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The current I in the circuit shown below is: (All diodes are ideal and identical)
A
$\frac{5}{3}$ A
B
$\frac{15}{2}$ A
C
$\frac{1}{3}$ A
D
$\frac{5}{9}$ A
Detailed Solution
Here $D_1$ and $D_3$ are forward biased (act as closed switches), while $D_2$ and $D_4$ are reverse biased (open circuit).
So only the $4\ \Omega$ and $2\ \Omega$ branches conduct, and they are in parallel:
$R_{eq} = \frac{4 \times 2}{4 + 2} = \frac{4}{3}\ \Omega$
$I = \frac{V}{R_{eq}} = \frac{10}{4/3} = \frac{15}{2}$ A