In the circuit shown in figure (1), the `V_(0),I_(1),I_(D_(1))`, and `I_(D_(3))` are respectively.
(A)

Options

A. `0.5V,25mA,15mA`
B. `0.7V,28 mA,14.09 mA`
C. `0.4V,15mA,20mA`
D. `0.3V,15.06mA,20.18mA`

Answer

B

Explanation

Since applied voltage is greater than `0.7 V` and direction of current is same as direction of arrow head, both the diodes are in ''on'' state (see above figure(2)). Now `V_(0)` is parallel to `D_(1)` and `D_(2)`.
Hence `V_(0) = 0.7 V`
Current `I_(1)=(V_(R))/(R)=(E-V_(D))/(R)=(10 V-0.7V)/(0.33kOmega) = 28.18 mA`
Both the silecon diodes have similar characteristics.
Hence `I_(D_(1))=I_(D_(2)) =(I_(1))/(2) = 14.09 mA`.