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The current in the $20 Ω$ resistor is

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0.2A

0.1A

0.4A

Zero

answer is A.

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Detailed Solution

$E = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} : E = \frac{100}{20} = 5 V$ and
$r = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{10}{2} = 5$

$I = \frac{E}{R + r} = \frac{5}{5 + 20} = 0.2 A$

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