Explanation

(b) Consider the case when $Ge$ and $Si$ diodes are connected as show in the given figure.
Equivalent voltage drop across the combination $Ge$ and $Si$ diode $= 0.3\ V$
$\Rightarrow$ Current $i = \frac{12 - 0.3}{5\ k\Omega} = 2.34\ mA$
$\therefore$ Out put voltage $V_{0} = Ri = 5\ k\Omega \times 2.34\ mA = 11.7\ V$
Now consider the case when diode connection are reversed. In this case voltage drop across the diode’s combination $= 0.7\ V$
$\Rightarrow$ Current $i = \frac{12 - 0.7}{5\ k\Omega} = 2.26\ mA$
$\therefore V_{0} = iR = 2.26\ mA \times 5\ k\Omega = 11.3\ V$
Hence charge in the value of $V_{0} = 11.7 - 11.3 = 0.4\ V$