A $5\, V$ battery with internal resistance $2\, \Omega$ and a $2\,V$ battery internal resistance $1\, \Omega$ are connected to a $10\, \Omega$ resistor as shown in the figure. The current in the $10\, \Omega$ resistor is :-
$\mathrm{i}=\frac{\mathrm{E}_{\mathrm{eq}}}{10+\mathrm{r}_{\mathrm{eq}}}=0.03 \mathrm{\,A}$ from $\mathrm{P}_{2}$ to $ \mathrm{P}_{1}$
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