Question

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

• A. $0.5V$
• B. $1.9V$
• C. $1.95V$
• D. $72V$

Answer 1

Answer: (C)

$1.95V$

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 ontaking more and more current from the cell.
$\because$ $V=E-ir$
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
$E=2V,r=0.1\Omega ,R=3.9\Omega$
$V=2-\left(\frac{2}{3.9+0.1}\right)\times 0.1$
$V=1.95$ volt