Calculate the ratio of current `I_(1)` to current `I_(2)` as depicted in the following diode circuits.
(Assume both the diodes to be ideal)

Answer

`01.65`

Explanation

In figure a, diode D is forward biased.
Hence the current will flow directly through the diode and equivalent circuit would be

`:.I_(1)=V/R=5/(10xx10^(3))=1/2mA`……………
In figure b. diode D is reverse biased. Hence the current will flow through the branch containgn `10kOmega` resistor.

`:.I_(2)=5/((6.5+10)xx10^(3))`
`:.I_(2)=1/3.3 mA`
(ii)
From i and ii
`(I_(1))/(I_(2))=3.3/2=1.65`