1.

In figure, assuming the diodes to be ideal,

Answer»

`D_(1)` is FORWARD biased and `D_(2)` is reversebiased and hence current flows from A to B
`D_(2)` is forward biased and `D_(1)` is reverse biased and henceno current flows from B to A and vice versa.
`D_(1)` and `D_(2)` are both forward 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.

Solution :`D_(2)` is forward biased and `D_(1)` is reverse biased and henceno 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) ` isreverse 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 itsequivalent 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.


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