In the figure shown, the current in the $10\, V$ battery is close to
JEE MAIN 2020, Medium
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$E _{ eq }=\frac{20 \times 10}{17}=\frac{200}{17}$
and $R _{ eq }=\frac{7 \times 10}{17}=\frac{70}{17}$
$\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 A$
from $+ve$ to $-ve$ terminal
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