the formula for internal resistance.
\( r = R \left( \frac{10}{9} - 1 \right) \).
\( r = R \left( \frac{10 - 9}{9} \right) = R \left( \frac{1}{9} \right) = \frac{R}{9} \).
The internal resistance of the cell is R/9.