Physics - Wave Optics Question with Solution | TestHub

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