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 :-
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$\mathrm{E}_{\mathrm{eq}}=\left(\frac{\frac{5}{2}-\frac{2}{1}}{\frac{1}{2}+1}\right) \mathrm{V}^{\mathrm{r}_{\mathrm{eq}}}=\frac{2 \times 1}{2+1}=\frac{2}{3}\, \Omega$

$\mathrm{i}=\frac{\mathrm{E}_{\mathrm{eq}}}{10+\mathrm{r}_{\mathrm{eq}}}=0.03 \mathrm{\,A}$ from $\mathrm{P}_{2}$ to $ \mathrm{P}_{1}$

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