In the circuit shown, the cell has $emf = 10\,V$ and internal resistance $= 1\, \Omega$ :-
Diffcult
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Equivalent resistance $\mathrm{R}_{\mathrm{eq}}=10 \,\Omega$ so curent passing through battery and $3\, \Omega$ resistance is
$\mathrm{i}=\frac{10}{10}=\mathrm{A}$
and current passing through $4\, \Omega$ is $0.25 \mathrm{\,A}$
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