Question

In a potentiometer experiment two cells of emf $E_{1}$ and $E_{2}$ are used in series and in conjunction and the balancing length is found to be $58~\mathrm{cm}$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes $29~\mathrm{cm}$. The ratio $\frac{E_1}{E_2}$ of the emfs of the two cells is}

• A. 1:1
• B. 2:1
• C. 3:1
• D. 4:1

Hint:

In a potentiometer, the balancing length is directly proportional to the emf.

Answer 1

The Correct Option is C

In this potentiometer experiment, we have two scenarios to consider: one where the cells are in series with their emfs reinforcing each other (\(E_1 + E_2\)) and another where the cells are in series with the emf of the second cell (\(E_2\)) reversed (\(E_1 - E_2\)).

We know that the balancing length of the potentiometer wire is directly proportional to the total emf of the cell combination. Hence, for the first case:

\(L_1 = 58\, \text{cm}\) (with emfs in conjunction)

For the reversed polarity:

\(L_2 = 29\, \text{cm}\) (with emfs opposing each other)

According to the principle of potentiometry,

\(\frac{L_1}{L_2} = \frac{E_1 + E_2}{E_1 - E_2}\)

Given, \(L_1 = 58\, \text{cm}\) and \(L_2 = 29\, \text{cm}\), substituting these values:

\(\frac{58}{29} = \frac{E_1 + E_2}{E_1 - E_2}\)

Simplifying, \(58 = 2 \times 29\), so:

\(2 = \frac{E_1 + E_2}{E_1 - E_2}\)

Cross-multiplying gives:

\(2(E_1 - E_2) = E_1 + E_2\)

Expanding and rearranging terms:

\(2E_1 - 2E_2 = E_1 + E_2\)

\(2E_1 - E_1 = E_2 + 2E_2\)

\(E_1 = 3E_2\)

Thus, the ratio of \(\frac{E_1}{E_2}\) is \(3:1\), which matches the given correct answer option.