When a cell of e.m.f 3.06V is connected, the balance of a potentiometer is 75cm, Calculate the new balance of a cell of e.m.f 2.295V

Assessment: JAMB UTME - Physics - 2024 Subject: Physics

Question 1 Report

When a cell of e.m.f 3.06V is connected, the balance of a potentiometer is 75cm, Calculate the new balance of a cell of e.m.f  2.295V
Answer Details

To solve this problem, we first need to understand the principle behind a potentiometer. A potentiometer is a device used to measure the electromotive force (e.m.f) of a cell by comparing it with a known voltage. The balance length on a potentiometer corresponds to a proportional measurement of the e.m.f.


Let's denote:

- \( V_1 \): the e.m.f of the first cell = 3.06V

- \( l_1 \): the balance length for the first cell = 75 cm

- \( V_2 \): the e.m.f of the second cell = 2.295V

- \( l_2 \): the balance length for the second cell (which we need to find)


The basic relationship for a potentiometer is given by:

\( V_1 / V_2 = l_1 / l_2 \)


Substituting the given values:

\( 3.06 / 2.295 = 75 / l_2 \)


We need to solve for \( l_2 \):

\( l_2 = (2.295 \times 75) / 3.06 \)


Now, calculating the above expression:

\( l_2 = 171.975 / 3.06 \approx 56.26 \) cm


Therefore, the new balance length for the cell with an e.m.f of 2.295V is approximately 56.26 cm.

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