Ge and Si diodes conduct at 0.3 V and 0.7 V, respectively. In  the  following  figure,  if  Ge  diode  connection  are reversed, the valve of ${\mathrm{V}}_0$  changes by

Consider   the  case  when  Ge  and  Si  diodes   are connected as shown in the given figure. Equivalent voltage drop across the combination G and Si diode = 0.3 V

$\Rightarrow \mathrm{Current} \mathrm{i} =\frac{12-0.3}{5\mathrm{k\Omega }}=2.34 \mathrm{mA}$

$\therefore$Output voltage, ${\mathrm{V}}_{\mathrm{o}}=\mathrm{Ri}=5 \mathrm{k\Omega } \times 2.34 \mathrm{mA}=11.7 \mathrm{V}$

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

$\Rightarrow \mathrm{Current} \mathrm{i} =\frac{12-0.7}{5\mathrm{k\Omega }}=2.26 \mathrm{mA}$

$\therefore {\mathrm{V}}_{\mathrm{o}}-\mathrm{iR}=2.26 \mathrm{mA} \times 5 \mathrm{k\Omega }=11.3 \mathrm{V}$

hence, change in the value of ${\mathrm{V}}_{\mathrm{o}}=11.7 \mathrm{V}-11.3=0.4 \mathrm{V}$