Question

The cell has an emf of $ 2\, V $ and the internal resistance of this cell is $ 0.1 \Omega $ , it is connected to resistance of $ 3.9 \Omega $ , the voltage across the cell will be

• A. $ 1.95\,V $
• B. $ 1.5\,V $
• C. $ 2\,V $
• D. $ 1.8\,V $

Answer 1

Answer: (A)

Key Idea When cell is giving current then the potential difference across its plates is less than its emf as potential drop across internal resistance occurs.
When a cell of emf $E$ is connected to a resistance of $3.9 \Omega,$ then the $emf E$ of the cell remains constant, while voltage $V$ goes on decreasing on taking more and more current from the cell.

$\therefore V=E-ir$
where $r$ is internal resistance.
Also current $ i=\frac{E}{R+r} $
$\therefore V=E-\left(\frac{E}{R+r}\right) r$
Putting the numerical values, we have
$E=2 V, r =0.1 \Omega, R=3.9 \Omega $
$V =2-\left(\frac{2}{3.9+0.1}\right) \times 0.1$
$V=1.95$ volt