(A) Calculate the value of `V_(0)` and `i` if the silicon and germanium diode start conducting at 0.7 V and 0.3 V respectively.
(B) If the Ge diode connection is now reversed, what will be the new values of `V_(0)` and `i` ?

Explanation

Here cut-in voltage (thereshold voltge) for Si diode is 0.7V and cut-in voltage for Ge diode is 0.3V.
Battery voltage = 12V
Load resistance `R_(L)=5kOmega=5xx10^(3)Omega`
(A) Current will flow earlier in Ge then Si diode.
Forward bias voltage `=12.0-0.3`
`=11.7V`
`therefore ` Current `I_(F)` in `R_(L)=(11.7)/("Load resistance")`
`=(11.7)/(5xx10^(3))`
`=2.34xx10^(-3)A`
`therefore I=2.34mA`
(B) In reverse bias connection current will flow earlier in Si diode.
`therefore ` Reverse bias voltage `=12.0-0.7`
`=11.3V`
`therefore` Current `I_(R )` in `R_(L )=(1.3)/("Load resistance")`
`=(11.3)/(5xx10^(3))`
`=2.26xx10^(-3)A`
`therefore I_(R )=2.26 mA`