Question: Q. 2. Two cells of emf’s approximately $5 \mathrm{~V}$ and $10 \mathrm{~V}$ are to be accurately compared using a potentiometer of length $400 \mathrm{~cm}$ :
(a) The battery that runs the potentiometer should have voltage of $8 \mathrm{~V}$.
(b) The battery of potentiometer can have a voltage of $15 \mathrm{~V}$ and $R$ adjusted so that the potential drop across the wire slightly exceeds $10 \mathrm{~V}$.
(c) The first portion of $50 \mathrm{~cm}$ of wire itself should have a potential drop of $10 \mathrm{~V}$.
(d) Potentioneter is usually used for comparing resistances and not voltages.
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Solution:
Ans. Correct option : (b)
Explanation: Given that,
emf of primary cells are $5 \mathrm{~V}$ and $10 \mathrm{~V}$
The potential drop across potentiometer wire must be slightly more than that larger emf $10 \mathrm{~V}$.
So, the battery should be of $15 \mathrm{~V}$ and about $4 \mathrm{~V}$ potential is dropped by using rheostat or resistances.
Very Short Answer Type Question
(1 marks)
[AI Q. 1. Why is potentiometer preferred over a voltmeter for determining the emf of a cell?
] [Delhi Comptt., 2016]
Ans. Potentiometer does not draw any (net) current from the cell. Voltmeter draws some current from cell, when connected across it, hence it measures terminal voltage.
[CBSE Marking Scheme, 2016] Detailed Answer :
Potentiometer is preferred over voltmeter because voltmeter needs a current to pass through it so as to measure the potential difference. The potential difference being measured by the voltmeter is given as $\quad V=E-I R \quad 1 / 2$ Hence the voltage $V$ measured is less than emf of a battery $E$. Potentiometer does not draw any current from the cell and hence will not cause such problem and can measure the emf directly.
