In the electrical circuit shown in the figure, the current $i$ through the side $AB$ is
NEET 2017, Medium
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$R_{eq}=25 \Omega$
$I=\frac{10}{25}$
$I=0.4$
$I_{(20\;\Omega)}=\frac{30 \times \frac{4}{10}}{20+30}=\frac{12}{50}=\frac{6}{25} A$
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