The cell has an emf of $2 \mathrm{~V}$ 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

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 $\operatorname{emf} E$ 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.
$$ \therefore \quad V=E-i r $$ where $r$ is internal resistance. $$ \begin{array}{ll} \text { Also current } & i=\frac{E}{R+r} \\ \therefore & V=E-\left(\frac{E}{R+r}\right) r \end{array} $$ Putting the numerical values, we have $$ \begin{aligned} E=2 V, r &=0.1 \Omega, R=3.9 \Omega \\ V &=2-\left(\frac{2}{3.9+0.1}\right) \times 0.1 \\ V &=1.95 \mathrm{volt} \end{aligned} $$