Question 215 of 380

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

A$0.71A$ from positive to negative terminalB$0.42A$ from positive to negative terminalC$0.21A$ from positive to negative terminalD$0.36A$ from negative to postive terminal

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Answer: (C) $0.21A$ from positive to negative terminal

$-5{i}_{2}-10\left({i}_{1}+{i}_{2}\right)-2{i}_{2}+20=0$
$-10{i}_{1}-17{i}_{2}+20=0$
$-10+4{i}_{1}+10\left({i}_{1}+{i}_{2}\right)=0$
$14{i}_{1}+10{i}_{2}+10=0$
$10{i}_{1}+17{i}_{2}=20\rightarrow \times 10$
$14{i}_{1}+10{i}_{2}=10\rightarrow 17$
$-138{i}_{1}=30$
${i}_{1}=-\dfrac{30}{138}=-0.217$
${i}_{1}$ is negative it means current flows from positive to negative terminal

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