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The current $i$ in the network is:
(1) 0.2 A
(2) 0.6 A
(3) 0.3 A
(4) 0 A
Both diodes are reverse biased, so no current flows through $5,\Omega$ and $20,\Omega$. Two resistors of $10,\Omega$ and two of $5,\Omega$ are in series. $I = \frac{V}{R_{\text{eq}}} = \frac{9}{10+5+10+5} = \frac{9}{30} = 0.3$ A.
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