In the figure shown, the current in the 10 V battery is close to :

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In the figure shown, the current in the 10 V battery is close to :

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0.42 A from positive to negative terminal

0.21 A from positive to negative terminal

0.36 A from negative positive terminal

0.71 A from positive to negative terminal

answer is B.

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Detailed Solution

$\begin{array}{l} K J R, i = i_{1} + i_{2} \\ K V L i n l e f t l o o p, \\ 20 - 5 i - 10 i_{1} - 2 i = 0 \\ \Rightarrow i_{1} = \frac{20 - 7 i}{10} \\ K V L i n o u t e r m o s t l o o p, \\ 20 - 5 i - 10 - 4 i_{2} - 2 i = 0 \\ \Rightarrow i_{2} = \frac{10 - 7 i}{4} \\ u \sin g f i r s t e q u a t i o n, \\ i = \frac{20 - 7 i}{10} + \frac{10 - 7 i}{4} \\ \Rightarrow 40 i = 80 - 28 i + 100 - 70 i \\ \Rightarrow i = \frac{180}{138} A \end{array}$