When two cells of emfs $E_1$ and $E_2$ and different internal resistances are connected in series with an…
When two cells of emfs $E_1$ and $E_2$ and different internal resistances are connected in series with an external load resistor, the current through the load is $5 \mathrm{~A}$. If the polarity of cell of emf $E_2$ is reversed then the current through the load is $2 \mathrm{~A}$. Then $\frac{E_1}{E_2}=$