Q. In the circuit shown, the currents $i_1$ and $i_2$ are ___________

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Solution:

$R\:=\:\frac{12\:\times 4}{12\:+\:4}\:+\:2\:=\:5\:\Omega \:$
$\:\:I\:=\:\frac{E}{R\:+\:r}=\:\frac{12}{6}\:=\:2\:A\:$
$\:\:\:I_1\:+\:I_2\:=\:2A\:$
$\:I\:\propto \frac{1}{R}\:\therefore I_1\:=\:0.5A,\:I_{2\:}=\:1.5A$

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