In figure, assuming the diodes to be ideal,

Options

A. `D_(1)` is forward biased and `D_(2)` is reverse biased and hence current flows from A to B
B. `D_(2)` is forward biased and `D_(1)` is reverse biased and hence no current flows from B to A and vice versa.
C. `D_(1)` and `D_(2)` are both forward biased and hence current flows from A to B.
D. `D_(1) and D_(2)` are both reverse biased and hence no current flows from A to B and vice versa.

Answer

B

Explanation

`D_(2)` is forward biased and `D_(1)` is reverse biased and hence no current flows from B to A and vice versa.
For diode `D_(1), V_(A)= -10V and V_(K)=0 rArr V_(A) lt V_(R ) rArr D_(1) ` is reverse biased. Hence its resistance would be `R_(1) = oo (because D_(1)` is an ideal diode)
For diode `D_(2), V_(A)=0 and V_(K) =-10V rArr V_(A) gt V_(K) rArr D_(2) ` is forward biased. Hence its resistance would be `R_(2)=0. (because D_(2)` is an ideal diode)
Here, `D_(1) and D_(2)` are connected in series and so its equivalent resistance `R=R_(1)+R_(2)=oo +0=oo` and so current will not flow, neither from A to B nor from B to A.
`rArr` Option (B) is correct.