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In figure, assuming the diodes to be ideal,

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$D_{1}$ is forward biased and $D_{2}$ is reverse biased and hence current flows from A to B.

$D_{1}$ and $D_{2}$ are both reverse biased and hence no current flows from A to B and vice-versa.

$D_{1}$ and $D_{2}$ are both forward biased and hence current flows from A to B.

$D_{2}$ is forward biased and $D_{1}$ is reverse biased and hence no current flows from B to A and vice-versa.

answer is B.

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Detailed Solution

For diode $D_{1}$ ,

$V p - V n = - 10 - 0 = - 10 V$

So, $D_{1}$ is in reverse bias.

For diode $D_{2}$ ,

$V p - V n = 0 - (- 10) = 10 V$

So, $D_{2}$ is in forward bias.

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