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Determine the current in the branch AB of the network shown in figure

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$\frac{4}{17} A$

$\frac{6}{17} A$

$\frac{2}{17} A$

$\frac{8}{17} A$

answer is A.

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Detailed Solution

ABD 100P

$- 10 i_{1} + 5 (i - 2 i_{1}) + 5 (i - i_{1}) = 0$

$2 i - 5 i_{1} (i)$

$100 p$ ADCEFA

$- 5 (i - i_{1}) 10 i_{1} + 10 - 10 i = 0$

$5 i_{1} + 15 i = 10 (i i)$

From (i) and (ii)

Current in AB branch $\Rightarrow i_{1} = \frac{4}{17}$

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