Ge and Si diodes conduct at $0.3\text{\hspace{0.17em}V}$ and $0.7\text{\hspace{0.17em}V}$ respectively.  In the following figure, if Ge diode connection are reversed, the value of ${\text{V}}_{\text{0}}$ changes by

 

Consider the case when Ge and Si diodes are connected as show in the given figure.

Equivalent voltage drop across the combination Fe and Si diode  $=0.3V$

Current  $i=\frac{10-03}{2k\Omega }=4.85mA$

Output voltage  $V_0=Ri=2\mathrm{k\Omega }\times 4.85\mathrm{mA}=9.7\mathrm{V}$

Now consider the case when diode connection is reversed.  In this case voltage drop across the diode’s combination  $=0.7V$

Current $i=\frac{10-0.7}{2k\Omega }=4.65mA$

    $\therefore V_0=iR=4.65\mathrm{mA}\times 2\mathrm{k\Omega }=9.3\mathrm{V}$        

Hence charge in the value of $V_0=9.7-9.3=0.4\mathrm{V}$